Che Ban-Calculus 2 - NVH

HANOI UNIVERSITY OF TECHNOLOGY

NGUYỄN VĂN HỘ

A COURSE

IN

CALCULUS2

2009

Calculus 2 – Chapter 1: Vector and Geometry of Space Nguyen Van Ho - 2009

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Chapter 1Vector and Geometry of Space1.1 VECTORS

In this chapter we introduce vectors and coordinate system for three-dimensional space. Wewill see that vectors provide simple descriptions of lines and planes in space.

We first choose in space a fixed point O (the origin) and three directed lines through O thatare perpendicular to each other, called the coordinate axes, labeled the x-axis, y-axis, and z-axis. The direction of z-axis is determined by the right-hand-rule: If you curl the fingers ofyour right hand around the z-axis in the direction of a 90o counterclockwise rotation from thepositive x-axis to the positive y-axis, then your thumb points the positive direction of the z-axis. The three coordinate planes divide space into 8 octants. There is a one-to-onecorrespondence between any point P in space and a triple ( ), ,P P Px y z of real numbers, the

coordinates of P. See Figure 1.1.1

Figure 1.1.1: The coordinate axes Figure 1.1.2: Vector a = AB

A. VECTORThe term vector is used to indicate a quantity that has both magnitude and direction (velocity,force,...). A vector is used to be denoted by AB

or a ... See Figure 1.1.2.

DEFINITION 1.1.1 VECTOR AND COMPONENTSA three-dimensional vector is an ordered triple 1 2 3, ,a = a a a of real numbers. The umbers

1 2 3, ,a a a are called the components of a.

PROPERTY 1.1.1

a. Given ( ) ( ), , , , ,A A A B B BA x y z B x y z , then

, ,B A B A B AAB x x y y z z= − − −

(1.1.1)

b. The length of the vector 1 2 3, ,a = a a a is

2 2 21 2 3| |a = + +a a a (1.1.2)


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DEFINITION 1.1.2 VECTOR ADDITION

If 1 2 3, ,a = a a a , 1 2 3, ,=b b b b , then 1 1 2 2 3 3, ,a b+ = + + +a b a b a b

DEFINITION 1.1.3 MULTIPLICATION A VECTOR BY A SCALAR

If 1 2 3, ,a = a a a , and c is a scalar (number), then 1 2 3, ,a =c ca ca ca

PROPERTY 1.1.2

If a, b, and c are vectors in space, k, l are scalars (numbers), thena. a b b a+ = +b. ( ) ( )a b c a b c+ + = + +c. a 0 a+ = , where 0 is the zero vectord. ( )a a 0+ − =e. ( )a b a b+ = +k k k

f. ( )a a a+ = +k l k l

g. ( ) ( )a a=kl k l

h. 1 =a a

DEFINITION 1.1.4 STANDARD BASIC VECTORS

The vectors 1,0,0 , 0,1,0 , 0,0,1i j k= = = are called the standard basic vectors

PROPERTY 1.1.3

If 1 2 3, ,a = a a a , then 1 2 3a i j k= + +a a a

EXAMPLE 1.1.1

a. Given ( ) ( )1,2, 2 , 2, 3,0A B− − − , find , and | |AB=a a

Solution: 3, 5, 2AB= = −a

, 2 2 2| | 3 ( 5) 2 38AB= = + − + =a

.

b. A 100-kG weight hangs from two wires as shown in Figure 1.1.3. Find the tensions(forces) T1 and T2 in both wires and their magnitudes.

Figure 1.1.3: Figure 1.1.4:Solution:Look at Figure 1.1.4. Express first the forces in terms of their components, then equate thetotal components to zero (balance the forces).

0 0 311 1 1 1 12 2( | | cos 60 ) (| | cos30 ) ( | |) ( | |)= − + = − +T T i T j T i T j ,

0 0 3 12 2 2 2 22 2(| | cos30 ) (| | cos 60 ) ( | |) ( | |)= + = +T T i T j T i T j ,


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1 2 1 225 3 75 and 25 3 25+ + = ⇒ = − + = +T T W 0 T i j T i j

B. DOT PRODUCT

DEFINITION 1.1.5 DOT PRODUCT

If 1 2 3, ,a = a a a and 1 2 3, ,b b b=b , then the dot product of a and b is the number ⋅a b given

by

1 1 2 2 3 3a b a b a b⋅ = + +a b (1.1.3)

PROPERTY 1.1.4

If a, b, and c are vectors in space and k is a constant, then

a. 2| |⋅ =a a a

b. ⋅ = ⋅a b b ac. ( )⋅ + = ⋅ + ⋅a b c a b a c

d. ( ) ( ) ( )k k k⋅ = ⋅ = ⋅a b a b a b

e. 0⋅ =0 a

THEOREM 1.1.1

a. If , 0 , ≤ ≤ is the angle between a, b, then

| | | | cos⋅ =a b a b (1.1.4)

1 1

1 1 2 2 3 3

2 2 2 2 2 22 3 2 3

cos| | | |

a b a b a b

a a a b b b + +⋅= =

+ + + +a ba b

(1.1.5)

b. Vectors a and b are orthogonal if and only if

1 1 2 2 3 30 i.e. a a a 0b b b⋅ = + + =a b (1.1.6)

Proof

Apply the Law of Cosines to a triangle:2 2 2| | | | | | 2 | | | |cos= + −b - a a b a b .

On the other hand:2 2 2| | ( ) ( ) 2 | | | | 2= ⋅ = ⋅ ⋅ ⋅ + ⋅ = ⋅ ⋅ + ⋅ = + ⋅b - a b - a b - a b b - b a - a b a a b b - a b a a a b - a b .

Therefore we obtain (1.1.4).

Figure 1.1.5


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DEFINITION 1.1.6 DIRECTION ANGLES AND DIRECTION COSINES

a. The direction angles of a nonzero vector a are the angles , , and in the interval [0,π]that the vector a makes with the positive x- , y- , and z- axes.b. The cosines of these direction angles, cos, cos, and cos are called the directioncosines of the vector a.

It follows from (1.1.5) that

31 2cos ; cos ; cos ;| | | | | | | | | | | | | | | | | |

aa a ⋅ ⋅ ⋅= = = = = =a i a j a ka i a a j a a k a

(1.1.7)

1 2 3| | cos ; | | cos ; | | cos ;a a a = = =a a a (1.1.8)

Therefore | | cos , cos , cos =a a (1.1.9)

The ofunit vector a : cos , cos , cos| |

=aa

(1.1.10)

2 2 2cos cos cos 1 + + = (1.1.11)

DEFINITION 1.1.7 SCALAR PROJECTION AND VECTOR PROJECTION

a. The scalar projection of a onto b (also called the component of a along b):

OP = comp | | cos| |

⋅= =b

a ba a

b(1.1.12)

b. The vector projection of a onto b:

2= proj

| | | | | |OP

⋅ ⋅= =

b

a b b a ba b

b b b

(1.1.13)

OP 0> , = , 0OP k k >b

OP 0< , = , 0OP k k <b

Figure 1.1.6: Projection

EXAMPLE 1.1.2

a. Find the scalar projection of 1, 2, 3= − −a onto 2,3, 1= − −b .

b. The force (in newtons) 3,2,5=F moves a particle from the point A(2, 1, 4) to the

point B(6, 3, 5) , (the length unit is meter). Find the work done.

c. Find the direction cosines of 2,3,1=a


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c. Find the angle between 1,3, 2= − −a and 1, 2, 1= − −b .

d. Find the unit vector of 2,4, 4= −a .

e. Are 1,3, 7= − −a and 1, 2, 1= − −b orthogonal?

f. Let 1,4,3= −a and 1, 1, 2=b . Find the scalar and vector projections of b onto a and

of a onto b.

Answer: comp 9 / 6=ba ;9 9 18

proj , ,6 6 6

=ba ; comp 9 / 26=ab ;9 36 27

proj , ,26 26 26

−=ab

C. C. CROSS PRODUCT

DEFINITION 1.1.8 CROSS PRODUCT

The cross product of a and b is a vector, denoted by ×a b , that satisfies

(i) ×a b is orthogonal to both vectors a and b.

(ii) The direction of ×a b is given by the right-hand rule (the fingers of the right hand curlin the direction of a rotation through an angle less than 180o from a to b, then the thumbpoints the direction of ×a b )

(iii) | | | || | sin× =a b a b , where , 0 , ≤ ≤ is the angle between a, b (1.1.14)

Note that the condition (1.1.14) means that the magnitude (length) of ×a b is equal to thearea of the parallelogram determined by a and b. See Figure 1.1.7

Figure 1.1.7

PROPERTIES 1.1.5 CROSS PRODUCT

a. , ,× = × = × =i j k j k i k i j

b. × =a b 0 if and only if a and b are parallel.

c. ( )× = − ×b a a b

d. ( ) ( ) ( )× + = × + ×a b c a b a c

e. ( ) ( )k k× = ×a b a b

f. ( ) ( ) ( )( )k l kl× = ×a b a b , where k and l are numbers

From these properties, it is easy to obtain the following component expression for the crossproduct of two vectors 1 2 3, ,a = a a a and 1 2 3, ,b b b=b .


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THEOREM 1.1.2 MATRIX EXPRESSION OF THE CROSS PRODUCT

If 1 2 3, ,a = a a a and 1 2 3, ,b b b=b , then the cross product of a and b is determined by

2 3 1 3 1 21 2 3

2 3 1 3 1 21 2 3

a a a a a aa a a

b b b b b bb b b

× = = − +i j k

a b i j k

2 3 3 2 3 1 1 3 1 2 2 1( ) ( ) ( )a b a b a b a b a b a b= − + − + −i j k

2 3 3 2 3 1 1 3 1 2 2 1, ,a b a b a b a b a b a b= − − − (1.1.15)

Proof

1 2 3 1 2 3 1 2 3 1 2 3, , , , ( ) ( )a a a b b b a a a b b b× = × = + + × + +a b i j k i j k . Apply properties 1.1.5.

EXAMPLE 1.1.3

a. Given 1,3, 2= −a , 2,4,1= −b , find ×a b , | |×a b .

Answer: 11,3,10 , 230 .

b. Find the area of the triangle ABC, A(2,8,12), B(4,5,8), C(1,4,10).

Answer: 285 / 2 .

c. Find the height AH of the triangle ABC, A(1,6,4), B(2,5,8), C(-1,4,0).

Answer:| | 176 88

74 37| |

BA BCAH

BC

×= = =

d. Prove that

( ) ( ) ( )× × = ⋅ − ⋅a b c a c b a b c (1.1.16)

Hint: Apply (1.1.3) and (1.1.15).

D. D. SCALAR TRIPLE PRODUCT

DEFINITION 1.1.9 SCALAR TRIPLE PRODUCT

The scalar triple product of three vectors a, b, and c, denoted by ( , , )a b c , is a number thatis defined by the scalar product of a and ×a b :

( , , )a b c = ( )⋅ ×a b c (1.1.17)

THEOREM 1.1.3 EXPRESSION OF THE SCALAR TRIPLE PRODUCT

a. Let 1 2 3, ,a = a a a , 1 2 3, ,b b b=b , and 1 2 3, ,c c c=c .

Then the scalar triple product of three vectors a, b, and c is determined by thedeterminant


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1 2 3

1 2 3

1 2 3

( , , ) ( )

a a a

b b b

c c c

= ⋅ × =a b c a b c (1.1.18)

b. ( , , ) ( , , )= −b a c a b c , i.e., .( ) ( )× = − ⋅ ×b a c a b c (1.1.19)

c. ( , , ) ( , , ) ( , , ), i.e, ( ) ( ) ( )= = ⋅ × = ⋅ × = ⋅ ×a b c b c a c a b a b c b c a c a b (1.1.20)

Proof

The conclusion a. follows directly from (1.1.3), (1.1.15), and (1.1.16). The conclusions b. andc. follow from the determinant properties.

EXAMPLE 1.1.4

Let 2, 1, 2= − −a , 1,3,1= −b , and 3,2,2= −c . Find ( , , )a b c , ( , , )b c a , ( , , )b a c .

Answer: -5, -5, 5. These results justify (1.1.18) and (1.1.19).

PROPERTIES 1.1.6 SCALAR TRIPLE PRODUCT

a. The volume of the parallelepiped determined by the vectors a, b, and c is the magnitude(absolute value) of their scalar triple product:

| ( , , ) |V = a b c (1.1.21)

b. The vectors a, b, and c are coplanar if and only if

( , , )a b c = 0 (1.1.22)Proof

a. (1.1.20) follows directly from (1.1.4) and (1.1.14):( , , ) ( ) | | | ( ) |cos= ⋅ × = ×a b c a b c a b c , where , 0 , ≤ ≤ is the angle between a and ×b c .The length | |×b c is equal to the area of the parallelogram determined by b and c; the heighth of the parallelepiped is the magnitude of | | cosa . Figure 1.1.8 shows that when0 / 2 ≤ ≤ : | | cos 0 ( ) 0 ≥ ⇒ ≥a a,b,c and when / 2 < ≤ : | | cos 0 ( ) 0 < ⇒ <a a,b,c .

c. It is the consequence of a.

a) 0 / 2 ≤ ≤ : ( , , ) ( ) 0= ⋅ × ≥a b c a b c b) / 2 < ≤ : ( , , ) ( ) 0= ⋅ × <a b c a b c

Figure 1.1.8: The parallelepiped determined by the vectors a, b, and c


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EXAMPLE 1.1.5

Let 3, 3, 4= − −a , 1, 3, 1= − −b , and , 2, 4k= −c .

a. Find the volume of the parallelepiped determined by a, b, and c, if k = 5.

b. Determine k so that a, b, and c are coplanar.

c. Determine whether the points A(1, 0, 1), B(2, 4, 6), C(3, -1, 2), and D(6, 2, 8) lie in thesame plane.

Answer: a. 67; b. -22/9; c. Yes.

1.2 EQUATIONS OF LINES AND PLANES

A. EQUATIONS OF LINES

A line L is determined by a point 0 0 0 0( , , )P x y z on it and its direction vector , ,a b c=v .

Let ( , , )P x y z be an arbitrary point on L . Let , ,OP x y z= =r

, and 0 0 0 0 0, ,OP x y z= =r

be

the position vectors of P and P0 . Vector 0 0P P = r - r

is parallel to v. Look at Figure 1.2.1.

Figure 1.2.1

The vector equations of L :

0 t= +r r v (1.2.1)

The parametric equations of L :

0 0 0, ,x x at y y bt z z ct= + = + = + (1.2.2)

(t is the parameter, t ∈ )

The symmetric equations of L :

0 0 0x x y y z z

a b c

− − −= = (1.2.3)

If one of a, b, or c = 0, these equations become, for instance if a = 0:(The line lies in the plane 0x x= that is parallel to the xy-plane)

0 00

y y z zx x

b c

− −= = (1.2.4)


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If two among a, b, or c = 0, these equations become, for instance if a = 0 and b = 0:

0 0x x y y= = (1.2.5)

EXAMPLE 1.2.1

a. Find the vector equation and the parametric equations for the line that passes throughthe point (3, 2, -1) and is parallel to the vector 2, 5, 1= − −v .

At what points does the line passes through the xy-plane?

Solution:

The vector equation:

3 2 ,2 5 , 1 (3 2 ) ( 2 5 ) ( 1 )t t t t t t= − − − + = − + − + − +r i j k

The parametric equations:

3 2 , 2 5 , 1x t y t z t= − = − = − + .

Put z = 0 1 1, 3, 0t x y z⇒ = ⇒ = = − = ⇒

the line passes through the xy-plane at the point P(1, -3, 0).

b. Find the vector equation, the parametric equation, and the symmetric equation for theline that passes through the points A(4, 3, 6), B(2, -5, 6). Is this line parallel to the line inquestion a?

B. EQUATIONS OF PLANES

A plane P is determined by a point 0 0 0 0( , , )P x y z in it and its normal vector , ,a b c=n .

Let ( , , )P x y z be an arbitrary point in P . Let , ,OP x y z= =r

, and 0 0 0 0 0, ,OP x y z= =r

be

the position vectors of P and P0 . Vector 0 0P P = r - r

is orthogonal to , ,a b c=n and so we

have

The vector equations of P : 0 0( ) 0 or⋅ − = ⋅ = ⋅n r r n r n r (1.2.6)

The scalar equation of P : 0 0 0( ) ( ) ( ) 0a x x b y y c z z− + − + − = (1.2.7)

or : 0ax by c z d+ + + = (1.2.8)

where 0 0 0( )d ax by cz= − + +

EXAMPLE 1.2.2

a. Find an equation of the plane that passes through A(1, 3, -4) and is orthogonal to BC

,where B(-1, -3, 1), C(3, 4, -2).

Solution:


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4,7, 3 4( 1) 7( 3) 3( 1) 0 or 4 7 3 28 0BC x y z x y z= = − ⇒ + + + − − = + − + =n

b. Find an equation of the plane P that passes through points A(2, 1, -2), B(-2, -1, 2), and

C(3, 4, 1). Show that P passes through the origin.

Solution:

Let P(x, y, z) be an arbitrary point in the plane. Four points A, B, C, and P are coplanar, or

three vectors ,BA

,AC

and AP

are coplanar. It follows from (1.1.21) that the equation of P is

9 8 5 0x y z− + = . The coordinates of the origin O(0, 0, 0) satisfy the equation of P. It means

that P passes through the origin O.

c. Find the point P at which the line 4 2 , 2 3 , 1 2x t y t z t= − = − = − + intersects the plane8 6 9 0x y z+ − + = .

Answer: P(1, -5/2, 2).

d. Fìnd the formula for the distance d from a point 1 1 1 1( , , )P x y z to the plane P0ax by c z d+ + + = .

Solution:

Let 0 0 0 0( , , )P x y z be a point in the plane P. The normal vector of P is , ,a b c=n . The

distance d is equal to the magnitude of the scalar projection of 0 1 1 0 1 0 1 0, ,P P x x y y z z= − − −

onto , ,a b c=n :

0 1 0 1 0 1 00 2 2 2

| | | ( ) ( ) ( ) |

| |

P P a x x b y y c z zd comp P P

a b c

⋅ − + − + −= = =+ +

n

nn

.

Since, 0 0 0 0 0P ax by c z d∈ ⇒ + + + =P , then

1 1 1

2 2 2

| |ax by cz dd

a b c

+ + +=+ +

(1.2.9)

1.3 CYLINDERS AND QUADRATIC SURFACES

DEFINITION 1.3.1 CYLINDERS

A cylinder is a surface that consists of all lines (called rulings) that are parallel to a givenline and pass through a given plane curve.

By rotation it can be supposed that the rulings are parallel to one axis.

EXAMPLE 1.3.1

a. The surface 2z x= (the equation does not involve y) is a parabolic cylinder. The rulingsare parallel to the y-axis and pass through the parabola 2z x= in the xz-plane


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b. The surface 2 2 4x y+ = (the equation does not involve z) is a circular cylinder. The

rulings are parallel to the z-axis and pass through the circle 2 2 4x y+ = in the xy-plane

c. Determine the surface 2 24 1z x+ =

DEFINITION 1.3.2 QUADRATIC SURFACES

A quadratic surface is the graph of a second-degree equation2 2 2 0Ax By Cz Dxy Exz Fyz Gx Hy Iz J+ + + + + + + + + = .

By translation and rotation it can be brought into one of two standard forms2 2 2 0Ax By Cz J+ + + = or 2 2 0Ax By Iz+ + =

EXAMPLE 1.3.2

a. Ellipsoid2 2 2

2 2 21

x y z

a b c+ + = (1.3.1)

b. Elliptic paraboloid2 2

2 2

z x y

c a b= + (1.3.2)

Circular paraboloid: If a = b

b. Hyperbolic paraboloid2 2

2 2

z x y

c a b= − (1.3.3)

d. Hyperboloid of one sheet2 2 2

2 2 21

x y z

a b c+ − = (1.3.4)

e. Hyperboloid of two sheets2 2 2

2 2 21

x y z

a b c+ − = − (1.3.5)

f. Cone2 2 2

2 2 2

z x y

c a b= + (1.3.6)

1.4 CYLINDRICAL AND SPHERICAL COORDINATES

A. CYLINDRICAL COORDINATES

DEFINITION 1.4.1 CYLINDRICAL COORDINATES

A point ( , , )P x y z in the space can be represented by the ordered triple ( , , )r z , where( , )r are polar coordinates of the projection of P onto the xy-plane. The ordered triple( , , )r z is called the cylindrical coordinates of P.


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Figure 1.4.1 Cylindrical coordinates

To convert from cylindrical to rectangular coordinates we use the equations:

cos , sin ,x r y r z z = = = (1.4.1)

To convert from rectangular to cylindrical coordinates we use the equations:

2 2 2, tan ,y

r x y z zx

= + = = (1.4.2)

EXAMPLE 1.4.1

a. Equation z = kr represents a cone 2 2 2 2( )z k x y= + .

b. Equation r = k represents a cylinder 2 2 2x y k+ = .

c. Equation = k represents a plane tany x = .

d. Find the rectangular coordinates of the point with cylindrical coordinates (3, 2 / 3, 5)

Answer: ( 3 / 2, 3 3 / 2, 5)−

B. SPHERICAL COORDINATES

DEFINITION 1.4.1 SPHERICAL COORDINATES

A point ( , , )P x y z in the space can be represented by the ordered triple ( , , ) , where isthe same angle as in cylindrical coordinates, , 0 , ≤ ≤ is the angle between the positive z-

axis and OP

, and , 0, ≥ s the distance from the origin O to P. The ordered triple ( , , ) is called spherical coordinates of P.

Figure 1.4.2 Spherical coordinates

To convert from spherical to rectangular coordinates we use the equations:


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sin cos , sin sin , cosx y z= = = (1.4.3)

To convert from rectangular to spherical coordinates we use the equations:

2 2 2 2, tan , cosy z

x y zx

= + + = =

(1.4.4)

EXAMPLE 1.4.2

a. Equation , 0k k = > , represents a sphere 2 2 2 2x y z k+ + = .

b. Equation , 0 2 ,k k= ≤ ≤ represents a plane tany x = .

c. What does the equation , 0 ,k k= < < represent?

d. What does the equation sin , 0 ,k k= < < represent?

d. Find the rectangular coordinates of the point with spherical coordinates (4, / 3, / 6)

Answer: ( )1, 3, 2 3

EXERCISES

1.1 Find the angle between a diagonal of a cube and a diagonal of one of its faces.

1.2 Find the angle between a diagonal of a cube and one of its edges.

1.3 Provea. The Cauchy-Schwarz Inequality: | | | || |⋅ ≤a b a bb. The Triangle Inequality for vectors: | | | | | |+ ≤ +a b a b

c. The Parallelogram Law: ( )2 2 2 2| | | | 2 | | | |+ + − = +a b a b a b

1.4 Suppose that a, b, and c are all nonzero vectors such that | | | |= +c a b b a . Show that cbisects the angle between a and b.

1.5 Given the points A(1, 0, 0), B(0, 1, 0), C(0, 0, 1), S(1, 1, 1), and H(½, ½, ½). Show thatSABC is a regular tetrahedron and H is its center.

1.6 A molecule of methane, CH4, is structured with the four hydrogen atoms at the vertices ofa regular tetrahedron and the carbon atom at the center. Show that the angle between the linesthat joint the carbon atom to two of the hydrogen atoms is about 109.5o.

1.7 Let P be a point not on the line L that passes through the points A and B. Show that thedistance from P to L is

| |

| |

AP ABd

AB

×=

.

Calculate the distance, if P(1,1,1), A(0,6,8), B(-1,4,7).


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1.8 Let D be a point not on the plane P that passes through the points A, B, and C. Show that

the distance from D to P is( )AD AB AC

dAB AC

⋅ ×=

×

Calculate this distance, if D(1,2,2), A(0,4,-1), B(-1,2,3), C(-3,2,5).

1.9 Given M(0, 1, -2), N(1, 3, 4), A(-2, 4, -1), B(-3, 1, 3), and C(3, 2, 5).a. Find the equation of the plane P that contains A, B, and C.

b. Find the equation of the line that goes through M and is perpendicular to P .c. Find the equation of the line L1 that goes through A and B.d. Find the equation of the line L2 that goes through M and is parallel to L1.e. Determine the distance from M to Pf. Determine the distance between L1 and L2.g. Find the volume of the parallelepiped determined from M, A, B, and Ch. Find the volume of the tetrahedron MABC

i. Find the distance between L1 and L3, the line goes through M and N.1.10 Prove that

a. ( ) ( ) 2( )− × + = ×a b a b a bb. ( ) ( ) ( )× × = ⋅ − ⋅a b c a c b a b cc. ( ) ( ) ( )× × + × × + × × =a b c b c a c a b 0

1.11 Plot the point whose cylindrical coordinates are given. Then find the rectangularcoordinates of the point.

a. (4, / 3, 6)b. (3, / 3, 5)−

1.12 Plot the point whose spherical coordinates are given. Then find the rectangularcoordinates of the point.

a. (5, / 3, / 4) −b. (7, 2 / 3, / 6) −

1.13 Plot the point whose rectangular coordinates are given. Then find the cylindricalcoordinates of the point.

a. (1, 1, 4)−b. (3 / 2, 3 / 2, 1)−

1.14 Plot the point whose rectangular coordinates are given. Then find the sphericalcoordinates of the point.

a. (1, 1, 2)−b. ( 3, 3, 2)− −

1.15 Show that the equation 1 1 2 2 3 3( )( ) ( )( ) ( )( ) 0x a x b y a y b z a z b− − + − − + − − = represents a

sphere and find its center and radius, where 1 2 3 1 2 3( , , ) ( , , )a a a b b b≠ are fixed.


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ANSWERS

1.1 cos 2 / 6 and cos 0= = .

1.2 cos 1/ 3 =1.7 97 / 3

1.8 2 / 17

1.9 a. 10 26 17 107 0x y z− − + = b.1 2

10 26 17

x y z− += =− −

c.2 4 1

1 3 4

x y z+ − += =−

d.1 2

1 3 4

x y z− += =−

e. 115 / 1065 f. 355 / 26

i. L3:1 2

1 2 6

x y z− += = ; 83 / 777d =

1.15 11 2 3 1 2 32Radius | |, ( , , ), ( , , )AB A a a a B b b b=

Calculus 2 – Chapter 2: Vector Functions Nguyen Van Ho

18

Chapter 2Vector Functions2.1 VECTOR FUNCTIONS AND SPACE CURVES

DEFINITION 2.1.1 VECTOR FUNCTIONS

A vector-valued function, or vector function, is a function whose domain is a set D ofnumbers, D ⊂ , and whose range is a set of vectors.A vector function can be expressed by

( ) ( ), ( ), ( ) ( ) ( ) ( )t f t g t h t f t g t h t= = + +r i j k (2.1.1)

where ( ), ( ), and ( )f t g t h t are real-valued functions called component functions, and t is realvariable taking values in the domain of r.

DEFINITION 2.1.2 LIMIT OF VECTOR FUNCTIONS

If r is defined by (2.1.1), then

lim ( ) lim ( ), lim ( ), lim ( ) lim ( ) lim ( ) lim ( )t a t a t a t a t a t a t a

t f t g t h t f t g t h t→ → → → → → →

= = + +r i j k (2.1.2)

DEFINITION 2.1.3 CONTINUITY OF VECTOR FUNCTIONS

Let r be defined by (2.1.1).

a. r is called to be continuous at t = a if ( ), ( ), and ( )f t g t h t are continuous at t = a.

b. r is called to be continuous on an interval I if ( ), ( ), and ( )f t g t h t are continuous on I.

We can consider the vector function r(t) as the position vector of the point

( )( ), ( ), ( )P f t g t h t , i.e. OP=r

. When the parameter t varies in I, the tip point P draws a

space curve C . Thus, any continuous vector function r(t) defines a continuous space curve C(see Figure 2.1.1) with parametric equations

( ), ( ), ( )x f t y g t z h t= = = (2.1.3)

Figure 2.1.1 A space curve traced out by the tip of a position vector r(t)


19

EXAMPLE 2.1.1

a. Describe the curve defined by the vector function ( ) 1 2 , 2 , 2 3t t t t= + − − +r

Answer: The curve is a line that passes through the point (1, -2, 2) and is parallel to thevector 2, 1, 3= −v

b. Describe the curve defined by the vector function ( ) cos , sin , 2t t t=r

Answer: The curve is a circle 2 2 1x y+ = in the plane z = 2.

c. Describe the curve defined by the vector function ( ) 2cos , 3sin ,t t t t=r

Answer: The curve spirals upward around the cylinder2 2

14 9

x y+ = as t increases.

2.2 DERIVATIVES OF VECTOR FUNCTIONS

DEFINITION 2.2.1 DERIVATIVES OF VECTOR FUNCTIONS

The derivative of a vector function r is defined by

0

( ) ( )( ) lim

h

d t h tt

dt h→

+ −′= =r r rr (2.2.1)

if the limit exists. (See Figure 2.2.1)

As 0h → , the vector ( ) ( )PQ t h t= + −r r

approaches a vector that lies on the tangent line of

C at P. So does ( )t′r .

Figure 2.2.1: ( ) ( )PQ t h t= + −r r

DEFINITION 2.2.2 TANGENT VECTOR

If the vector function r(t) is differentiable, i.e. ( )t′r exists, then

a. The vector ( )t′r is called the tangent vector to the curve C , given by r(t), at P.

b. The unit tangent vector is defined by( )

( )( )

tt

t

′=

′r

Tr

(2.2.2)


20

THEOREM 2.2.1

If ( ) ( ), ( ), ( ) ( ) ( ) ( )t f t g t h t f t g t h t= = + +r i j k , then the limit (2.2.1) exists if and

only if the function ( ), ( ), and ( )f t g t h t are differentiable and

( ) ( ), ( ), ( ) ( ) ( ) ( )d

t f t g t h t f t g t h tdt

′ ′ ′ ′ ′ ′ ′= = = + +rr i j k (2.2.3)

Proof : It follows directly from the definition 2.2.1 and the definition of the derivative of realfunctions.

EXAMPLE 2.2.1

a. Find ( )t′r and (0)′r if 2 2( ) (3 ) sin 4 ln( 2 1)tt t e t t t−= + + + + +r i j k .

Answer: (0) 4 2′ = + +r i j k

b. Find the unit tangent vector at t = 1, if 1( ) cos 2 sin 2 (1 )tt t t e −= + + −r i j k

Answer:( 2sin 2) (2cos 2)

5

− + −i j k

THEOREM 2.2.2 DIFFERENTIATION RULES

Suppose u(t) and v(t) are differentiable vector functions, c is a scalar, and f(t) isdifferentiable real-valued function. Then

a. [ ]( ) ( ) ( ) ( )d

t t t tdt

′ ′+ = +u v u v

b. [ ]( ) ( )d

c t c tdt

′=u u

c. [ ]( ) ( ) ( ) ( ) ( ) ( )d

f t t f t t f t tdt

′ ′= +u u u

d. [ ]( ) ( ) ( ) ( ) ( ) ( )d

t t t t t tdt

′ ′⋅ = ⋅ + ⋅u v u v u v

e. [ ]( ) ( ) ( ) ( ) ( ) ( )d

t t t t t tdt

′ ′× = × + ×u v u v u v

f. [ ]( ( )) ( ) ( ( ))d

f t f t f tdt

′ ′=u u (Chain Rule)

EXAMPLE 2.2.2

a. Let 2 3 3 2( ) 2 , ( ) (1 ) sintt t t e t t t t= − + = − + +u i j k v i j k . Find [ ]( ) ( )d

t tdt

⋅u v by two ways:

apply Theorem 2.2.2.d. and direct calculation (calculate ( ) ( )t t⋅u v , then find its derivative) .

Answer: 3 32 12 (3sin cos )tt t e t t+ + +

b. Show that if ( )t c=r (constant), then ( )t′r is orthogonal to ( )tr .


21

Proof:2 2( ) ( ) ( )t t t c⋅ = =r r r = constant. Then Theorem 2.2.2.d. implies 2 ( ) ( ) 0t t′ ⋅ =r r , thus

( ) ( )t t′ ⋅r r = 0, i.e. ( )t′r is orthogonal to ( )tr .

Geometrically, this result says that if a curve lies on a sphere, then the tangent vector ( )t′r isalways perpendicular to the position vector ( )tr .

2.3 INTEGRALS OF VECTOR FUNCTIONS

If r(t) is continuous on [a, b], we can define its integral in the same way as for real-valuedfunction.

DEFINITION 2.3.1 INTEGRAL OF VECTOR FUNCTIONS

Suppose ( ) ( ), ( ), ( ) ( ) ( ) ( )t f t g t h t f t g t h t= = + +r i j k is continuous on [a, b], i.e. the

component functions ( ), ( ), and ( )f t g t h t are continuous on [a, b]. Then we can define as for

scalar function: subdivide [a, b] in to n subintervals 0 1by points ... .na t t t b= < < < = Denote

the 1[ , ], 1 .i i it t t i n−∆ = ≤ ≤ Use the same notations , 1 ,it i n∆ ≤ ≤ for the lengths of the

subintervals. Choose * , 1 . Denote max{ ,1 }.i i it t i n t i n∈ ∆ ≤ ≤ = ∆ ≤ ≤ Then

* * * *

0 01 1 1 1

( ) lim ( ) lim ( ) ( ) ( )b n n n n

i i i i i i i ii i i ia

t dt t t f t t g t t h t t → →= = = =

= ∆ = ∆ + ∆ + ∆

∑ ∑ ∑ ∑∫r r i j k

( ) ( ) ( )b b b

a a a

f t dt g t dt h t dt

= + + ∫ ∫ ∫i j k (2.3.1)

if the limit on the right hand side of (2.3.1) exists and does not depend on the division of theinterval [a, b] and on the choosing points * , 1 .i it t i n∈ ∆ ≤ ≤

We can extend the Fundamental Theorem of Calculus to continuous vector functions asfollows:

( ) ( ) ( ) ( )b

a

bt dt t b a

a= = −∫r R R R (2.3.2)

where ( )tR is an anti-derivative of r(t), that is, ( ) ( )t t′ =R r .

EXAMPLE 2.3.1

a. Evaluate1

0

( )t dt∫r if 4 3/ 2 2( ) 2 tt t t e−= − +r i j k

Answer:22 2 1

5 5 2

e−−− +i j k


22

b. Evaluate/ 2

0

( )t dt∫ r

if ( ) ( cos3 ) sin 4 ( )tt t t t t e= + + + −r i j k .

Answer:2 2

/ 211

8 3 8e

− + − −

i k

c. Find 2 3( ) if ( ) ( ) 4 ( ) and (0) 2t t t t t t t′ = + + + − = +r r i j k r i j

Answer:2 3 2 4

21 2( 1)2 3 2 4

t t t tt

+ + + + + −

i j k

2.4 ARC LENGTH OF SPACE CURVES

The length of a space curve is defined in the same way as for a plane curve.

DEFINITION 2.4.1 ARC LENGTH

Suppose the continuous space curve C is defined by a differentiable vector function on

[a,b], ( ) ( ), ( ), ( ) ( ) ( ) ( )t f t g t h t f t g t h t= = + +r i j k . Moreover, ( )f t′ , ( )g t′ , and ( )h t′ are

continuous on [a,b] and the curve is traversed exactly once when t increases from a to b. Thenthe length of C is

[ ] [ ] [ ]2 2 2( ) ( ) ( ) ( )

b b

a a

L f t g t h t dt t dt′ ′ ′ ′= + + =∫ ∫ r (2.4.1)

PROPERTY 2.4.1

If we denote the arc length function of the curve C by the length of the part of C betweenr(a) and r(t), a t b≤ ≤ , by

[ ] [ ] [ ]2 2 2( ) ( ) ( ) ( ) ( )

t t

a a

s t f t g t h t dt t dt′ ′ ′ ′= + + =∫ ∫ r (2.4.2)

then

[ ] [ ] [ ]2 2 2( )( ) ( ) ( ) ( )

ds tf t g t h t t

dt′ ′ ′ ′= + + = r (2.4.3)

EXAMPLE 2.4.1

a. Find the length of the circular helix ( ) cos , sin , , 0, 0 2 .t a t a t t a t = > ≤ ≤r

Answer: 22 1 a+

b. Find the arc length function of the circular helix ( ) cos , sin , , 0, 0.t a t a t t t a= ≥ >r

Answer: 21t a+


23

c. Find the length of the curve 2( ) sin cos , cos sin , , 0 2 .t t t t t t t t t = − + ≤ ≤r

Answer: 22 5

2.5 CURVATURE

Suppose that the space curve C, defined by a vector function ( )tr , is smooth, that is r(t) isdifferentiable and ( )t′ ≠r 0 . The unit tangent vector T(t), defined in (2.2.2), indicates thedirection of the curve:

( )( )

( )

tt

t

′=

′r

Tr

(2.5.1)

T(t) changes its direction less or more if C is fairly straight or twists more sharply. The

curvature of C at a given point is a measure of how quickly the curve changes direction atthat point. Thus we can define

DEFINITION 2.5.1 CURVATURE

The curvature of a curve is defined by

d

ds= T

(2.5.2)

where T is the unit tangent vector and s is the arc length function

The formula (2.5.1) can be expressed by

//

/ /

d dtd d dt

ds ds dt ds dt= = =

TT T

Noting (2.4.3), we obtain

( )

( )

t

t

′=

′

T

r(2.5.3)

EXAMPLE 2.5.1

a. Show that the curvature of the circle of radius a is 1/a.

Solution:

We can take the circle with equation defined by

( ) cos , sin , 0 , 0 2 .t a t a t t = ≤ ≤r Then

( ) sin , cos , 0 ; ( ) ;t a t a t t a′ ′= − =r r so( )

( ) sin , cos , 0 ;( )

tt t t

t

′= = −

′r

Tr


24

( ) cos , sin , 0 ; ( ) 1.t t t t′ ′= − − =T T Therefore ( ) / ( ) 1/t t a ′ ′= =T r .

b. Determine the curvature of the circular helix ( ) cos , sin , , 0 2 .t a t a t t t= ≤ ≤r

Answer:2 1

a

a +

The following Theorem is more convenient for calculating the curvature.

THEOREM 2.5.1

The curvature of the curve given by the vector function r(t) is calculated from

3

( ) ( )( )

( )

t tt

t

′ ′′×=

′r r

r(2.5.4)

Proof

It follows from (2.4.3) and (2.5.1) that( ) ( ) ( ) ( ) ( )t t t s t t′ ′ ′= =r r T T

( ) ( ) ( ) ( ) ( )t s t t s t t′′ ′′ ′ ′⇒ = +r T T

[ ] [ ]( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )t t s t t s t t s t t′ ′′ ′ ′′ ′ ′⇒ × = × +r r T T T

[ ] [ ]2( ) ( ) ( ) , becauseofs t t t′ ′= × × =T T T T 0

[ ] [ ] [ ]2 2 2( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )t t s t t t s t t t s t t′ ′′ ′ ′ ′ ′ ′ ′⇒ × = × = =r r T T T T T ,

because 1=T and then ′⊥T T , by Example 2.2.2.b.

Now, (2.5.4) follows from (2.5.3).

COROLLARY OF THEOREM 2.5.1

The curvature of a plane curve with equation ( )y f x= is calculated from

[ ]3/ 22

( )( )

1 ( )

f xx

f x

′′=

′+ (2.5.5)

Proof

Apply (2.5.4), noting that the plane curve ( )y f x= can be given by a vector function

( ) , ( ), 0 .x x f x=r

EXAMPLE 2.5.2

a. Show that the curvature of the circle of radius a is 1/a, applying (i) (2.5.4) , (ii) (2.5.5) .

b. Find the curvature of the curve siny x= at the point with 0x = and at / 2x = .

Answer: (0) 0; ( / 2) 1= =


25

d. Find the curvature of the curve 2 1y x= + at A(0, 1), B(1, 2), and C(2, 5).

Answer: 3/ 2 3/ 2(0) 2, (1) 2 / 5 , and (2) 2 /17= = = .

2.6 NORMAL AND BINORMAL VECTORS

Suppose that the space curve C, defined by a vector function ( )tr , is smooth. Note that ( )tT

is the unit tangent vector of the curve C. Then ( )tT and ( )t′T are orthogonal, ′⊥T T , byExample 2.2.2.b.

DEFINITION 2.6.1 NORMAL AND BINORMAL VECTORS. NORMAL PLANE

a. The unit vector of ( )t′T , denoted by N(t),

( )( )

| ( ) |

tt

t

′=

′T

NT

(2.6.1)

The vector N(t) is called the unit normal vector.

b. The vector

( ) ( ) ( )t t t= ×B T N (2.6.2)

is perpendicular to both ( )tT and ( )tN . ( )tB is called the binormal vector.

DEFINITION 2.6.2 NORMAL PLANE AND OSCULATING PLANE

a. The plane determined by N(t) and ( )tB at the point ( )P t ∈C , is called the normal

plane of C at ( )P t .

b. The plane determined by N(t) and ( )tT at the point ( )P t ∈C , is called the osculating

plane of C at ( )P t .

c. The circle, that lies in the osculating plane of C at ( )P t , has the same unit tangent vector

( )tT , lies on the concave side of C , and has radius 1/ ( )t = , it means that has the same

curvature ( )t as C at ( )P t , is called the osculating circle of C at ( )P t . (See Figure 2.6.1)

Figure 2.6.1

The normal plane is orthogonal to the tangent vector ( )tT , while the osculating plane is

the one that comes closest to containing the part of C near the point ( )P t .

EXAMPLE 2.6.1

Find the radius and graph the osculating circle of the curve 2siny x= at ( / 2, 2)P .


26

Solution: 3/ 22

2sin( )

1 4cos

xx

x

−= ⇒

+ ( / 2) 2 = and 1/ 2 = . See Figure 2.6.2

-π -π/2 π/2 π

-2

-1.5

-1

-0.5

0.5

1

1.5

2

x

y

Figure 2.6.2

2.7 MOTION IN SPACE: VELOCITY AND ACCELERATION

Suppose a particle moves on a smooth space curve C, defined by a vector function ( )tr . Thevelocity vector v(t) is defined from

0

( ) ( )( ) lim ( )

t

t t tt t

t∆ →

+ ∆ − ′= =∆

r rv r (2.7.1)

The speed is, see (2.4.3),

| ( ) | | ( ) |ds

t tdt

′= =v r (2.7.2)

The acceleration vector a(t) is( ) ( ) ( )t t t′ ′′= =a v r (2.7.3)

EXAMPLE 2.7.1

a. Find the velocity and acceleration vectors and speed of a particle with position vector

3 2( ) 2 , , 3tt t e t=r

Answer: 3 3 6 2( ) 2, 3 , 6 , ( ) 0, 9 , 6 , | ( ) | 4 9 36t t tt e t t e t e t= = = + +v a v .

b. A moving particle starts at an initial position (0) 2, 3, 4= −r with initial velocity

(0) 1, 5, 4= −v . Find its velocity and position at time t if 2 2( ) 2, , tt t e=a . Find its

speed at t = 1.

Answer:

3 2

( ) 2 1, 5, 4 ,3 2

tt et t= + + −v

4 22( ) 2, 3, 4 ,

12 4

tt et t t= + + − +r

2256 ( 8)| (1) | 9

9 4

e −= + +v

c. An object with mass m and that moves in an elliptical path in the plane xOy with constantangular speed has position vector ( ) cos sint a t b t = +r i j . Find the force acting on


27

the object and show that it is directed toward the origin. (Such a force is called acentripetal force).

EXERCISES

1. Show that the curve with parametric equations cos , sin ,x t t y t t z t= = = lies on the

cone 2 2 2x y z+ =

2. Show that the curve with parametric equation 2 21, 1 2 , 1 3x t y t z t= − = + = − passesthrough the points A(0, 3, -2), B(8, 7, -26), and C(0, -1, -2) but not D(3, -3, -2).

3. Try to sketch the curve of intersection of the circular cylinder 2 2 1x y+ = and parabolic

cylinder 2z x= . Then find the parametric equations for this curve.

4. Suppose that u and v are vector functions that possess limits as 0t t→ . Prove

a.0 0 0

lim ( ) ( ) lim ( ) lim ( )t t t t t t

t t t t→ → →

⋅ = ⋅ u v u v

b.0 0 0

lim ( ) ( ) lim ( ) lim ( )t t t t t t

t t t t→ → →

× = × u v u v

5. Find the unit tangent vector, the normal vector, and the binormal vector at the point withthe given value of t.

a. 2 323( ) , ,t t t t=r , t = 1. b. 2 2 2( ) sin , cos , , 0t t tt e t e t e t= =r

6. The curves 1( ) sin , cos 1, 2t t t t= −r and 2 22 ( ) , , 2t t t t t= +r intersect at the origin.

Find the angle of intersection.

7. Find the derivative of ( ) ( )t t⋅u v and ( ) ( )t t×u v if

2 2 2( ) , , , ( ) sin , cos 2 ,t t tt t e e e t t t t− −= + =u v

8. Prove that

a. [ ] [ ]( ) ( ) ( ) ( )d

t t t tdt

′ ′′× = ×r r r r

b. [ ]( ) [ ]( ) ( ) ( ) ( ) ( ) ( )d

t t t t t tdt

′ ′′ ′ ′′′⋅ × = ⋅ ×r r r r r r

c. [ ]1( ) ( ) ( )

( )

dt t t

dt t′= ⋅r r r

r

9. Find the curvature and the equation of the osculating circle of the ellipse 2 29 4 36x y+ =at the points (2, 0) and (0, 3).


28

10. Find the velocity vector, acceleration vector, and speed of a moving particle with thegiven position function.

a. ( ) 2 , ,t tt t e e−=r . b. ( ) cos , sin ,tt e t t t=r

11. A force 20 N acts directly upward from the xy-plane on an object with mass 4 kg. Theobject starts at the origin with initial velocity (0) 1, 2, 2= −v . Find its position function

and its speed at time t.

12. Find the curvature of the trajectory in Exercise11.

13. A projectile is fired from the ground with angle of elevation , initial position at theorigin, and initial speed 0v . Suppose that air resistance is negligible and the only external

force is due to gravity, find the position vector ( ) ( ) ( )t x t y t= +r i j of the projectile.

What value of maximizes the horizontal distance traveled and what is the maximumheight in this case? Determine these values if 0 98 /v m s= .

Answers3. 2cos , sin , cosx t y t z t= = = .

5. a. 2 2 1 1 2 2 2 1 23 3 3 3 3 3 3 3 3, , ; , , ; , ,− − − .

b. 51 2 2 2 1 2 43 3 3 5 5 3 5 3 5 3 5, , ; , , 0 ; , , −−

9. 2 2 2 281 3 5 1629 4 4 3 9, ( 2.5) ; , ( )x y x y+ + = + − = .

11. 252( ) , 2 , 2t t t t t= − +r ; 2| ( ) | 25 20 9t t t= + +v

◙ ◙ ◙ ◙ ◙ ◙ ◙ ◙

Calculus 2 – Chapter 3: Double Integrals Nguyen Van Ho - 2009

29

Chapter 3DOUBLE INTEGRALS3.1 DEFINITIONS AND PROPERTIES

We extend the definite integral of one variable functions to double integral of two variablefunctions. Double integrals give us to compute areas, masses,..., and center of plane regions.

DEFINITION 3.1.1 DOUBLE INTEGRAL

Let ( , )f x y be defined on the domain D, 2[ , ] [ , ]D G a b c d⊂ = × ⊂ . Divide the rectangle Ginto sub-domains (sub-rectangles) by lines

0 1 2 ... ma x x x x b= < < < < = and 0 1 2 ... nc y y y y d= < < < < =Denote 1,1 ,i i ix x x i m−∆ = − ≤ ≤ 1,1 ,j j jy y y j n−∆ = − ≤ ≤ and

* *,

1 1

( , )m n

m n i j i ji j

S f x y x y= =

= ∆ ∆∑∑ (3.1.1)

where *1[ , ], 1 ,i i ix x x i m−∈ ≤ ≤ *

1[ , ], 1 ,j j jy y y j n−∈ ≤ ≤

and with convention * * * *( , ) 0, if ( , )i j i jf x y x y D= ∉Denote , max{ , :1 ,1 }m n i jx y i m j n = ∆ ∆ ≤ ≤ ≤ ≤If there exists a unique and finite limit

, ,

* *,0 0

1 1

lim lim ( , )m n m n

m n

m n i j i ji j

S f x y x y L → → = =

= ∆ ∆ =∑∑ (3.1.2)

then the limit is called double integral of ( , )f x y on the domain D and denoted by

( , ) or ( , )D D

f x y dA f x y dxdy∫∫ ∫∫Therefore we can write

( , ) = ( , )D D

f x y dA f x y dxdy∫∫ ∫∫ =,

* *

01 1

lim ( , )m n

m n

i j i ji j

f x y x y → = =

∆ ∆∑∑ (3.1.3)

THEOREM 3.1.1 EXISTENCE CONDITION FOR DOUBLE INTEGRAL

If ( , )f x y is continuous on the domain D, 2[ , ] [ , ]D G a b c d⊂ = × ⊂ , then the doubleintegral defined in (3.1.3) exists.

PROPERTY 3.1.1 DOUBLE INTEGRALS

Suppose double integrals of ( , )f x y , ( , )g x y exist on the domains D, 1 2,D D , where 1 2,D D do

not overlap except perhaps on their boundary, and 1 2D D D∪ = .k is a constant.

a. ( , ) = ( , )D D

k f x y dxdy k f x y dxdy∫∫ ∫∫ , (k is a constant). (3.1.4)


30

b. [ ( , ) ( , )] = ( , ) ( , )D D D

f x y g x y dxdy f x y dxdy g x y dxdy+ +∫∫ ∫∫ ∫∫ (3.1.5)

c. ( , ) ( , ) ( , ) ( , )D D

f x y g x y f x y dxdy g x y dxdy≥ ⇒ ≥∫∫ ∫∫ (3.1.6)

d.1 2

( , ) = ( , ) ( , )D D D

f x y dxdy f x y dxdy f x y dxdy+∫∫ ∫∫ ∫∫ (3.1.7)

e. = Area( )D

dxdy D∫∫ (3.1.8)

f. ( , ) on .Area( ) ( , ) .Area( )D

m f x y M D m D f x y dxdy M D≤ ≤ ⇒ ≤ ≤∫∫ (3.1.9)

PROPERTY 3.1.2 VOLUMES AND AREA - DOUBLE INTEGRALS

a. Let S be a solid that lies above domain D in the xy-plane and under a surface withequation ( , ), ( , ) 0 onz f x y f x y D= ≥ . Then the volume V of S can be expressed by doubleintegral

( , )D

V f x y dxdy= ∫∫ (3.1.10)

b. Let S be a solid that is determined by two surfaces: upper surface with equation

2( , ), ( , )z f x y x y D= ∈ and lower surface with equation 1( , ), ( , ) .z f x y x y D= ∈ Then the

volume V of S can be expressed by double integral

2 1[ ( , ) ( , )]D

V f x y f x y dxdy= −∫∫ (3.1.11)

c. In (3.1.10), if ( , ) 1 onf x y D= then we obtain the area A of D:

D

A dxdy= ∫∫ (3.1.12)

From the definition and Property 3.1.1.f of double integral we obtain the following properties.

PROPERTY 3.1.3 MIDPOINT RULE FOR DOUBLE INTEGRALS

If the domain D is closed and ( , )f x y is continuous on the domain D, then there exist a point

0 0( , )x y D∈ so that

0 0( , ) ( , ).Area( )D

f x y dxdy f x y D=∫∫ (3.1.13)

PROPERTY 3.1.4 APPROXIMATE CALCULATION

The double integral can be approximately calculated by

* *

1 1

( , ) ( , )m n

i j i ji jD

f x y dxdy f x y x y= =

≈ ∆ ∆∑∑∫∫ (3.1.14)


31

3.2 ITERATED INTEGRALS

Suppose that ( , )f x y is continuous on the domain D. Double integral can be evaluated bycalculating two single integrals, as shown below.

CASE 1 If D is a rectangle, 2[ , ] [ , ]D a b c d= × ⊂ , then

( , ) ( , ) ( , )b d d b

D a c c a

f x y dxdy f x y dy dx f x y dx dy

= =

∫∫ ∫ ∫ ∫ ∫ (3.2.1)

EXAMPLE 3.2.1

a. Evaluate I = 2( ) where [0,1] [0,2]D

x x y dxdy D+ = ×∫∫ in two ways:

1 22

0 0

( )x x y dy dx

+

∫ ∫ and2 1

2

0 0

( )x x y dx dy

+

∫ ∫

Answer: 7/6.

b. Find the volume of the solid S bounded by three coordinates planes, the ellipticparaboloid 2 232 ( 4 )z x y= − + , and the planes x = 1, y = 1.

Answer:1 1

2 2

0 0

91(32 4 )

3V x y dy dx

= − − =

∫ ∫

CASE 2 If { } 21 2( , ) : ( ) ( ),D x y x y x a x b = ≤ ≤ ≤ ≤ ⊂ , then

2

1

( )

( )

( , ) ( , )xb

D a x

f x y dxdy f x y dy dx

=

∫∫ ∫ ∫ (3.2.2)

EXAMPLE 3.2.2

a. Find the mass of a plane lamina D that is bounded by two parabolas2 23 , 8 ,y x y x= = + and has mass density function ( , ) 1x y y = + .

Solution:

Two parabolas 2 23 and 8y x y x= = + intersect at 1 2( 2, 12) and (2, 12)M M− . That is

{ }2 2( , ) : 3 8 , 2 2D x y x y x x= ≤ ≤ + − ≤ ≤ . Therefore2

2

82

2 3

704( , ) (1 )

5

x

D x

M x y dxdy y dy dx+

−

= = + =

∫∫ ∫ ∫

b. Find the volume of the tetrahedron T bounded by the coordinate planes and the plane3 3x y z+ + = .

Solution:


32

The base D of T in the xy-plane is3

( , ) : 0 ,0 33

xD x y y x

− = ≤ ≤ ≤ ≤

.

The equation of the upper surface is 3 3z x y= − − . Therefore (3 3 ) 3 / 2D

V x y dxdy= − − =∫∫

CASE 3 If { } 21 2( , ) : ( ) ( ),D x y y x y c y d = ≤ ≤ ≤ ≤ ⊂ , then

2

1

( )

( )

( , ) ( , )yd

D c y

f x y dxdy f x y dx dy

=

∫∫ ∫ ∫ (3.2.3)

EXAMPLE 3.2.3

Evaluate I = 2

D

x y dxdy∫∫ , where D is bounded by 1, 1, 2, and 0xy y y x= = = = .

Solution:

{ }1( , ) : 0 , 1 2yD x y x y= ≤ ≤ ≤ ≤ . 2 1

6D

I x y dxdy= =∫∫Note: If the domain D is complicated, we divide D into sub-domains that belong to one of the

cases 1, 2, 3, and apply (3.1.7) of Property 3.1.1.d.

3.3 CHANGE OF VARIABLES IN DOUBLE INTEGRALS

In Chapter 7 (Calculus 1) we have a formula for change variable: if f is continuous on aninterval [ , ]a b , ( )x t′ is continuous on [ , ] , the range of x(t) belongs to [ , ]a b , and

( ) , ( )x a x b = = , then

( )( ) ( ) ( )b

a

f x dx f x t x t dt

′=∫ ∫ (3.3.1)

How a change of variables ( , )x y to ( , )u v affects a double integral?

Suppose a 1-1 transformation T from uvD in uv-plane to xyD in xy-plane is defined by

differentiable functions ( , ), ( , )x x u v y y u v= = , or in vector function: 0x y= + +r i j k . A smallrectangle S of sides ,u v∆ ∆ in uv-plane becomes a small parallelogram S’ in xy-plane withsecant vectors:

0 , 0u v

x y x yu u u v v v

u u u v v v

∂ ∂ ∂ ∂ ∂ ∂ ∆ ≈ ∆ = ∆ = ∆ + + ∆ ≈ ∆ = ∆ = ∆ + + ∂ ∂ ∂ ∂ ∂ ∂ r r

a r i j k b r i j k


33

The area of S: ( )A S u v= ∆ ∆ . The area of S’: ( ) | | | |u vA S u v′ ≈ ∆ × ∆ = × ∆ ∆a b r r , where

/ / 0

/ / 0u v

x y x x

u u u vx u y ux y y y

x v y vv v u v

∂ ∂ ∂ ∂∂ ∂ ∂ ∂× = ∂ ∂ ∂ ∂ = =∂ ∂ ∂ ∂

∂ ∂ ∂ ∂∂ ∂ ∂ ∂

i j k

r r k k

The Jacobian of the transformation T is defined by

( , )

( , )

x xx y x y x yu vJ

y yu v u v v u

u v

∂ ∂∂ ∂ ∂ ∂ ∂∂ ∂= = = −

∂ ∂∂ ∂ ∂ ∂ ∂∂ ∂

(3.3.2)

Therefore

( , )( ) | |

( , )

x y x y x yA S J u v u v u v

u v u v v u

∂ ∂ ∂ ∂ ∂′ = ∆ ∆ = ∆ ∆ = − ∆ ∆∂ ∂ ∂ ∂ ∂

(3.3.3)

PROPERTY 3.5.1 CHANGE OF VARIABLE IN A DOUBLE INTEGRAL

Suppose a 1-1 transformation T from uvD in the uv-plane to xyD in the xy-plane is defined by

differentiable functions ( , ), ( , )x x u v y y u v= = . Suppose ( , )f x y is continuous on xyD . Then

( )

( )

( , ) ( , ), ( , ) | |

( , )( , ), ( , )

( , )

xy uv

uv

D D

D

f x y dx dy f x u v y u v J du dv

x yf x u v y u v du dv

u v

=

∂=∂

∫∫ ∫∫

∫∫(3.3.4)

EXAMPLE 3.3.1

a. Evaluate the plane area limited by the ellipse2 2

2 21, 0, 0.

x ya b

a b+ = > >

Solution:

Change variables: 2 20, ; 1

0 uv

ax yu v J ab D D u v

ba b= = ⇒ = = ↔ = + ≤

( )uvD D

A D dxdy ab dudv ab⇒ = = =∫∫ ∫∫

b. Evaluate ( ) /( )x y x y

D

e dxdy+ −∫∫ , where D is the trapezium with vertices (0, -1), (1, 0), (0, -2)

and (2, 0). (Hint: Use the change of variable ,u x y v x y= + = − )

Solution:


34

Use the change of variable ,u x y v x y= + = − . Then , whereuv uvD D D↔ is the trapezium

with vertices (-2, 2), (2, 2), (1, 1) and (-1, 1).

1, ; | |

2 2 2

u v u vu x y v x y x y J

+ −= + = − ⇒ = = ⇒ =

( )2

( ) /( ) / / 1

1

1 3| | | |

2 4uv

vx y x y u v u v

D D v

e dxdy e J dudv dv e J dudv e e+ − −

−

⇒ = = = −∫∫ ∫∫ ∫ ∫

c. EvaluateD

y dxdy∫∫ , where D is bounded by the x-axis, and the parabolas 2 4 4y x= −

and 2 4 4y x= + such that 0.y ≥ (Hint: Use the change of variable 2 2 , 2x u v y uv= − = )

Solution:

[0, 1] [0, 1]uvD D↔ = × .

( )2 2 2 2, 2 4x u v y uv J u v= − = ⇒ = + 2 | | 2uvD D

y dxdy uv J dudv⇒ = =∫∫ ∫∫

3.4 DOUBLE INTEGRALS IN POLAR COORDINATES

Remind that cos , sinx r y r= = cos sin( , )

sin cos( , )

rx yJ r

rr

−∂⇒ = = =∂

(3.4.1)

Therefore, as given in Section 3.3,

( )( , ) cos , sinxy rD D

f x y dx dy f r r rdrd=∫∫ ∫∫

(3.4.2)

CASE 1 If the domain D is a polar rectangle, { }( , ) : ,D r a r b = ≤ ≤ ≤ ≤ , then

( , ) ( cos , sin )b

D a

f x y dxdy f r r r dr d

=

∫∫ ∫ ∫ ( cos , sin )b

a

f r r d r dr

=

∫ ∫(3.4.3)

CASE 2 If { }1 2( , ) : ( ) ( ),D r r r r = ≤ ≤ ≤ ≤ , then

2

1

( )

( )

( , ) ( cos , sin )r

D r

f x y dxdy f r r r dr d

=

∫∫ ∫ ∫ (3.4.4)

EXAMPLE 3.4.1

a. Evaluate I = 2 2( 2 )D

x y dxdy+∫∫ , where D is the domain in the first quadrant and bounded

by two circles 2 2 1x y+ = and 2 2 16x y+ = .


35

Solution:

I =/ 2 4

2 2 2 2

0 1

( 2 ) (1 sin )D

x y dxdy r r dr d

+ = +

∫∫ ∫ ∫

/ 2 43

0 1

3 cos 2 765

2 16r dr d

−= =

∫ ∫ .

b. Evaluate I = ( )2 22D

x y dxdy+ +∫∫ , where D is the domain bounded by the circle

2 2 2x y y+ = .

Solution:

The polar equation of the circle 2 2 2x y y+ = is 2sin , 0 .r = ≤ ≤

I = ( )2sin 3 2sin

2 2 2

00 0 0

162 (2 ) 2

3 9D

rx y dxdy r r dr d r d

+ + = + = + = + ∫∫ ∫ ∫ ∫

3.5 APPLICATION OF DOUBLE INTEGRALS

3.5.1. AREA AND VOLUME

See (3.1.10)-(3.1.12), Property 3.1.2.

EXAMPLE 3.5.1

a. Find the volume of the solid inside both the cylinder 2 2 4x y+ = and the ellipsoid2 2 24 4 64x y z+ + =

Solution :

{ }( , ) : 2, 0 2D r r = ≤ ≤ ≤

( )2 2

2 2 2

0 0

642 64 4( ) 4 16 8 3 3

3D

V x y dx dy r rdr d

= − + = − = −

∫∫ ∫ ∫ .

b. Find the volume of the solid bounded by two paraboloids 2 23z x y= − − and2 2 1z x y= + − .

Solution:2 2

2 2 2 2 2

0 0

[(3 ) ( 1)] (4 2 ) 4D

V x y x y dx dy r r dr d

= − − − + − = − =

∫∫ ∫ ∫

.

c. Find the area of one loop of the four-leaved rose cos 2r =

Solution:


36

-1 -0.5 0.5 1

-1

-0.5

0.5

1

x

y

r = 1

r = cos(2θ)

{ }( , ) : 0 cos 2 , / 4 / 4D r r = ≤ ≤ − ≤ ≤ ./ 4 cos2

/ 4 0

/ 8D

A dx dy r dr d

−

= = =

∫∫ ∫ ∫

d. Find the volume of the solid between two paraboloids 2 23( )z x y= + , 2 2z x y= + , and

inside the cylinder 2 22y x y= + .Solution :

{ }( , ) : 0 2sin ,0D r r = ≤ ≤ ≤ ≤2sin

2 2 2 2 2

0 0

[(3( ) ( )] 2 . 3D

V x y x y dx dy r r dr d

= + − + = =

∫∫ ∫ ∫ .

e. Find the volume of the solid above the cone 2 2z x y= + and below the sphere2 2 2 1x y z+ + = .

Answer:

{ }( , ) : 2 / 2, 0 2D r r = = ≤ ≤ ;2 1

13 2

V = −

3.5.2. DENSITY AND MASS

Suppose a thin plate or a lamina occupies a region D on the xy-plane and its mass density at apoint (x, y) in D is given by ( , )x y . Suppose ( , )x y is continuous in D. This means that

0( , ) lim

mx y

A

→

∆=∆

(3.5.1)

where A∆ is the area of an –neighborhood G of (x, y) , and m∆ is the mass of G .

The total mass of the lamina is

( , )D

m x y dx dy= ∫∫ (3.5.2)


37

EXAMPLE 3.5.2

a. Find the mass of the lamina that occupies the region D bounded by 2 , 2,x y y x= = − andhas the density ( , )x y x =

Solution

2

22

1

y

D y

m x dx dy x dx dy+

−

= =

∫∫ ∫ ∫ ( )

22 4

1

1 364 4

2 5y y y dy

−

= + + − =∫

b. Find the mass of the lamina that occupies the part of the disk 2 2 1x y+ = in the firstquadrant if the density at ( , )x y is proportional to its distance from the x-axis.

Answer: k/3, where k is the proportional coefficient.

c. Find the mass of a triangular lamina with vertices (0, 0), (1, 0), and (0, 2) if the densityis ( , ) 1 3x y x y = + +

Answer: 8 / 3m =

3.5.3. MOMENTS AND CENTER OF MASSSuppose a thin plate or a lamina occupies a region D on xy-plane. Its mass density at a point(x, y) in D is given by ( , )x y .

The moment about the x-axis of the lamina is determined by

. ( , )x

D

M y x y dx dy= ∫∫ (3.5.3)

The moment about the y-axis of the lamina is determined by

. ( , )y

D

M x x y dx dy= ∫∫ (3.5.4)

The center of mass ( , )x y of the lamina is determined by

( , ) . ( , )

,( , ) ( , )

y xD D

D D

x x y dx dy y x y dx dyM M

x ym mx y dx dy x y dx dy

= = = =∫∫ ∫∫∫∫ ∫∫

(3.5.5)

Note:

The moment Mx (My) measures the tendency of the lamina to rotate about the x-axis (aboutthe y-axis). The center of mass is the point where a single particle of mass m would have thesame moments as the mass lamina.

EXAMPLE 3.5.3

Find the center of mass of the laminas given in Example 3.5.2


38

a. D is bounded by 2 , 2,x y y x= = − and has the density ( , ) ; 36 / 5x y x m= = :

2

22

1

45.

8

y

x

D y

M y x dx dy y x dx dy+

−

= = =

∫∫ ∫ ∫ ;

423.

28y

D

M x x dx dy= =∫∫ ; ( ) 2115 225, ,

1008 288x y

=

d. The lamina occupies the part of the disk 2 2 1x y+ = in the first quadrant and thedensity at ( , )x y is proportional to its distance from the x-axis. / 3m k= .

Answer: ( , )16x

D

kM y x y dx dy= =∫∫

; ( , )8y

D

kM x x y dx dy= =∫∫ ; ( ) 3 3

, ,8 16

x y =

c. The lamina is a triangle with vertices (0, 0), (1, 0), and (0, 2) and the density is( , ) 1 3x y x y = + + . 8 / 3m =

Answer:3 11

( , ) ,8 16

x y =

3.5.4. MOMENT OF INERTIA

The moment of inertia about the x-axis of the lamina is determined by

2 ( , )x

D

I y x y dx dy= ∫∫ (3.5.6)

The moment of inertia about the y-axis of the lamina is determined by

2 ( , )y

D

I x x y dx dy= ∫∫ (3.5.7)

The moment of inertia about the origin of the lamina is determined by

( )2 20 ( , ) x y

D

I x y x y dx dy I I= + = +∫∫ (3.5.8)

EXAMPLE 3.5.4

Find the moments of inertia 0, , andx yI I I of a disk D with density 2 2( , )x y x y = + , center

the origin, and radius a.

Answer:

By the symmetry,5 5

00

2, then ;

5 2 5x y x y

Ia aI I I I I= = = = =

.

3.5.5. SURFACE AREA

Let S be a surface 2( , ), ( , ) ,z f x y x y D= ∈ ⊂ where ( , )f x y has continuous partialderivatives.


39

How can we define and determine the area of S? As usually, we divide S into patches i jS by

the planes perpendicular to the x-axis and the y-axis. See Figure below. The area ( )i jA S of i jS

is approximated by the area ( )i jA T of a parallelogram i jT on the tangent plane at some point

( ), , ( , )i j i j i j i jP x y f x y S∈ with two sides determined by tangent vectors ai and b j in the

direction of the x-axis and the y-axis, respectively:

, 0, ( , )i i x i j ix f x y x= ∆ ∆a , 0, , ( , )j j y i j jy f x y y= ∆ ∆b

Clearly,

0 ( , ) ( , ), ( , ), 1

0 ( , )i j i x i j i i j x i j y i j

j y i j j

x f x y x x y f x y f x y

y f x y y

× = ∆ ∆ = ∆ ∆ − −∆ ∆

i j k

a b

Therefore2 2

, ,( ) ( ) 1 ( , ) ( , )i j i j i j x i j y i j i jA S A T f x y f x y x y ≈ = × = + + ∆ ∆ a b

and the surface area, A(S), of S is determined from

[ ] 22( ) 1 ( , ) ( , )x y

D

A S f x y f x y dx dy = + + ∫∫ (3.5.9)

EXAMPLE 3.5.5

a. Evaluate the surface area of the plane z ax by c= + + that is limited by the cylinder2 2

2 21, 0, 0.

x yp q

p q+ = > >

Answer: 2 2 2 2( ) 1 1D

A S a b dx dy pq a b= + + = + +∫∫

b. Show that the surface area of the sphere of radius a is 2( ) 4A S a= .

c. Evaluate the surface area of the paraboloid 2 2( ), 0 , 0, 0.z a x y z h a h= + ≤ ≤ > >

Answer: ( )3/ 2

21 4 1

6ah

a + −


40

EXERCISES1. Evaluate the double integral of the function f (x, y) on the domain D. Find the average

value of f on the domain D.a. 2 22f x y= + , { }( , ) : 0 1, 1 2D x y x y= ≤ ≤ ≤ ≤

b. 23f x y= − , D is the triangle ABC, A(1, 3), B(3, 5), C(5, 0) .

c. 2f x y= − , D is the quadrilateral ABEF, A(0, 3), B(2, 5), E(4, 3) and F(3, 1) .

d. 2f x y= − , D is bounded by x- and y-axis and the line going through A(1, 3) andB(4, 1).

2. Find the volume of the solid S bounded by2

282

xz y= − − , the planes 2x = and 2y = .

and the three coordinate planes.

3. Find the volume of the solid S in the first octant bounded by the cylinder 29z y= − andthe plane 2x = .

4. Find the volume of the solid S lying under the elliptic paraboloid 2 2/ 4 / 9 1x y z+ + =and above the square [ 1, 1] [ 2, 2]− × − .

5. Find the volume of the solid S lying under the paraboloid 2 22z x y= + and above the

region in the xy-plane bounded by 22 andy x y x= = .

6. Evaluate the double integral of the function f (x, y) on the domain D.

a. 2f x y= − , { }2 2( , ) : 4D x y x y= + ≤ .

b. cosf x y= , D is bounded by 20, , and 1.y y x x= = =Answer: a. 0; b. (1 cos1) / 2− ;

7. Find the volume of the given solid.a. Under the paraboloid 2 2z x y= + and above the region in the xy-plane bounded by

2 2,y x x y= = .

b. Bounded by the cylinder 2 2 9x z+ = , the plane 2 2x y+ = , and the three coordinateplanes.

8. Find the center of mass of the lamina that occupies the region D and has the givendensity function .

a. { } 2( , ) : 1 1, 0 1 ; ( , )D x y x y x y x= − ≤ ≤ ≤ ≤ =

b. { }2 2( , ) : 1, 0, 0 ; ( , )D x y x y x y x y ky= + ≤ ≥ ≥ = , k is a positive constant.

9. Find the area of the region D in the xy-plane, where { }2 2( , ) : 1 2D x y x y y= ≤ + ≤ .

10. Find the surface area of the part of the surface 23 4z x y= + that lies above the triangleABC in the xy-plane, O(0, 0), A(1, 2), B(1, 0) .

11. Find the surface area of the part of the plane 3 4 2z x y= + + that lies above the rectangle[0, 5] [1, 4]D = × in the xy-plane.


41

12. Evaluate (3 4 )D

x y dxdy+∫∫ , where D is bounded by ,y x= 2,y x= − 2 ,y x= − 3 2y x= −

using the transformation 1 13 3( ), ( 2 )x u v y v u= + = − .

13. Evaluate the surface area of the ellipsoid2 2 2

2 21, 0, 0, .

x y za b a b

a b

+ + = > > ≠

ANSWERS

2. 24; 3. 36; 4. 166/27;

5. 108/35; 6. a. 0; b. (1 cos1) / 2− ;

7. a. 6/35; b.9 1

arcsin(2 / 3) (11 5 27)2 6

+ − ;

8. a. (0, 1/2); b. (3 / 8, 3 /16) ;

9.3

3 2+ ; 10. 3/ 253 / 54 ;

11. 15 26 ; 12. 11/3

13. Case 1: b a> ⇒ 2

2

22

2

1 1( ) 4 arctan ; where 1 0

2ba

aA S b

b

= + − = − >

Case 2: b a< ⇒ 2

2

22

2

1 1( ) 4 ln , where 1 ,0 1

2 1ba

aA S b

b

+= + = − < < −

Calculus 2 – Chapter 4: Triple Integrals Nguyen Van Ho - 2009

42

Chapter 4TRIPLE INTEGRALS4.1 DEFINITIONS AND PROPERTIES

We extend the double integral of two-variable functions to the triple integral of three-variablefunctions. Triple integrals give us to compute volumes, masses,..., and center of solids.

DEFINITION 4.1.1 TRIPLE INTEGRAL

Let ( , , )f x y z be defined on the domain D, 3[ , ] [ , ] [ , ]D G a b c d s t⊂ = × × ⊂ . Divide therectangular box G into sub-boxes by planes

0 1 2 ... ma x x x x b= < < < < = , 0 1 2 ... nc y y y y d= < < < < = and 0 1 2 ... ps z z z z t= < < < < =

Denote

1,1 ,i i ix x x i m−∆ = − ≤ ≤ 1,1 ,j j jy y y j n−∆ = − ≤ ≤ 1,1 ,k k kz z z k p−∆ = − ≤ ≤

* * *, ,

1 1 1

( , , )pm n

m n p i j k i j kji j k

S f x y z x y z= = =

= ∆ ∆ ∆∑∑∑ (4.1.1)

where*

1[ , ], 1 ,i i ix x x i m−∈ ≤ ≤ *1[ , ], 1 ,j j jy y y j n−∈ ≤ ≤ *

1[ , ], 1 ,k k kz z x k p−∈ ≤ ≤* * * *( , ) 0, if ( , )i j i jf x y x y D= ∉ , in convention.

Denote

, , max{ , , :1 ,1 ,1 }m n p i j kx y z i m j n k p = ∆ ∆ ∆ ≤ ≤ ≤ ≤ ≤ ≤

If there exists an unique and finite limit

, , ,

* * *, ,0 0

1 1 1

lim lim ( , , )m n p m n

pm n

m n p i j k i j ki j k

S f x y z x y z L → → = = =

= ∆ ∆ ∆ =∑∑∑ (4.1.2)

then the limit is called triple integral of ( , , )f x y z on the domain D and denoted by

( , , ) or ( , , )D D

f x y z dV f x y z dx dy dz∫∫∫ ∫∫∫Therefore we can write

,

* * *

01 1 1

( , , ) ( , , ) lim ( , , )m n

pm n

i j k i j ki j kD D

f x y z dV f x y z dx dy dz f x y z x y z → = = =

= = ∆ ∆ ∆∑∑∑∫∫∫ ∫∫∫ (4.1.3)

THEOREM 4.1.1 EXISTENCE CONDITION FOR DOUBLE INTEGRAL

If ( , , )f x y z is continuous on the domain D, 3[ , ] [ , ] [ , ]D G a b c d s t⊂ = × × ⊂ , then the tripleintegral defined in (4.1.3) exists.


43

PROPERTY 4.1.1 TRIPLE INTEGRALS

Suppose that triple integrals of ( , , )f x y z , ( , , )g x y z exist on the domains D, 1 2,D D ,

where 1 2,D D do not overlap except perhaps on their boundary, and 1 2D D D∪ = , and k is a

constant.

a. . ( , , ) ( , , )D D

k f x y z dxdydz k f x y z dxdydz=∫∫∫ ∫∫∫ , (k is a constant). (4.1.4)

b. [ ( , , ) ( , , )] ( , , ) ( , , )D D D

f x y z g x y z dxdydz f x y z dxdydz g x y z dxdydz+ = +∫∫∫ ∫∫∫ ∫∫∫ (4.1.5)

c. ( , , ) ( , , ) on ( , , ) ( , , )D D

f x y z g x y z D f x y z dxdydz g x y z dxdydz≥ ⇒ ≥∫∫∫ ∫∫∫ (4.1.6)

d.1 2

( , , ) ( , , ) ( , , )D D D

f x y z dxdydz f x y z dxdydz f x y z dxdydz= +∫∫∫ ∫∫∫ ∫∫∫ (4.1.7)

e. V( )D

dxdydz D=∫∫∫ (4.1.8)

f. ( , , ) on .V( ) ( , , ) .V( )D

m f x y z M D m D f x y z dxdydz M D≤ ≤ ⇒ ≤ ≤∫∫∫ (4.1.9)

PROPERTY 4.1.2 MIDPOINT RULE FOR TRIPLE INTEGRALS

If the domain D is closed and ( , , )f x y z is continuous on the domain D, then there exist a

point 0 0 0( , , )x y z D∈ so that

0 0 0( , , ) ( , , ). ( )D

f x y z dxdydz f x y z V D=∫∫∫ (4.1.10)

PROPERTY 4.1.3 APPROXIMATE CALCULATION

The triple integral can be approximately calculated by

( , , )D

f x y z dx dy dz ≈∫∫∫ * * *

1 1 1

( , , )pm n

i j k i j ki j k

f x y z x y z= = =

∆ ∆ ∆∑∑∑ (4.1.11)

4.2 ITERATED INTEGRALS

Suppose that ( , , )f x y z is continuous on the domain D. The triple integral (4.1.3) can beevaluated by calculating three single integrals, as shown below.

CASE 1 If D is a rectangular box, 3[ , ] [ , ] [ , ]D a b c d s t= × × ⊂ , then

( , , ) ( , , ) ( , , )b d t b d t

D a c s a c s

f x y z dxdydz f x y z dz dy dx f x y z dz dy dx

= =

∫∫∫ ∫ ∫ ∫ ∫ ∫ ∫ (4.2.1)


44

CASE 2 If { }2 31 2( , , ) : ( , ) ( , ), ( , )D x y z z x y z z x y x y G= ≤ ≤ ∈ ⊂ ⊂ , then

2

1

( , )

( , )

( , , ) ( , , )z x y

D G z x y

f x y z dxdydz f x y z dz dxdy

=

∫∫∫ ∫∫ ∫ (4.2.2)

a. If { } 21 2( , ) : ( ) ( ),G x y y x y y x a x b= ≤ ≤ ≤ ≤ ⊂ then (4.2.2) can be written in detail as

2 1

1 1

( ) ( , )

( ) ( , )

( , , ) ( , , )y x z x yb

D a y x z x y

f x y z dxdydz f x y z dz dy dx=∫∫∫ ∫ ∫ ∫ (4.2.3)

b. If { } 21 2( , ) : ( ) ( ),G x y x y x x y c y d= ≤ ≤ ≤ ≤ ⊂ then (4.2.2) can be written in detail as

2 1

1 1

( ) ( , )

( ) ( , )

( , , ) ( , , )x y z x yd

D c x y z x y

f x y z dxdydz f x y z dz dx dy=∫∫∫ ∫ ∫ ∫ (4.2.4)

EXAMPLE 4.2.1

a. Evaluate I = ( ) where [0,1] [0,2] [0,3]D

z x y dxdydz D+ = × ×∫∫∫

Answer:1 2 3

0 0 0

27( ) ( )

2D

I z x y dxdydz x y zdzdydx= + = + =∫∫∫ ∫ ∫ ∫

b. Evaluate I = { }where ( , , ) : 0, 0,0 1D

xy dxdydz D x y z x y z x y= ≥ ≥ ≤ ≤ − −∫∫∫

Answer:11 1

0 0 0

1

120

x yx

D

I xydxdydz xy dzdydx− −−

= = =∫∫∫ ∫ ∫ ∫

c. Evaluate I = ( )D

x y dxdydz+∫∫∫ , where

{ }( , , ) : 4 5 ( ),0 1,0 2)D x y z x y z x y x y= + − ≤ ≤ − + ≤ ≤ ≤ ≤ .

Answer: { }5 ( )

4

49( ) ( ) , ( , ) : 0 1,0 2

3

x y

D G x y

x y dxdydz x y dz dxdy G x y x y− +

+ −

+ = + = = ≤ ≤ ≤ ≤

∫∫∫ ∫∫ ∫

4.3 CHANGE OF VARIABLES IN TRIPLE INTEGRALS

In Chapter 3 we have seen how a change of variables ( , )x y to ( , )u v affects a double integral.

Suppose a 1-1 transformation T from uvwD in uvw-space to xyzD in xyz-space is defined by

differentiable functions

T: ( , , ), ( , , ), ( , , )x x u v w y y u v w z z u v w= = = (4.3.1)


45

Similarly as in Section 3.3 of Chapter 3, The Jacobian of the transformation T is defined by

( , , )

( , , )

x x x

u v wx y z y y y

Ju v w u v w

z z z

u v w

∂ ∂ ∂∂ ∂ ∂

∂ ∂ ∂ ∂= =∂ ∂ ∂ ∂

∂ ∂ ∂∂ ∂ ∂

(4.3.2)

Therefore

( )( , , ) ( , , ), ( , , ), ( , , ) | |xyz uvwD D

f x y z dx dydz f x u v w y u v w z u v w J du dvdw=∫∫∫ ∫∫∫ (4.3.3)

4.4 TRIPLE INTEGRALS IN CYLINDRICALAND SPHERICAL COORDINATES

Look at Section 1.4 in Chapter 1 for definitions of cylindrical coordinates and sphericalcoordinates. See Figures 5.1.1 – 5.1.2.

To convert from rectangular to cylindrical coordinates and inversely, we use

2 2 2 , tan ,y

r x y z zx

= + = = (4.4.1)

cos , sin ,x r y r z z = = = (4.4.2)

Figure 5.1.1 Cylindrical coordinates

Note that the Jacobian, determined from (4.3.2) is

( , , )

( , , )

x y zJ r

z r ∂= =∂

(4.4.3)

Therefore

dxdydz r drd dz= (4.4.4)

To convert from rectangular to spherical coordinates and inversely, we use


46

2 2 2

2 2 2, tan , cos

y zx y z

x x y z = + + = =

+ +(4.4.5)

sin cos , sin sin , cosx y z = = = (4.4.6)

Figure 5.1.2 Spherical coordinates

Note that the Jacobian, determined from (4.3.2) is

2( , , )sin

( , , )

x y zJ

∂= =∂

(4.4.7)

Therefore2 sindx dy dz d d d = (4.4.8)

EXAMPLE 4.4.1

a. Evaluate 2 2

D

I x y dxdydz= +∫∫∫ , where { }2 2 2 2( , , ) : 1, 1 5D x y z x y x y z= + ≤ − − ≤ ≤

Answer:2

2 1 52

0 0 1

46

15r

I r dzdrd

−

= =∫ ∫ ∫

b. Evaluate ( )2 2

D

I x y dxdydz= +∫∫∫ , where { }2 2( , , ) : 3D x y z x y z= + ≤ ≤ .

Answer:2 3 3

3

0 0

243

10r

I r dz dr d = =∫ ∫ ∫

c. Evaluate2 2 2 3 / 2( )x y z

D

I e dxdydz− + += ∫∫∫ , where D is the ball of center (0, 0, 0) and radius a.

Answer: ( )341

3ae−−

4.5 APPLICATION OF TRIPLE INTEGRALS

4.5.1. VOLUME( )

D

V D dx dy dz= ∫∫∫ (4.5.1)


47

EXAMPLE 4.5.1

a. Find the volume of the solid: { }2 2 2 2( , , ) :1 4,0 1D x y z x y z x y= ≤ + ≤ ≤ ≤ + +

Answer:

22 2 1

0 1 0

21

2

r

V r dzdrd

+

= =∫ ∫ ∫

b. Find the volume of the solid: { }2 2 2 2 2 2( , , ) : 1, ( 1) 1D x y z x y z x y z= + + ≤ + + − ≤

Answer:

3 22

2

2 1

0 0 1 1

5

12

r

r

V r dzdrd

−

− −

= =∫ ∫ ∫

c. Use spherical coordinates to find the volume of the solid that lies above the cone2 2z x y= + and below the sphere 2 2 2z x y z= + +

Answer:2 cos/ 4

2

0 0 0

sin8

V d d d = =∫ ∫ ∫

4.5.2. DENSITY AND MASS

Suppose a solid occupies a region D on the space and its mass density at a point (x, y, z) in Dis given by ( , , )x y z . Suppose ( , , )x y z is continuous in D. It means that

0( , , ) lim

mx y z

V

→

∆=∆

(4.5.2)

where V∆ is the volume of an –neighborhood G of (x, y, z), and m∆ is the mass of G .

The total mass of the solid is

( , , )D

m x y z dx dy dz= ∫∫∫ (4.5.3)

EXAMPLE 4.5.2

a. Find the mass of the solid D = { }2( , , ) : 0 2,0 4 ,0x y z y x y z y≤ ≤ ≤ ≤ − ≤ ≤ if the

density ( , , )x y z x = .

Answer: 2

b. Find the mass of the cube D = [0, ] [0, ] [0, ]a a a× × if the density is 2 2 2x y z = + + .

Answer: 5a

4.5.3. MOMENTS, CENTER OF MASS, AND MOMENTS OF INERTIA

Suppose a solid occupies a region D on the space. Its mass density at a point (x, y, z) in D isgiven by ( , , )x y z .


48

The moment about the yz-plane, xz-plane, and xy-plane, of the solid is determined by

. ( , , )yz

D

M x x y z dx dy dz= ∫∫∫ (4.5.4)

. ( , , )xz

D

M y x y z dx dy dz= ∫∫∫ (4.5.5)

. ( , , )xy

D

M z x y z dx dy dz= ∫∫∫ (4.5.6)

The center of mass ( , , )x y z of the lamina is determined by

, ,yz xyxzM MM

x y zm m m

= = = (4.5.7)

If the density is constant, the center of mass is called the centroid of D.

The moment of inertia about the x-axis, y-axis, and z-axis are determined by

( )2 2 ( , , )x

D

I y z x y z dx dy dz= +∫∫∫ (4.5.8)

( )2 2 ( , , )y

D

I x z x y z dx dy dz= +∫∫∫ (4.5.9)

( )2 2 ( , , )z

D

I x y x y z dx dy dz= +∫∫∫ (4.5.10)

EXAMPLE 4.5.3

Find the center of mass of the cube given in Example 4.5.2.b

Answer: (7 /12, 7 /12, 7 /12)a a a


49

EXERCISES

4.1 Find the volume of the solid S bounded by four cylinders2 23 , 8 ,y x y x= = + 2 2, and (1 )z y z y= = + .

4.2 Find the volume of the solid, in the first octant, bounded by the coordinate planes andbetween the planes 12x y z+ + = , and 3 4 5 60x y z+ + = .

4.3 Given11 1

0 0

( , , )y

x

f x y z dz dy dx−

∫ ∫ ∫a. Rewrite the integral in other orders.

b. Evaluate the integral, if ( , , ) 1 2f x y z x z= + +

4.4 Find the moments of inertia of the cube D = [0, ] [0, ] [0, ]a a a× × if the density is k(constant).

4.5 Find the mass and the moments of inertia of the solid bounded by the cylinder2 2 2x y a+ = and planes 0,z z h= = , (a >0, h > 0), if the mass density at a point M is

proportional to the distance from the xy-plane to M.

4.6 Find the mass and the moments of inertia of a ball of center 0 and radius a if the massdensity at a point M is proportional to the distance from the origin to M.

4.7 Find the volume of a solid bounded by the ellipsoid2 2 2

2 2 21

x y z

a b c+ + =

4.8 Find the volume of the solid that lies above the cone / 3 = and below the sphere4cos . = Determine its centroid.

4.9 Find the mass and center of mass of the solid S bounded by the paraboloid 2 24( )z x y= +

and the plane , 0,z a a= > if S has the constant density K.

4.10 a. Find the centroid of a solid homogeneous hemisphere of radius a, and

b. Find the moment of inertia of the solid about a diameter of its base.

ANSWERS

4.14

26015

+ ; 4.2 312; 4.3 2/15; 4.4 523x y zI I I ka= = =

4.522 2 2 2 2 4 21 1 1

2 4 2 4, ( ),ax y zm k a h I I k a h h I k a h = = = + = .

4.6 4 649, x y zm k a I I I k a = = = = . 4.7

4

3abc ; 4.8 10 ; (0, 0, 2.1);

4.9 2 / 8; (0, 0, 2 / 3);K a a 4.10 a.3

0,0, ;8

a

b. 54 /15K a ;

Calculus 2 – Chapter 5: Line Integrals Nguyen Van Ho

50

Chapter 5

LINE INTEGRALS5.1 LINE INTEGRALS

We extend the single integral over a curve C. Such integrals are called line integrals.

DEFINITION 5.1.1 LINE INTEGRALS ON PLANE

Let ( , )f x y be defined on an open domain D that includes the curve C. Divide C into sub-

arcs that are labeled together with their lengths by , 1 ,is i n∆ ≤ ≤ and choose any points* * *( , )i i i iP x y s∈ ∆ ,1 ,j n≤ ≤ see Figure 5.1.1, and form the sum

* *

1

( , )n

n i j ii

S f x y s=

= ∆∑ (5.1.1)

Figure 5.1.1

Denote max{ :1 }n is i n = ∆ ≤ ≤ . If there exists an unique and finite limit L

* *

0 01

lim lim ( , )n n

n

n i j ii

S f x y s L → → =

= ∆ =∑ (5.1.2)

then the limit is called line integral of ( , )f x y on the curve C and denoted by

( , )f x y ds∫C

(5.1.3)

Therefore

* *

0 01

( , ) lim lim ( , )n n

n

n i j ii

f x y ds S f x y s → → =

= = ∆∑∫C

(5.1.4)

THEOREM 5.1.1 EXISTENCE CONDITION FOR LINE INTEGRALS

a. If the equation of the curve C is ( ), ,y x a x b= ≤ ≤ where ( )x ′ is continuous on theinterval [a, b], (we say that the curve is smooth) and ( , )f x y is continuous on an open

domain D that contains C. Then the line integral defined by (5.1.4) exists and it isevaluated from


51

( ) [ ]2( , ) , ( ) 1 ( )

b

a

f x y ds f x x x dx ′= +∫ ∫C

(5.1.5)

b. If the parametric equations of the curve C is ( ), ( ), ,x x t y y t t = = ≤ ≤ where ( )x t′and ( )y t′ are continuous on [, ], (we say that the curve is smooth), and ( , )f x y is

continuous on the domain D that contains C. Then the line integral defined by (5.1.4)exists and it is evaluated from

( ) [ ] [ ]2 2( , ) ( ), ( ) ( ) ( )f x y ds f x t y t x t y t dt

′ ′= +∫ ∫C

(5.1.6)

Note: We do not care about the direction of the curve. But we suppose always that a < b in(5.1.5) and < in (5.1.6).

PROPERTY 5.1.1 LINE INTEGRALS

a. If C is a piecewise-smooth curve, that is, C is a union of a finite number of smooth

curves Ci , 1 i m≤ ≤ . Then

1 2 m

( , ) ( , ) ( , ) ( , )f x y ds f x y ds f x y ds f x y ds= + + +∫ ∫ ∫ ∫C C C C

(5.1.7)

b. . ( , ) . ( , )k f x y ds k f x y ds=∫ ∫C C

, (k is a constant). (5.1.8)

c. [ ]( , ) ( ) ( , ) ( , )f x y g xy ds f x y ds g x y ds+ = +∫ ∫ ∫C C C

(5.1.9)

d. ( , ) ( , ) on ( , ) ( , )f x y g x y D f x y ds g x y ds≥ ⊃ ⇒ ≥∫ ∫C C

C (5.1.10)

e. ( , ) on . ( ) ( , ) . ( )m f x y M D m L f x y ds M L≤ ≤ ⊃ ⇒ ≤ ≤∫C

C C C (5.1.11)

( )L C = length of C .

f. If ( , )x y is the linear mass density of a thin wire shaped like the curve C . Then themass of the wire is determined from

( , )m x y ds∫C

= (5.1.12)

The center of mass of the wire is located at the point ( , )x y , where

1 1( , ) , ( , )x x x y ds y y x y ds

m m ∫ ∫

C C

= = (5.1.13)

g. The length of C is determined from

(L ds∫C

C) = (5.1.14)


52

EXAMPLE 5.1.1

a. Find the length, mass, and mass center of a thin wire shaped like the parabola2 ,0 1y x x= ≤ ≤ , with linear mass density ( , )x y x = .

Solution:

i.( )1

2

0

2 5 ln 2 5( 1 4

4L ds x dx

+ += + =∫ ∫

C

C) = , by integration by parts. (5.1.15)

ii.1

2

0

5 5 11 4

12m x ds x x dx

−= + =∫ ∫C

=

iii.( )

( )1

2 2

0

3 18 5 ln 2 51 1 1( , ) 1 4

16 5 5 1x x x y ds x x dx J

m m m

− + = + = =−∫ ∫

C

=

( )12 2

0

18 5 ln 2 51 4 , by integration by parts and (5.1.15)

64J x x dx

− + = + =

∫

iv. ( )1

2 2

0

1 1 1 25 5 1( , ) . 1 4

10 5 5 1y y x y ds x x x dx K

m m m

++ =−∫ ∫

C

= = =

12 2

0

25 5 1. 1 4 , by integration by parts

120K x x x dx

+ = + =

∫

b. Evaluate 2(1 )I x x y ds+ +∫C

= , C : the circumference of the upper half of the unit disk.

Solution:

C = C1 + C2, C1 = AOB, C2 = ACB,

1

21 (1 ) 2I x x y ds+ +∫

C

= = ,2

2 22 3(1 )I x x y ds= + + = +∫

C

, 83I = +

DEFINITION 5.1.2 LINE INTEGRALS IN SPACE

Let ( , , )f x y z be defined on an open domain D that contains a curve C in space. Divide Cinto sub-arcs that are labeled together with their lengths by , 1 ,is i n∆ ≤ ≤ and choose any

points * * * *( , , )i i i i iM x y z s∈ ∆ ,1 ,j n≤ ≤ and form the sum

* * *

1

( , , )n

n i j i ii

S f x y z s=

= ∆∑ (5.1.16)


53

Denote max{ :1 }n is i n = ∆ ≤ ≤ . If there exists an unique and finite limit

* * *

0 01

lim lim ( , , )n n

n

n i j j ii

S f x y z s → → =

= ∆∑ (5.1.17)

then the limit is called line integral of ( , , )f x y z on the curve C and denoted by

( , , )f x y z ds∫C

(5.1.18)

Therefore we can write

* * *

0 01

( , , ) lim lim ( , , )n n

n

n i j j ii

f x y z ds S f x y z s → → =

= = ∆∑∫C

(5.1.19)


If the parametric equations of the curve C is ( ), ( ), ( ), ,x x t y y t z z t t = = = ≤ ≤where ( )x t′ , ( )y t′ and ( )z t′ are continuous on [, ], (we say that the curve is smooth),

and ( , , )f x y z is continuous on an open domain D that contains C. Then the lineintegral defined by (5.1.19) exists and it is evaluated from

( ) [ ] [ ] [ ]2 2 2( , , ) ( ), ( ), ( ) ( ) ( ) ( )f x y z ds f x t y t z t x t y t z t dt

′ ′ ′= + +∫ ∫C

(5.1.20)

EXAMPLE 5.1.2

a. Find the mass of a thin wire shaped like the circular helix given by the equationscos , sin , , 0 2x a t y a t z t t = = = ≤ ≤ , with linear mass density 2( , , )x y z x z = .

Solution:2

2 2 2 2 2 2 2 2 2 2

0

( , , ) cos sin cos 1 1m x y z ds x z ds a t t a t a t dt a a

= = = + + = +∫ ∫ ∫C C

b. Evaluate ( )xy z ds+∫C

, where C is the triangle OAB, (1,1,0), (1,1,1)A B .

Solution:

9 2 2 5 3( ) ( ) ( ) ( )

6OA AB OB

I xy z ds xy z ds xy z ds xy z ds+ += + = + + + + + =∫ ∫ ∫ ∫

C

5.2 LINE INTEGRALS OF VECTOR FIELDS

Firstly, let us explain the notions and terminologies of regions in 2

Simple region type I, see Figure 1. Simple region type II, see Figure 2.

Simple region: both of type I and type II, see Figure 5.4.1 and 5.4.2, Section 5.4

Union of finite simple regions type I and type II, see Figure 3, 1 2 3D D D D= ∪ ∪ .


54

Simply-connected region, see Figure 1, 2, and 3.

Not simply-connected region, see Figure 4.

Connected region, see Figure 1, 2, 3, and 4.

DEFINITION 5.2.1 VECTOR FIELD ON 2

Let 2D ⊂ . A vector field on D is a function F that assign to each point ( , )x y D∈ a two-dimensional vector

( , ) ( , ), ( , ) ( , ) ( , )x y P x y Q x y P x y Q x y= = +F i j (5.2.1)

DEFINITION 5.2.2 VECTOR FIELD ON 3

Let 3D ⊂ . A vector field on D is a function F that assign to each point ( , , )x y z D∈ a three-dimensional vector

( , , ) ( , , ), ( , , ), ( , , ) ( , , ) ( , , ) ( , , )x y z P x y z Q x y z R x y z P x y z Q x y z R x y z= = + +F i j k (5.2.2)

The air velocity vectors that indicate the wind speed and direction at points on D; the forcefield that associates a force vector with each point in a region D (e.g. the gravitational forcefield); the magnetic field... are examples for vector fields.

DEFINITION 5.2.3 THE WORK DONE BY A FORCE FIELD

Let 3D ⊂ . Let F be a force field, as given in (5.2.2), defined on an open 3D ⊂ thatcontains a curve C given by a vector function

( ) ( ), ( ), ( ) ,t x t y t z t t= ≤ ≤r .

The end points of C are ( ) ( )( ), ( ), ( ) and ( ), ( ), ( )M x y z M x y z= = , according

to andt t = = , respectively. How can we determine the work W done by the force field F

in moving a particle along the curve C from toM M ?

Divide C into small sub-arcs that are labeled together with their lengths by , 1 ,is i n∆ ≤ ≤ and

choose any points * * * *( , , )i i i i iM x y z s∈ ∆ , corresponding to the parameter value *it ,1 j n≤ ≤ .

The work done by the force F in moving a particle along the sub-arc is∆ , is approximately* * * *( , , ). ( )

ii j i ix y z t s∆F T , where *( )i

tT is the unit tangent vector at *iM , 1 i n≤ ≤ . Form the sum

* * * *

1

( , , ). ( )i

n

i j i ii

x y z t s=

∆∑F T (5.2.3)


55

Denote max{ :1 }n is i n = ∆ ≤ ≤ . If there exists an unique and finite limit

* * * *

01

lim ( , , ). ( )i

n

i j i ii

x y z t s→ =

∆∑F T = I (5.2.4)

then the work W is defined as the limit I .

DEFINITION 5.2.4 LINE INTEGRALS OF VECTOR FIELD ON 3

Let 3D ⊂ and let F be a vector field defined on 3D ⊂ that contains a curve C given by avector function ( ), .t t ≤ ≤r Use the same method as for the above work problem, wearrive to the limit (5.2.4) and define this limit as in (5.2.5) and call it as the line integral of Falong C from A to B and denoted by

* * * *

01

lim ( , , ). ( )i

n

i j i ii

d ds x y z t s→ =

⋅ = ⋅ = ∆∑∫ ∫ F r F T F T

C C

(5.2.5)


If the curve C is smooth and F is continuous on an open domain D that contains C,then the line integral defined by (5.2.5) exists.

PROPERTY 5.2.1 EXPRESSION FOR LINE INTEGRALS OF VECTOR FIELD ON 3

a. ( , , ), ( , , ), ( , , ) , ,d P x y z Q x y z R x y z dz dx dy dz⋅ = ⋅∫ ∫F rC C

( , , ) ( , , ) ( , , )P x y z dx Q x y z dy R x y z dz= + +∫C

(5.2.6)

b. If the curve C from A to B is given by a vector function

( ) ( ), ( ), ( )t x t y t z t=r , t ≤ ≤ ,

then

d⋅∫ F rC

( ) ( ) ( )( ), ( ), ( ) , ( ), ( ), ( ) , ( ), ( ), ( ) ( ), ( ), ( )P x t y t z t Q x t y t z t R x t y t z t x t y t z t dt

′ ′ ′= ⋅∫

( ) ( ) ( )( ), ( ), ( ) ( ) ( ), ( ), ( ) ( ) ( ), ( ), ( ) ( )P x t y t z t x t Q x t y t z t y t R x t y t z t z t dt

′ ′ ′ = + + ∫ (5.2.7)

c. Denote the curve C from toA B by ABC and the curve C from toB A by BAC . Then

BA AB

d d⋅ = − ⋅∫ ∫F r F rC C

(5.2.8)


56

EXAMPLE 5.2.1

a. Evaluate d⋅∫ F rC

, where 1xe xy−= +F i j andC is given by 2 3 , 0 1.t t t= + ≤ ≤r i j

Solution:1 11 1

8xd e dx xydy

e−⋅ = + = −∫ ∫F r

C C

b. Evaluate d⋅∫ F rC

, where3 1 ( )xye x y−= + +F i j andC is given by 2 , 1 2.y x x= − ≤ ≤

Solution:3

7 21 27

( )3 2

x e ed ye dx x y dy

−− −⋅ = + + = +∫ ∫F r

C C

c. Find the work done by the force field 2( , , )x y z x xy yz= + +F i j k in moving a particlealong the elliptic helix given by the equations cos , sin , 2 ,x a t y b t z t t= = = ∈ , a, b > 0,

i. From ( ,0,0)A a to ( ,0, 2 )B a − .

ii. From ( ,0, 2 )B a − to ( ,0,0)A a .

Solution:

i. ( ,0,0)A a and ( ,0, 2 )B a − are corresponding to 0t and t = = , respectively.

2 22( ) 4

3AB

d a b a b⋅ = − +∫ F r C

ii. 2 22( ) 4

3BA

d a b a b ⋅ = − − + ∫ F rC

d. The force exerted by an electric charge Q at the origin on a charge q at a point( , , )x y z with position vector , ,x y z=r is 3/ | |Qq=F r r , where is a constant. Find

the work done by the force as the particle moves from (1, 0, 0)A to (0, 1, / 2)B .i. Along a straight line.ii. Along the helix 2: cos , sin , , 0x t y t z t t= = = ≤ ≤ C .

Solution:

i. 2: 1 , , , 0 1x t y t z t t= − = = ≤ ≤C ;

( ) 23/ 22 2 24

11

1

xdx ydy zdzW d Qq Qq

x y z

+ + = ⋅ = = − ++ +

∫ ∫F r

C C

ii. 2: cos , sin , , 0x t y t z t t= = = ≤ ≤ C

2

4

11

1d Qq

⋅ = − +

∫ F r

C


57

5.3 THE FUNDAMENTAL THEOREMFOR LINE INTEGRALS

The Fundamental Theorem (see Chapter 7, Calculus 1) shows that:If ( )f x′ is continuous on [a, b], then

( ) ( ) ( ) ( )b

a

b

af x dx f x f b f a′ = = −∫ (5.3.1)

We have the same result for line integral

DEFINITION 5.3.1 GRADIENT VECTOR FIELD

a. The gradient vector of ( , )f f x y= is defined by

f ff

x y

∂ ∂∇ = +∂ ∂

i j (5.3.2)

b. The gradient vector of ( , , )f f x y z= is defined by

f f ff

x y z

∂ ∂ ∂∇ = + +∂ ∂ ∂

i j k (5.3.3)

THEOREM 5.3.1 THE FUNDAMENTAL THEOREM FOR LINE INTEGRALS

Let C be a smooth curve, in plane (or in space), given by ( ), .t t ≤ ≤r Let f be adifferentiable function of two (or three) variables whose gradient vector f∇ is

continuous on an open domain D that contains C. Then

( ) ( )( ) ( )f d f f ∇ ⋅ = −∫ r r rC

(5.3.4)

Proof:

( ) ( ) ( )f f f

f d x t y t z t dtx y y

∂ ∂ ∂′ ′ ′∇ ⋅ = + + ∂ ∂ ∂ ∫ ∫r

C

( ) ( ) ( )( )( ) ( )

df tdt f f

dt= = −∫

rr r

.

DEFINITION 5.3.2 CONSERVATIVE VECTOR FIELD

A vector field F is called a conservative vector field if there is a scalar function f so that F isthe gradient of f

f= ∇F (5.3.5)


58

EXAMPLE 5.3.1

a. The vector field 3/ | |K=F r r in Example 5.2.1.c is conservative because (5.3.3) is

satisfied with 2 2 2( , , ) /f x y z K x y z= − + + .

b. Show that the vector field2| |e −= rF r is conservative: determine f such that f= ∇F ?

◙®◙ 2 2 2 2 2 2 2( ) ( ) | |1( , , ) , ,

2x y z x y zf x y z e f x y z e e− + + − + + −= − ⇒ = ∇ = = rF r ◙®◙

DEFINITION 5.3.3 INDEPENDENT OF PATH

If F is a continuous vector field on a domain D, we say that the line integral d⋅∫ F rC

is

independent of path if1 2

d d⋅ = ⋅∫ ∫F r F rC C

for any two curves C1 and C2 in D that have the

same initial and terminal points.

THEOREM 5.3.2

d⋅∫ F rC

is independent of path in D if and only if

0d⋅ =∫ F rC *

for any closed path C * in D.

Proof: Clearly.

THEOREM 5.3.3

Let ( , ) ( , )P x y Q x y= +F i j be a vector field in an open connected domain D.

Then d⋅∫ F rC

is independent of path in D if and only if F is conservative .

Proof:

“If” : It follows from (5.3.4), Theorem 5.3.1.

“Only if”: Take the case of 2 . Suppose the line integral of ( , ) ( , )P x y Q x y= +F i j isindependent of path.

Denote for fixed point ( , )A a b .


59

( , )AB BC

f x y d d d= ⋅ = ⋅ + ⋅∫ ∫ ∫F r F r F rC C C

Consider firstly the curve C as ABC in Figure 5.3.1, where ( , )A a b , 1( , )B x y , ( , )C x y . Because

AB

d⋅∫ F rC

does not depend on x, then

( , )BC

f x yd d

x x x

∂ ∂ ∂= ⋅ = ⋅∂ ∂ ∂∫ ∫F r F r

C C

1

( , ) ( , )BC

x

x

Pdx Qdy P x y dx P x yx x

∂ ∂= + = = ∂ ∂ ∫ ∫C

.

Consider secondly the curve C as ABC in Figure 5.3.2, where ( , )A a b , 1( , )B x y , ( , )C x y .

BecauseAB

d⋅∫ F rC

does not depend on y, then

( , )BC

f x yd d

y y y

∂ ∂ ∂= ⋅ = ⋅∂ ∂ ∂∫ ∫F r F r

C C

1

( , ) ( , )BC

y

y

Pdx Qdy Q x y dy Q x yy x

∂ ∂= + = = ∂ ∂ ∫ ∫C

.

Therefore f= ∇F .

Figure 5.3.1 Figure 5.3.2

5.4 GREEN’S THEOREMConsider a domain D in the xy-plane that is bounded by a simple closed curve C. The positive

direction of the curve is chosen so that D is always on the left when a point traverses C in

this direction. Denote by C +the positively oriented.

THEOREM 5.4.1 GREEN’S THEOREM

Let ( , ) ( , )P x y Q x y= +F i j , where ( , )P x y and ( , )Q x y have continuous partialderivatives on an open region G that contains D, where D is a closed region bounded bya piecewise-smooth and simple closed curve C. Then

( , ) ( , )( , ) ( , )

D

Q x y P x yd P x y dx Q x y dy dxdy

x y

∂ ∂⋅ = + = − ∂ ∂ ∫ ∫ ∫∫F r+ +C C(5.4.1)

Proof :

(For the special case when D is a simple region, i.e. D is both of type I and type II, seeFigures 5.4.1 and 5.4.2 below.)


60

4

3

( )

( )

( , ) ( , )x yd

Dc x y

Q x y Q x ydxdy dx dy

x x

∂ ∂= ∂ ∂ ∫∫ ∫ ∫ , see Figure 5.4.2, 3 3 4 4: ( ), : ( )x x y x x y= =C C ,

[ ] [ ]( )4 3( ), ( ),d

c

Q x y y Q x y y dy= −∫ [ ] [ ]4 3( ), ( ), ( , )d c

c d

Q x y y dy Q x y y dy Q x y dy= + =∫ ∫ ∫+C

Then( , )

D

Q x ydxdy

x

∂ =∂∫∫ ( , )Q x y dy∫

+C

(5.4.2)

( , )D

P x ydxdy

y

∂−∂∫∫

2

1

( )

( )

( , )y xb

a y x

P x ydy dx

y

∂= − ∂ ∫ ∫ , see Figure 5.4.1, 1 1 2 2: ( ), : ( )y y x y y x= =C C ,

[ ] [ ]( )2 1, ( ) , ( )b

a

P x y x P x y x dx= − −∫

[ ] [ ]2 1, ( ) , ( ) ( , )a b

b a

P x y x dx P x y x dx P x y dx= + =∫ ∫ ∫+C

Then( , )

D

P x ydxdy

y

∂− =∂∫∫ ( , )P x y dx∫

+C

(5.4.3)

Adding (5.4.2) and (5.4.3), we obtain (5.4.1).

Figure 5.4.1 Figure 5.4.2 Figure 5.4.3

NOTE: The Green’s Theorem can be extended to the case where D is not simply-connectedregion as given in Figure 5.4.3. In this case, the closed curve C is not simple.

COROLLARY 5.4.1 OF GREEN’S THEOREM

a. The area of a domain D in G is determined by (5.4.1) if( , ) ( , )

1Q x y P x y

x y

∂ ∂− =∂ ∂

.

For example

1( )

2A D xdy ydx xdy ydx= = − = −∫ ∫ ∫+ + +C C C

(5.4.4)

b. Let ( , ) ( , )P x y Q x y= +F i j , where ( , )P x y and ( , )Q x y have continuous partialderivatives on an open simply-connected region G and that

( , ) ( , )Q x y P x y

x y

∂ ∂=∂ ∂

throughout G (5.4.5)

Then


61

(i) For any simple closed path C in G that bounds a region D

( , ) ( , ) 0P x y dx Q x y dy+ =∫C(5.4.6)

(ii) ( , ) ( , )d P x y dx Q x y dy⋅ = +∫ ∫F rC C

is independent of path in G. (5.4.7)

(iii) F is a conservative vector field, it means that there is a scalar function f so that

f= ∇F (5.4.8)

EXAMPLE 5.4.1

a. Evaluate 2( sin )x x dx xy dy+ +∫ +Cwhere C is OABO , (1,0), (0,1).A B

Solution:

2( sin )x x dx xy dy+ +∫ +C=

1

6Dydxdy =∫∫

b. Evaluate 2 2( 2 ) (2 3 )xe yx dx xy y dy− + −∫ +Cwhere C is the circle 2 2 2.x y+ =

Solution:

2 2( 2 ) (2 3 )xe yx dx xy y dy− + −∫ +C= 2 22( ) 4

Dx y dxdy + =∫∫ .

c. Using the line integral to show that the area enclosed by the ellipse2 2

2 21 is

x yab

a b+ = .

d. Evaluate22 2( / 2) sin( ) ( / 2)x x yxy y e e dx x ye xy dy− + + + + + ∫

C

where : , 0 1.OA y x x= = ≤ ≤C

Answer: 3 12 2 cos(1) cos( )e e− + −

e. Evaluate 2(sin 2 ) (2 cos )y x dx y x y dy− + +∫ +Cwhere C is the curve 2 2 1x xy y+ + = .

Answer: 0.

We define now two operations that can be performed on vector field F and that haveimportant applications in vector calculus to fluid flow and electricity and magnetism:one, curlF , produces a vector field and the other, div F , produces a scalar field; both concernwith differentiation.


62

5.5 CURL

DEFINITION 5.5.1 VECTOR DIFFERENTIAL OPERATOR “DEL” ∇

The vector differential operator “del” ∇ denoted formally by

x y z

∂ ∂ ∂∇ = + +∂ ∂ ∂

i j k (5.5.1)

is defined as when it operates on a scalar function ( , , )f f x y z= it gives the gradient of f:

x y z

f f ff f f f f

x y z x y z

∂ ∂ ∂ ∂ ∂ ∂∇ = + + = + + = + + ∂ ∂ ∂ ∂ ∂ ∂ i j k i j k i j k (5.5.2)

DEFINITION 5.5.2 CURL OF VECTOR FIELDS

Let ( , , ) ( , , ) ( , , ) ( , , )x y z P x y z Q x y z R x y z= = + +F F i j k be a vector field. Define

curl = ∇×F F (5.5.3)

curl

R Q P R Q P

x y z y z z x x y

P Q R

= ∇×

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ = = − + − + − ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

F F

i j k

i j k(5.5.4)

EXAMPLE 5.5.1

a. 2 2 22 5 ; curl ?x z yxz y z= + −F i j k F

2 2 2

curl

2 5

x y z

x z xyz y z

∂ ∂ ∂= ∇× =∂ ∂ ∂

−

i j k

F F

( ) ( ) ( )2 2 2 210 4 0 2 0 2 (5 2 ) 2yz xyz x yz yz x x yz= − − + − + − = − + + +i j k i j k

b. 3 2 2 3( ) ( ) ( 2 ) ; curl ?x y z x yz x y= + + + + −F i j k F

Answer: 2 3 2curl (6 ), ( 2 ), (2 3 )y y y x x y z= − + − −F

c. 3 2 2 2 32 3 ; curl ?xy z x y z x y= + +F i j k F

Answer: curl =F 0 .


63

EXAMPLE 5.5.2 CONNECTION BETWEEN THE CURL VECTOR AND ROTATIONS

Let B be a rigid body rotating about the z-axis.

The rotation can be described by the vector =w k , where is the angular speed of B. Let

, ,x y z=r be the position vector of P. Let v be the tangential speed of P.

Let be the angle between the z-axis and vector , ,x y z OP= =r

.

sin sin , sin sin ,d OP d = = = = =r v r w r and .Oz OP⊥v

Therefore

= ×v w r

0 0 y x

x y z

= × = = − +i j k

v w r i j curl 2 2

0

x y z

y x

∂ ∂ ∂⇒ = = =∂ ∂ ∂

−

i j k

v k w

It shows that curl v and w have the same direction and | curl | 2 2 | |= =v w .

It means that the higher the angular speed the greater the curl of the tangential speed.

DEFINITION 5.5.3

If curl =F 0 , then F is called irrotational.

THEOREM 5.5.1

If ( , , )f f x y z= has continuous second-order partial derivatives, thencurl ( )f∇ = 0 (5.5.5)

Proof

Sincef f f

fx y z

∂ ∂ ∂∇ = + +∂ ∂ ∂

i j k , then by (5.5.4),


64

curl ( ) ( )

x y z

f fx y z

f f f

∂ ∂ ∂∇ = ∇× ∇ =∂ ∂ ∂

i j k

( ) ( ) ( )zy yz xz zx yx yxf f f f f f= − + − + − =i j k 0 ,

because , ,zy yz xz zx yx yxf f f f f f= = = , see Theorem in Section 8.4, Calculus 1.

COROLLARY 5.5.1 OF THEOREM 5.5.1

If ( , , ) ( , , ) ( , , )P x y z Q x y z R x y z= + +F i j k is a conservative vector field, where, ,P Q and R have continuous partial derivatives, then curl =F 0 , (F is irrotational).

Proof

Since F is a conservative vector field, f= ∇F , by (5.3.5). Then (5.5.5) implies that curl =F 0 .

The converse statement of the Corollary 5.5.1 is also true if D is simply-connected. We cancombine these statements in the following theorem.

THEOREM 5.5.2

Let ( , , ) ( , , ) ( , , )P x y z Q x y z R x y z= + +F i j k be defined on a simply-connected domainD. Let , ,P Q and R have continuous partial derivatives on D. Then

F is conservative ⇔ curl =F 0 (5.5.6)

EXAMPLE 5.5.3

a. Show that 2 3 2 3 2 2 2 2(2 2 ) (3 1/ )y y ye z xe z yz xe z y z= + + + + −F i j k is conservative.Determine f such that .f= ∇F

Solution

It is conservative, because

2 3 2 3 2 2 2 2

curl 0 0 0

2 2 3 1/y y y

x y z

e z xe z yz xe z y z

∂ ∂ ∂= ∇× = = + + =∂ ∂ ∂

+ + −

i j k

F F i j k 0 .

Then there exists a scalar function f such that .f= ∇F Thus2 3 2 3 2 2 2 2, 2 2 , 3 1/y y y

x y zf e z f xe z yz f xe z y z= = + = + −2 3 2 3( , , ) ( , , ) ( , )y y

xf x y z f x y z dx e z dx xe z g y z= = = +∫ ∫ 2 3 ( , )yxe z g y z= +


65

2 3 2 3( , , ) 2 ( , ) 2 2 ( , ) 2y yy y yf x y z xe z g y z xe z yz g y z yz⇒ = + = + ⇒ =

2( , ) ( , ) 2 ( )yg y z g y z dy yz dy y z h z⇒ = = = +∫ ∫2 3 2 3 2( , , ) ( , ) ( )y yf x y z xe z g y z xe z y z h z⇒ = + = + +

2 2 2 2 2 22

1( , , ) 3 ( ) 3y y

zf x y z xe z y h z xe z yz

′⇒ = + + = + −

2

1 1( ) ( )h z h z C

z z′⇒ = − ⇒ = +

Therefore2 3 2 2 3 2 1

( , , ) ( )y yf x y z xe z y z h z xe z y z Cz

= + + = + + + . You can verify that .f= ∇F

b. Verify wherther the vector fields F given in Example 5.5.1.a and b are conservative.

c. Show that the vector field F given in Example 5.5.1.c is conservative. Determine f suchthat f= ∇F ?

d. Show that the vector field F below is conservative. Determine f so that .f= ∇F

2 2 23

22 4 2 4

2 42 ; ; 2

1 1x y x y zy z

xye x e zey z y z

− −= − − + ++ + + +

F

5.6 DIVERGENCE

DEFINITION 5.6.1 DIVERGENCE

Let ( , , ) ( , , ) ( , , )P x y z Q x y z R x y z= + +F i j k , and / , / , /P x Q y R z∂ ∂ ∂ ∂ ∂ ∂ exist, then thedivergence of F is a scalar function defined by

divP Q R

x y z

∂ ∂ ∂= + +∂ ∂ ∂

F (5.6.1)

Using the symbol of the gradient operator “del” we can write symbolically the divergence asthe dot product

div .P Q R

x y z

∂ ∂ ∂= + + = ∇∂ ∂ ∂

F F (5.6.2)

EXAMPLE 5.6.1

a. 2 2 3 4(2 ) ( ) (2 4 )yx e y z xy xz x yz= + + − + +F i j k

Then 2 3div 4 3 16yxe xy yz= + +F

b. 2 3 2 2( ) ( ) ( )x x zxe x z e y yz zx ye− −= + + − + +F i j k

Then 2 2 2div (1 ) 2 3 2x x ze x xz e y z x ye− −= − + + − + −F


66

THEOREM 5.6.1

Let ( , , ) ( , , ) ( , , )P x y z Q x y z R x y z= + +F i j k , and P, Q, R have continuous second-orderpartial derivatives, then

div (curl ) 0=F (5.6.3)

Proof

Apply (5.5.3) and (5.6.1).

EXAMPLE 5.6.2

a. 2 210 , , 2zxy yz yz x e zx yx= − + −F . Evaluate div F .

Answer: div x y z= + +F

b. 2 210 4 , , 2yz xyz x yz= − −G . Evaluate divG .

c. 2 3 26 , 2 , 2 3y y y x x y z= − − − −G . Evaluate div F .

The reason for the name divergence is from the context of fluid (or gas) flow. If F is thevelocity of a fluid, then div F represents the net rate of change of the mass of the fluidflowing from the point ( , , )x y z per unit volume, i.e. div F measures the tendency of the fluidto diverge from the point ( , , )x y z . If div 0,=F then F is said to be incompressible.

5.7 VECTOR FORM OF GREEN’S THEOREM

Consider a vector field on the plane ( , ) ( , )P x y Q x y= +F i j . Then

curl ; curl

0

Q P Q P

x y z x y x y

P Q

∂ ∂ ∂ ∂ ∂ ∂ ∂= = − ⋅ = − ∂ ∂ ∂ ∂ ∂ ∂ ∂

i j k

F k F k (5.7.1)

Then the formula (5.4.1) of Green’s Theorem can be rewritten in the vector form

( ) ( )( , ) ( , ) curlD

d ds P x y dx Q x y dy dxdy⋅ = ⋅ = + = ⋅∫ ∫ ∫ ∫∫F r F T F k+ + +C C C(5.7.2)

Formula (5.7.2) expresses the integral of the tangential component of F along the closedcurve C + as the double integral of the vertical component of curlF over the domain D

enclosed by C +.

What about the integral of the normal component of F along the closed curve C +?


67

If the vector equation of C + is

( ) ( ) ( )t x t y t t b= + ≤ ≤r i j ,

then the unit tangent vector T is

( )( ) 1( ) ( ) ( ) ,

( ) ( )

tt x t y t

t t

′ ′ ′= = +′ ′

rT i j

r r

and the unit outward normal vector n is (note that n is perpendicular to T and to k)1

( ) ( ), ( ) , 0 0, 0, 1( )

t x t y tt

′ ′= = ×′

n T×kr

1 ( ) ( )( ) ( ) 0

( ) ( )0 0 1

y t x tx t y t

t t

′ ′−′ ′= =′ ′

i j ki j

r r

It is easy to provide the following calculations

[ ]( ) ( ) | ( ) |ds t t t dt

′⋅ = ⋅∫ ∫F n F n r+C

( ) ( )1( ), ( ) ( ) ( ), ( ) ( ) | ( ) |

( )P x t y t y t Q x t y t x t t dt

t

′ ′ ′ = − ′∫ rr

( ) ( )( ), ( ) ( ) ( ), ( ) ( )P x t y t y t Q x t y t x t dt

′ ′ = − ∫

, by Green's formulaD

P QQ dx P dy dxdy

x y

∂ ∂= − + = + ∂ ∂ ∫ ∫∫+C

.

Therefore

div ( , )D

ds x y dxdy⋅ =∫ ∫∫F n F+C

(5.7.3)

Formula (5.7.3) expresses the integral of the normal component of F along the closed curveC + in the xy-plane as the double integral of div F over the domain D enclosed by C +.


68

EXERCISES

5.1 Evaluate 4 ,xy ds∫C

where C is the right half of the circle 2 2 16x y+ = .

5.2 Find the mass and the center of mass of a thin wire of shape of the upper half of the unitcircle 2 2 1.x y+ = The density is 21 x y = + .

5.3 Show that the line integral d⋅∫ F rC

is independent of path and evaluate the integral.

a. 2 22 sin , cos 3x y x y y= −F , C is any path from ( 1,0) to (5,1).−

b. 22 cos cos , sin sinx y y x x y x= − − −F , C is any path from (0,0) to (1,1).

5.4 Find the work done by the force field in moving an object.

a. 2 3 3 2, ,x y x y=F from (0,0) to (2,1)A B , by any path.

b. , , ,y xy xyz=F from (0, 0, 0) to (1, 2, 3)A B on a straight line .

5.5 Evaluate the line integral by two methods: directly and using Green’s Theorem.

a. 2 3xy dx x dy+∫ +C, C is the rectangle with vertices (0,0), (2,0), (2,3),and (0,3)

b. 2y ye dx xe dy+∫ +C,C is the circumference of the square with vertices (0,0), (1,0),

(1,1), and (0,1)

c. 2( ) (2 cos )xy e dx x y dy+ + +∫ +C, C is the boundary of the region enclosed by the

parabolas 2 2andy x x y= =

5.6 Let2 2

( , ) ( , )y x

P x y Q x yx y

− += + =+i j

F i j .

a. Show that( , ) ( , )Q x y P x y

x y

∂ ∂=∂ ∂

in \{(0,0)}D =

b. Show that1

d⋅∫ F rC

and2

d⋅∫ F rC

, where C 1 and C 2 are the lower half and upper half

of the circle ( )2 2 1 from ( 1,0) to 1,0x y+ = − , are not the same. Does this contradict

(5.4.7)?

5.7 Find the curl and the divergence of the vector

a. ;xy yz zx= + +F i j k

b. sin cos ;x xe y e y z= + +F i j k

c.2 2 2 2 2 2 2 2 2

;x y z

x y z x y z x y z= + +

+ + + + + +F i j k


69

d. ;yz xz xyxe ye ze= + +F i j k

5.8 Determine whether or not the vector field is conservative. If it is conservative, find thefunction f such that .f= ∇F

a. ;xy yz zx= + +F i j k

b. ;x y z= + +F i j k

c. 2 22 ( 2 ) ;xy x yz y= + + +F i j k

d. ;x z ye e e= + +F i j k

e. ;xz xz xzyze e xye= + +F i j k

5.9 Evaluate the integrals ds⋅∫ F T+C

and ds⋅∫ F n+C

a. 2xy y= +F i j , C +: 2 2 2 0x y y+ − = .

b.2 2 2 2x y x yye xe+ += − +F i j , C +: 2 2 1x y+ = . Answer: e , 0.

c. 2 24 ( 2 )xy x y= + +F i j , C +: 2 2 2 0, 0, and 0, 0 2x y x y y x+ − = ≥ = ≤ ≤

5.10 Prove that

a. curl ( ) curl curl+ = +F G F G ;

b. curl ( ) curl ( )f f f= + ∇ ×F F F .

ANSWERS

5.1 624

5× ;

5.28 3(5 16)

; ( , ) 0,3 8(3 8)

m x y

+= + = + ;

5.3 a. 25sin1 1.− ; b. cos1 sin1.−5.4 a. 8/3; b. 41/6;

5.5 a. 6; b. e – 1; c. 1/3

5.6 b. and − . No (why?)

5.7 a. 0; 3 ; b. 2 ; 0;e c. ; 16 / 3;−

Calculus 2 – Chapter 6: Surface Integrals Nguyen Van Ho

70

Chapter 6

SURFACE INTEGRALSIn this chapter two kinds of surface integrals are taken into consideration: surface integrals ofscalar functions and surface integrals of vector fields

6.1 SURFACE INTEGRALS OF SCALAR FUNCTIONS

PROBLEM 6.1.1 SURFACE MASS

Suppose the differentiable function 2( , ), ( , ) ,z z x y x y D= ∈ ⊂ define a smooth surface S inspace. Suppose the mass density of the surface is determined by ( , , ), ( , , ) .Sx y z x y z = ∈How can we define and determine the mass of S? As usually, we divide S into patches i jS by the

planes perpendicular to the x-axis and the y-axis. See Figure 6.1.1. The area ( )i jA S of i jS is

approximated by the area ( )i jA T of a parallelogram i jT on the tangent plane at some point

( ), , ( , )i j i j i j i jP x y z x y S∈ with two sides determined by tangent vectors ai and b j in the direction

of the x-axis and the y-axis, respectively:

, 0, ( , )i i x i j ix z x y x= ∆ ∆a , 0, , ( , )j j y i j jy z x y y= ∆ ∆b

Figure 6.1.1

Clearly,

0 ( , ) ( , ), ( , ),1

0 ( , )i j i x i j i i j x i j y i j

j y i j j

x z x y x x y z x y z x y

y z x y y

× = ∆ ∆ = ∆ ∆ − −∆ ∆

i j k

a b (6.1.1)

Therefore2 2

, ,( ) ( ) 1 ( , ) ( , )i j i j i j x i j y i j i jA S A T z x y z x y x y ≈ = × = + + ∆ ∆ a b , (6.1.2)


71

the mass of i jS is approximated by

( ), ,( ) , , ( , ) ( )i j i j i j i jm S x y z x y A S≈ ⋅

( ) 2 2, , ( , ) 1 ( , ) ( , )i j i j x i j y i j i ix y z x y z x y z x y x y ≈ ⋅ + + ∆ ∆

and the mass of S is approximated by

( ) 2 2

,

, , ( , ) 1 ( , ) ( , )i j i j x i j y i j i ii j

x y z x y z x y z x y x y ⋅ + + ∆ ∆ ∑ (6.1.3)

Denote

max( , )i jx y = ∆ ∆ (6.1.4)

The mass of S is defined by the limit

( ) 2 2

0,

lim , , ( , ) 1 ( , ) ( , )i j i j x i j y i j i ii j

x y z x y z x y z x y x y

→

⋅ + + ∆ ∆ ∑ (6.1.5)

if the limit exists and is independent of the way of dividing S and choosing the points i jP .

DEFINITION 6.1.1 SURFACE INTEGRALS

Suppose the differentiable function 2( , ), ( , ) ,z z x y x y D= ∈ ⊂ define a smooth surface S in

space. Suppose ( , , )f f x y z= is defined on S. By the same way provided in Problem 6.1.1 with( , , )x y z replaced by ( , , )f f x y z= we arrive to the sum and limit

( ) 2 2

,

, , ( , ) 1 ( , ) ( , )i j i j x i j y i j i ii j

f x y z x y z x y z x y x y ⋅ + + ∆ ∆ ∑ (6.1.6)

( ) 2 2

0,

lim , , ( , ) 1 ( , ) ( , )i j i j x i j y i j i ii j

f x y z x y z x y z x y x y→

⋅ + + ∆ ∆ ∑ (6.1.7)

If the limit (6.1.5) exists and is independent of the way of dividing S and choosing the points i jP ,

then we denote the limit by

( , , )S

f x y z dS∫∫ (6.1.8)

and it is called the surface integral of ( , , )f f x y z= over the surface S.

Therefore

( ) 2 2

0,

( , , ) lim , , ( , ) 1 ( , ) ( , )S

i j i j x i j y i j i ii j

f x y z dS f x y z x y z x y z x y x y→

= ⋅ + + ∆ ∆ ∑∫∫ (6.1.9)

It follows from (6.1.7) that the surface integral can be expressed by the double integral


72

( ) [ ] 22( , , ) , , ( , ) 1 ( , ) ( , )x y

D

f x y z dS f x y z x y z x y z x y dx dy = ⋅ + + ∫∫ ∫∫S

(6.1.10)

The definition and (6.1.4) result to the expression for the mass of a surface S with mass density( , , )x y z :

( ) [ ] 22( ) ( , , ) , , ( , ) 1 ( , ) ( , )x y

D

m x y z dS x y z x y z x y z x y dx dy = ⋅ + + ∫∫ ∫∫ S

S = (6.1.11)

From the existence condition for double integrals we obtain the following Theorem:

THEOREM 6.1.1 EXISTENCE CONDITION FOR SURFACE INTEGRALS

Let 2( , ), ( , )z z x y x y D= ∈ ⊂ be the equation of the surface S . Suppose z has continuous

partial derivative on D. Suppose ( , , )f f x y z= is defined on S and ( ), , ( , )f f x y z x y= is

continuous on D. Then the surface integral (6.1.8) exists.

PROPERTY 6.1.1

1. ( )dS A=∫∫S

S

2. . ( , , ) . ( , , ) , constant.k f x y z dS k f x y z dS k= =∫∫ ∫∫S S

3. [ ]1 2 1 2( , , ) ( , , ) ( , , ) ( , , ) .f x y z f x y z dS f x y z dS f x y z dS+ = +∫∫ ∫∫ ∫∫S S S

4. If S is a union, 1 2∪S =S S , where 1 2 andS S intersect maybe only along their

boundaries then

1 2

( , , ) ( , , ) ( , , ) .f x y z dS f x y z dS f x y z dS= +∫∫ ∫∫ ∫∫S S S

5. If S is a piecewise-smooth surface, that is 1 ... k∪ ∪S =S S where 1 ,..., kS S are smooth

surfaces and intersect only along their boundaries then

1

( , , ) ( , , ) ( , , ) .k

f x y z dS f x y z dS f x y z dS= + +∫∫ ∫∫ ∫∫S S S

6. Suppose the mass density of the surface S is determined by ( , , ), ( , , ) .Sx y z x y z = ∈Then the mass m and the center of mass ( , , )x y z of the surface S are determined from

( , , )m x y z dS= ∫∫S

1 1 1. ( , , ) , . ( , , ) , . ( , , )x x x y z dS y y x y z dS z z x y z dS

m m m= = =∫∫ ∫∫ ∫∫

S S S

(6.1.12)


73

EXAMPLE 6.1.1

Evaluate the surface integral.

a. I = 2, where is the surface 2 6 3 , 0 2,0 2xy dS z x y x y= + ≤ ≤ ≤ ≤∫∫S

S

Solution:2 2

2

0 0

10361 24 36

27I xy dS xdx y y dy= = + + =∫∫ ∫ ∫

S

b. 2 2 2 2 2 2( ) , where is the surface , 4I x y dS z x y x y= + = + + ≤∫∫S

S

Solution:

( )2 2 2 21 391 17

( ) 1 4 460D

I x y x y dx dy+

= + + + =∫∫

c. I = 2 , where is the surface 1 2 3 , 0 3,0 2x yz dS z x y x y= + + ≤ ≤ ≤ ≤∫∫S

S

Solution:3 2

2 2 2 2

0 0

(1 2 3 ) 1 2 3 171 14I x yz dS dx x y x y dy= = + + + + =∫∫ ∫ ∫S

d. 2, where is the surface 4 , 0 2,0 2x dS y x z x z= + ≤ ≤ ≤ ≤∫∫S

S

Solution:

( )2 2

2 2

0 0

11 (2 ) 4 33 33 17 17

6x dS dx x x dz= + + = −∫∫ ∫ ∫

S

e. Find the center of mass of a thin hemisphere of radius a and constant mass density k.

Solution:

The surface can be determined by: 2 2 2 2 2 2( ) , ( , ) :z a x y x y D x y a= − + ∈ + ≤

( )2

2

2 2 2( , , ) 2

D

am k x y z dS k dS k dxdy ka

a x y= = = =

− +∫∫ ∫∫ ∫∫ S S

( , , ) 0y x y z dS =∫∫ S

, ( , , ) 0x x y z dS =∫∫ S

, 3( , , )z x y z dS ka=∫∫ S

, ( ) ( ), , 0, 0, / 2x y z a=

PARAMETRIC SURFACE

Suppose S is determined by a vector equation of two parameters u and v:

2( , ) ( , ), ( , ), ( , ) , ( , )u v x u v y u v z u v u v D= ∈ ⊂r (6.1.13)


74

The area ( )i jA S of i jS is approximated by the area ( )i jA T of a parallelogram i jT determined by

two vectors i uu∆ r and j vv∆ r :

, ,( ) ( ) | | | |i j i j i u j v u v i jA S A T u v u v≈ = ∆ × ∆ = × ∆ ∆r r r r (6.1.14)

where

, , , , ,u v

x y z x y z

u u u v v v

∂ ∂ ∂ ∂ ∂ ∂= =∂ ∂ ∂ ∂ ∂ ∂

r r (6.1.15)

Therefore

( )( , , ) ( , ), ( , ), ( , ) | |u v

D

f x y z dS f x u v y u v z u v du dv = ⋅ × ∫∫ ∫∫ r rS

(6.1.16)

EXAMPLE 6.1.2

a. Show that (6.1.10) is a particular form of (6.1.16)

Solution:

The equation 2( , ), ( , ) ,z z x y x y D= ∈ ⊂ of the surface S may be regarded as a parametric

equation with parameters x, y: 2( , ) , , ( , ) , ( , )x y x y z x y x y D= = ∈ ⊂r r . Then

2 21,0, , 0,1, , 1 0 , ,1 , | | 1

0 1

i j k

r r r r r rx x y y x y x x y x y x y

y

z z z z z z z

z

= = × = = − − × = + + .

b. Evaluate

( )I x yz dS= +∫∫S

, where S is determined by the parametric equations

2 2 , , , 1, 0, 0.x uv y u v z u v u v u v= = + = − + ≤ ≥ ≥Solution:

2 2, , ,1,1 , , , ,1, 1 , | | 2 ( ) 2u v u v

x y z x y zv u u v

u u u v v v

∂ ∂ ∂ ∂ ∂ ∂ = = = = − × = + + ∂ ∂ ∂ ∂ ∂ ∂r r r r

2 2 2 2 2 216 3 6( ) 2 ( ) 2 , where : 1, 0, 0.

30D

I uv u v u v dudv D u v u v− = + − + + = + ≤ ≥ ≥ ∫∫

c. Show that the surface area of a sphere of radius a is 24 a , using surface integral in two ways:

i. Equation of the sphere: 2 2 2 2x y z a+ + = .

ii. The Parametric Equation of the sphere:2( , ) sin cos , sin sin , cos ; [0, ], [0,2 ], | | sina a a a = = ∈ ∈ × =r r r r


75

6.2 SURFACE INTEGRALS OF VECTOR FIELDS

DEFINITION 6.2.1 ORIENTED SURFACES

Consider an orientable (two-sided) surface S that has a tangent plane at any point except atboundary points. There are two unit normal vectors at each points: 1n and 2 1= −n n . If it is

possible to choose a unit normal vector n at every such point ( , , )x y z so that it varies

continuously over S , then S is called an oriented surface. One choice of n provides S with anorientation.

For the closed surface (the boundary of a solid region) the positive orientation is defined as thenormal vectors point outward from the solid.

Note that the Mobius surface has only one side!

Mobius surface

DEFINITION 6.2.2 SURFACE INTEGRALS OF VECTOR FIELDS

Let ( , , ), ( , , ), ( , , )P x y z Q x y z R x y z=F be defined on an oriented surface S with unit normal

vector n, then surface integral of F over S is defined as the surface integral of the normalcomponent of F

d dS⋅ = ⋅∫∫ ∫∫F S F nS S

(6.2.1)

This integral is called the flux of F across the oriented surface S.

PROPERTY 6.2.1

1. d d− +

⋅ = −∫∫ ∫∫F S F SS S

, where S + and S − denote two different sides of S.

2. , constant.k d k d k⋅ = =∫∫ ∫∫F S F SS S

3. ( )1 2 1 2d d d+ ⋅ = ⋅ + ⋅∫∫ ∫∫ ∫∫F F S F S F SS S S

4. If S is a union of two surfaces, 1 2∪S =S S , where 1 2 andS S intersect maybe only

along their boundaries then

1 2

.d d d⋅ = ⋅ + ⋅∫∫ ∫∫ ∫∫F S F S F SS S S


76

GRADIENT, TANGENT PLANE, AND NORMAL LINE

Suppose that the smooth surface S in space is defined by the differentiable function2( , ), or, ( , , ) ( , ) 0, ( , )z z x y g x y z z z x y x y D= = − = ∈ ⊂ .

Suppose S is a smooth surface with equation ( , , )g x y z k= . Let 0 0 0( , , )P x y z be a point on S .

Let C be a curve on S and passes through the point P. The curve C is described by a continuous

vector function ( ) ( ), ( ), ( ) .t x t y t z t=r Let 0t be the parameter value corresponding to P:

( )0 0 0 0 0 0( ), ( ), ( ) ( , , )P x t y t z t x y z= = . Since C is on S , any point ( )( ), ( ), ( )x t y t z t must satisfy the

equation of S :( ( ), ( ), ( ))g x t y t z t k=

Then

0 ( ) 0g dx g dy g dz

g tx dt y dt z dt

∂ ∂ ∂ ′+ + = ⇒ ∇ ⋅ =∂ ∂ ∂

r

( )0 0 0 0 0 0 0 0 0 0at ( ), ( ), ( ) ( , , ) : ( , , ) ( )P x t y t z t x y z g x y z t′⇒ = = ∇ ⊥r

But 0( )t′r is the tangent vector to C at P, then the gradient vector 0 0 0( , , )g x y z∇ is perpendicular

to any curve C on S passing through P. Thus the gradient vector 0 0 0( , , )g x y z∇ is

perpendicular to the tangent plane to the surface S at P.

Therefore the equation of the tangent plane at the point 0 0 0( , , )P x y z is

0 0 0 0 0 0 0 0 0 0 0 0( , , )( ) ( , , )( ) ( , , )( ) 0x y zg x y z x x g x y z y x g x y z z x− + − + − = (6.2.2)

and the equation of the normal line to S at 0 0 0( , , )P x y z is

0 0 0

0 0 0 0 0 0 0 0 0

( ) ( ) ( )

( , , ) ( , , ) ( , , )x y z

x x y x z x

g x y z g x y z g x y z

− − −= = (6.2.3)

EXPRESSION 6.2.1 OF SURFACE INTEGRALS

a. Suppose that S is determined by ( , )z z x y= or ( , , ) ( , ) 0,g x y z z z x y= − = 2( , )x y D∈ ⊂ ,and is upward oriented. It means that the z-component of the unit normal vector n is positive.(see Figure 6.2.1 below). Let ( , , ), ( , , ), ( , , )P x y z Q x y z R x y z=F be defined on S. Then

( ) ( )22

( , ), ( , ), 1

| | 1 ( , ( ,

x y

x y

z x y z x yg

g z x y z x y

− −∇= =∇ + +

n

[ ] 221 ( , ) ( , )x ydS z x y z x y dx dy = + + (6.2.4)

Therefore (6.2.1) has the expression as the double integral below

( ) ( ) ( ), , ( , ) ( , ) , , ( , ) , , ( , )x y

D

dS P x y z x y z x y Q x y z x y z R x y z x y dx dy ⋅ = − − + ∫∫ ∫∫F nS

(6.2.5)


77

b. Suppose S is oriented by the downward unit normal vector n, (see Figure 6.2.2), then

( ) ( )22

( , ), ( , ), 1

| | 1 ( , ( ,

x y

x y

z x y z x y

z x y z x y

−×= − =× + +

a bn

a b,

[ ] 221 ( , ) ( , )x ydS z x y z x y dx dy = + + , see (6.2.4)

( ) ( ) ( ), , ( , ) ( , ) , , ( , ) , , ( , )x y

D

dS P x y z x y z x y Q x y z x y z R x y z x y dx dy ⋅ = + − ∫∫ ∫∫F nS

(6.2.6)

You can write similarly the expressions for the surface integrals if S is given by an equation ofthe form

2 2( , ), ( , ) or ( , ), ( , )x x y z y z D y y x z x z D= ∈ ⊂ = ∈ ⊂

EXPRESSION 6.2.2 OF SURFACE INTEGRALS

Let S be the oriented smooth surface in space.

Let ( , , ), ( , , ), ( , , )P x y z Q x y z R x y z=F be the vector field defined on S.

Let the unit normal vector of S be expressed by the direction cosines, (see 1.1, Chapter 1)

cos ,cos ,cos , =n (6.2.7)

Then

( , , ), ( , , ), ( , , ) cos ,cos ,cosP x y z Q x y z R x y z ⋅ = ⋅F n

( , , ) cos ( , , ) cos ( , , ) cosP x y z Q x y z R x y z= + + (6.2.8)

Therefore (6.2.1) can be expressed as below

( ) ( ) ( ), , cos , , cos , , cosd dS P x y z Q x y z R x y z dS ⋅ = ⋅ = + + ∫∫ ∫∫ ∫∫F S F n S S S

(6.2.9)

Remark

To evaluate the integral in the right hand side of (6.2.9), we need to divide the surface Sinto sub-surfaces that can be expressed by one of the following equations:

2 2 2( , ), ( , ) or ( , ), ( , ) or ( , ), ( , )xy yz xzz z x y x y D x x y z y z D y y x z x z D= ∈ ⊂ = ∈ ⊂ = ∈ ⊂


78

• If S has the equation ( , ), ( , ) y zx x y z y z D= ∈ then

( )

( )

( )

( , ), , if 02

, , cos 0 if2

( , ), , if2

yz

yz

D

D

P x y z y z dydz

P x y z dS

P x y z y z dydz

≤ <= = − < ≤

∫∫

∫∫

∫∫S

(6.2.10)

• If S has the equation ( , ), ( , ) x zy y x z x z D= ∈ then

( )

( )

( )

, ( , ), if 02

, , cos 0 if2

, ( , ), if2

xz

xz

D

D

Q x y x z z dxdz

Q x y z dS

Q x y x z z dxdz

≤ <= = − < ≤

∫∫

∫∫

∫∫S

(6.2.11)

• If S has the equation ( , ), ( , ) x yz z x y x y D= ∈ then

( )

( )

( )

, , ( , ) if 02

, , cos 0 if2

, , ( , ) if2

xy

xy

D

D

R x y z x y dxdy

R x y z dS

R x y z x y dxdy

≤ <= = − < ≤

∫∫

∫∫

∫∫S

(6.2.12)

EXAMPLE 6.2.1

a. Evaluate I = d⋅∫∫F SS

, where , ,x y z=F

and S is the upper half-sphere, upward oriented, 2 2 2 216 , 16z x y x y= − − + ≤

Solution

Apply (6.2.5), I ( ) ( )2 2 2 2( 2 ) ( 2 ) 16 16 384D D

x x y y x y dx dy x y dx dy = − − − − + − − = + + = ∫∫ ∫∫

b. Evaluate I = d⋅∫∫F SS

, where / , / , 2F x z y z z= −

and S is the upper surface, upward oriented, of 2 2 2 24 , 2z x y x y= − − + ≤


79

Solution

Apply (6.2.5),

I ( )2 22 2 2 2

( 2 ) ( 2 ) 4 2 (8ln 2 2)4 4D

x yx y x y dx dy

x y x y

= − − − − + − − − = − − − − −

∫∫

c. Find the flux of the vector field , ,x z y=F outward through the sphere 2 2 2 2x y z a+ + = .

Solution

For the upper half of the sphere, 2 2 2 2 2 21 : , ( , ) :z a x y x y D x y a= − − ∈ + ≤S , apply (6.2.5):

1

d⋅∫∫F SS

( )2 2 2

2 2 2 2 2 2D

x yx a x y y dx dy

a x y a x y

− − = − − − − + − − − −

∫∫

2

2 2 22

D

xy dx dy

a x y

= + − −

∫∫

For the lower half of the sphere, 2 2 2 2 2 22 : ,z a x y x y a= − − − + ≤S , apply (6.2.6):

2

d⋅∫∫F SS

( )2 2 2

2 2 2 2 2 2D

x yx a x y y dx dy

a x y a x y

= + − − − −

− − − − ∫∫

2

2 2 22

D

xy dx dy

a x y

= − − −

∫∫1 2

34

3

ad d d⇒ ⋅ = ⋅ + ⋅ =∫∫ ∫∫ ∫∫F S F S F S

S S S

d. A fluid with density k flows with velocity , 1,y z=v . Find the rate of flow upward through

the paraboloid S : 2 2 2 29 ( ) / 4, 36z x y x y= − + + ≤ .

Solution

The rate of flow upward through the paraboloid S is, applying (6.2.5),

R =2 2

1 9 1622 2 4D

x y x yk d k y dxdy k

− − + ⋅ = − − + − = ∫∫ ∫∫v S S

e. Find the flux of the vector field , 2 , 3x y z=F outward through the cube 3[ 1, 2]− .

Solution:

Apply (6.2.10) – (6.2.12) to six faces of the cube:

For the face S 1 : 212, ( , ) [ 1, 2] , / 2, 0,z x y D = ∈ = − = = =

1 1 1

3 3 2 6 9 54F SS S D

d z dxdy dxdy⋅ = = × = × =∫∫ ∫∫ ∫∫For the face S 2 : 2

11, ( , ) [ 1, 2] , / 2, ,z x y D = − ∈ = − = = =


80

2 2 1

3 3 ( 1) 3 9 27F SS S D

d z dxdy dxdy⋅ = − = − × − = × =∫∫ ∫∫ ∫∫

......i

6

1

162i

d d=

⇒ ⋅ = ⋅ =∑∫∫ ∫∫F S F SS S

f. Find the flux of the vector field 2 , 2 3 , 3x y y z z x= + + +F outward through the triangular

pyramid OABC, (0,0,0), (1,0,0), (0,1,0), (0,0,1)O A B C .

Answer:i

4

1

1 1 12 1

6 3 2i

d d=

⋅ = ⋅ = − − − + =∑∫∫ ∫∫F S F SS S

EXAMPLE 6.2.2 THE ELECTRIC FIELD

Suppose an electric charge Q is located at the origin. According to Coulomb’s Law, the electricforce F exerted by the charge Q on a charge q located at ( , , )M x y z is

3( ) , where , ,

| |

qQx y z

= =F r r rr

(6.2.13)

where is a constant.

The force exerted by the charge Q (located at the origin) on a unit chart, q = 1, located at anypoint 3( , , ), ( , , )M x y z x y z ∈ , is called the electric field of Q:

3( ) , where , ,

| |

Qx y z

= =E r r rr

(6.2.14)

Find the electric flux of E given in (6.2.14) outward through the sphere S : 2 2 2 2x y z a+ + = .

Solution

Method 1

On the sphere S : 2 2 2 23 3

( ) , , , whereQ Q

x y z x y z aa a

= = + + =E r r .

By the same way in calculating the flux in Example 6.2.1 b:

• For the upper half of the sphere, 2 2 2 2 2 21 : , ( , ) :z a x y x y D x y a= − − ∈ + ≤S ,

apply (6.2.5),

1

2 2 23 2 2 2 2 2 2

D

Q x yd x y a x y dx dy

a a x y a x y

− − ⋅ = − − + − − − − − −

∫∫ ∫∫E S

S

2 Q=

• For the lower half of the sphere, 2 2 2 2 2 22 : ,z a x y x y a= − − − + ≤S ,

apply (6.2.6),


81

( )2

2 2 23 2 2 2 2 2 2

D

Q x yd x y a x y dx dy

a a x y a x y

⋅ = + − − − −

− − − − ∫∫ ∫∫E S

S

2 Q= .

Therefore the electric flux of E given in (6.2.11) through the sphere S : 2 2 2 2x y z a+ + = is

4d Q⋅ =∫∫E S S

(6.2.15)

Method 2

Note that, E have the same direction as the unit outward normal vector n of the sphere S2 2 2 2x y z a+ + = , see (6.2.14).

Therefore

3 3 2 2( ) | | on the sphere

| | | | | |E r n r n r

r r rQ Q Q Q

a

⋅ = ⋅ = = = ,

and

2

Qd dS dS

a⋅ = ⋅ =∫∫ ∫∫ ∫∫E S E n

S S S

22 2

Area( ) 4 4Q Q

a Qa a

= = = S .

From (6.2.15), Q can be expressed by the electric flux as below

0

1

4Q d d= ⋅ = ⋅∫∫ ∫∫E S E S

S S

(6.2.16)

where2 2 2

0

18.8546 10 C / N m

4

−= ≈ × (6.2.17)

Formula (6.2.16) gives a result of Gauss’s Law, one of the important laws of electrostatics. TheLaw says that the net charge enclosed by any closed surface S is expressed by (6.2.16).

6.3 STOKES’ THEOREM

Looking back formula (5.7.2) in Chapter 5, for the vector field F in the plane

( ) ( )( , ) ( , ) curlD

d ds P x y dx Q x y dy dxdy⋅ = ⋅ = + = ⋅∫ ∫ ∫ ∫∫F r F T F k+ + +C C C(6.3.1)

Because k is the unit normal vector of the upper side of “surface” D on the xy-plane, the aboveformula can be rewrite in surface integral sense as below

( ) ( , ) ( , )d ds P x y dx Q x y dy⋅ = ⋅ = +∫ ∫ ∫F r F T+ + +C C C


82

( ) ( )curl curlD D

dxdy dS= ⋅ = ⋅∫∫ ∫∫F k F n (6.3.2)

This formula says that the line integral around the boundary C of D of the tangentialcomponent of F is equal to the surface integral of the normal component of the curl of Facross the D.

We have also the same conclusion for the vector field F in the space as given in the Theorem byStokes that we accept.

THEOREM 6.3.1 STOKES’ THEOREMLet S be an oriented piecewise-smooth surface that is bounded by a simple, closed,

piecwise-smooth curve C +, where the positive direction of C coincides with the directionof n in the right hand rule sense. Let F be a vector field with components havingcontinuous partial derivatives on an open region in 3 that contains S. Then

( )d ds⋅ = ⋅∫ ∫F r F T+ +C C

= ( , , ) ( , , ) ( , , )P x y z dx Q x y z dy R x y z dz+ +∫ +C

curl curld dS= ⋅ = ⋅∫∫ ∫∫F S F nS S

(6.3.3)

COROLLARY 6.3.1

If curl =F 0 , then F is conservative and 0d⋅ =∫ F rC

for any closed curve C.

Proof

If curl =F 0 , then by (6.3.3), 0d⋅ =∫ F r+Cfor any closed curve C. Therefore, by Corollary 5.4.1,

F is conservative.

Remark

This Corollary is the converse statement of Corollary 5.5.1, from which follows Theorem 5.5.2

EXAMPLE 6.3.1

a. Evaluate d⋅∫ F r+Cwhere 2 , ,F y x yz= − and C + is the intersection of the cylinder

2 2 1x y+ = and the plane 1x y z+ + = . The direction of C + coincides with the positive directionof the z-axis.

Solution


83

Method 1: Direct calculation of the line integral.

The parametric equations of C + : cos , sin , 1 cos sin , 0 2x t y t z t t t = = = − − ≤ ≤ .

( )2

2

0

( ) ( ) ( ) ( ) ( ) ( ) ( )d y t x t x t y t y t z t z t dt

′ ′ ′⋅ = − + +∫ ∫F r+C

23 2

0

sin cos sin (1 cos sin )(1 cos sin ) 2t t t t t t t dt

′ = + + − − − − = ∫ .

Method 2: Apply Stokes’Theorem’

2

2

, , curl / / / , 0, (1 2 )y x xz x y z z y

y x yz

= − ⇒ = ∂ ∂ ∂ ∂ ∂ ∂ = +−

i j k

F F

d⋅∫ F r+Ccurl curlF S F n

S S

d dS= ⋅ = ⋅∫∫ ∫∫ , apply (6.2.5) where 1z x y= − − ,

[ ](1 ) ( 1) 0( 1) (1 2 )D

x y y dxdy= − − − − + − + +∫∫2 1

0 0

(2 ) (2 cos sin ) 2D

x y dxdy d r r rdr

= − + = − + =∫∫ ∫ ∫ .

b. Evaluate curl dS⋅∫∫ F nS

where , ,yz xz xy= −F and S is the upper part, upward

oriented, of the sphere 2 2 2 25x y z+ + = that lies inside the cylinder 2 2 9x y+ = .

Solution

Method 1:

Direct calculation of the surface integral:

, , curl 0, 2 , 2yz xz xy y z= − ⇒ = −F F

2 2

2 2curl curl 0 ( 2 ) 2 25 72

25D

yd dS y x y dxdy

x y

−⋅ = ⋅ = − − + − − = − −

∫∫ ∫∫ ∫∫F S F n S S

Method 2: Apply Stokes’Theorem’

The parametric equations of the boundary C + of S : 3cos , 3sin , 4, 0 2x t y t z t = = = ≤ ≤2

0

curl [ ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )] 72d d y t z t x t x t z t y t x t y t z t dt′ ′ ′⋅ = ⋅ = − + + =∫∫ ∫ ∫F S F r

+CS

c. Use Stockes’ Theorem to evaluate d⋅∫ F r+Cwhere


84

2 2 2, ,x y y z z x= + + +F and C is the triangle with vertices (1, 0, 0), (0, 1, 0), (0, 0, 1),

oriented counterclockwise as viewed from above.

Solution:

2 2 2

curl / / / 2 , 2 , 2

i j k

F x y z z x y

x y y z z x

= ∂ ∂ ∂ ∂ ∂ ∂ = − − −+ + +

The equation of the plane containing the triangle:

{ }1 or 1 , ( , ) ( , ) : 0 1 ,0 1x y z z x y x y D x y y x x+ + = = − − ∈ = ≤ ≤ − ≤ ≤ .

Applying (6.2.4) and (6.3.3), we obtain

[ ]curl 2(1 ) 2 2 2 1D D

d dS x y x y dxdy dxdy⋅ = ⋅ = − − − − − = − = −∫ ∫∫ ∫∫ ∫∫F r F n+CS

6.4 THE DIVERGENCE THEOREM

Formula (5.7.3) in Chapter 5 expresses the integral of the normal component of F along theclosed curve C + in the xy-plane as the double integral of div F over the domain D enclosed bythe curve

div ( , )D

ds x y dxdy⋅ =∫ ∫∫F n F+C(6.4.1)

Note that C + is the positively oriented boundary of the domain D in the xy-plane. The followingtheorem gives the similar result for the vector field F in space, the domain D in space, and theboundary surface S of D with positive orientation, i.e. with outward orientation.

THEOREM 6.4.1 THE DIVERGENCE THEOREM

Suppose that D is a simple solid region (or D is a finite union of simple solid regions), andS is its boundary surface positively oriented, and ( , , ), ( , , ), ( , , )F P x y z Q x y z R x y z= is

a vector field where ( , , ), ( , , ), and ( , , )P x y z Q x y z R x y z have continuous partial derivativeson an open region that contains D. Then

divF S F n FS S D D

P Q Rd dS dxdydz dxdydz

x y z

∂ ∂ ∂⋅ = ⋅ = = + + ∂ ∂ ∂ ∫∫ ∫∫ ∫∫∫ ∫∫∫ (6.4.2)

We accept the Theorem.

It follows from the Midpoint Rule For Triple Integrals, see Property 4.1.2 of Chapter 4, and theDivergence Theorem that, there exist a point 0 0 0 0( , , )M x y z= in D so that the flux of F across

outward the surface S can be express by

0div div ( ) V( )F S F n F FS S D

d dS dxdydz M D⋅ = ⋅ = = ×∫∫ ∫∫ ∫∫∫ (6.4.3)

where V(D) denotes the volume of D.


85

If D is a ball of a given center ( , , )M x y z= and of a radius a then from (6.4.3) and the continuityof div F , by the hypotheses of the Divergence Theorem,

0

1div ( ) lim

V( )F F S

Sa

M dD→

= ⋅∫∫ (6.4.4)

This equation says that div F at a given point ( , , )M x y z= , div ( )F M , is the net rate of outwardflux per unit volume at M. If div ( ) 0F M > , M is called a source. If div ( ) 0F M < , M is called asink.

EXAMPLE 6.4.1

a. Find the flux of the vector field , ,x z y=F outward through the sphere 2 2 2 2x y z a+ + = .

Solution

We have done the direct calculation in Example 6.2.1b and obtained34

3F S

S

ad

⋅ =∫∫ .

Let us calculate it applying (6.4.2). Since , , then div 1F Fx z y= = and obtain easily that result:34

div Volume( )3

F n FS D D

adS dxdydz dxdydz D

⋅ = = = =∫∫ ∫∫∫ ∫∫∫

b Find the flux of the vector field , 2 , 3x y z=F outward through the cube 3[ 1, 2]− .

Solution

The direct calculation in Example 6.2.1d gives 162F SS

d⋅ =∫∫ .

Let us calculate it applying (6.4.2). Since , 2 , 3 then div 6F Fx y z= = and we obtain easily the

same result:3div 6 6 Volume( ) 6 3 162F n F

S D D

dS dxdydz dxdydz D⋅ = = = × = × =∫∫ ∫∫∫ ∫∫∫

c. Find the flux of the vector field 2 , 2 3 , 3x y y z z x= + + +H outward through the triangular

pyramid OABC, (0,0,0), (1,0,0), (0,1,0), (0,0,1)O A B C .

Solution

Example 6.2.1.f gives 1.d⋅ =∫∫F SS

We can obtain easily the result applying (6.4.2):

2 , 2 3 , 3x y y z z x= + + +F , then div 6=F , and then

1 1div 6 6 Volume( ) 6 1

3 2D D

dS dxdydz dxdydz D ⋅ = = = × = × = ∫∫ ∫∫∫ ∫∫∫F n F

S


86

EXERCISES

6.1 Find the area of

a. The ellipse cut from the plane z cx= by the cylinder 2 2 1x y+ = .

b. The surface 2 2ln 15 0x x y z− + − = above the square :1 2, 0 1,D x y≤ ≤ ≤ ≤ in thexy-plane.

c. The part of the paraboloid 2 2z x y= + which lies under the plane 6z = .

6.2 Evaluate the surface integral.

a. I = 2 2 2 2 2( ) , where is the surface (hemisphere) 4, 0x z y z dS x y z z+ + + = ≥∫∫S

S

b. , where is the surface with parametric equationsI xyz dS= ∫∫S

S

2 2 , , , ( , ) : 1, 0.x uv y u v z u v u v D u v u v= = − = + ∈ + ≤ ≥ ≥

c. 2 2 2 2 2, where :I y dS x y z a= + + =∫∫S

S .

d. ,I z dS= ∫∫S

where S is the closed surface that contains the cylinder 2 2 2x y a+ = and

the planes 0 and , 0.z z y a a= = + >

e. , where is the surface with parametric equationsI yz dS= ∫∫S

S

2 2 , , , 1x uv y u v z u v u v= = + = − + ≤ .

6.3 Use Stockes’ Theorem to evaluate

a. curl dS⋅∫∫ F nS

, where 2 2 2, ,yz xz xyx e y e z e=F and S is the hemisphere with equation

2 2 2 4, 0x y z z+ + = ≥ , oriented upward.

b. The work done by the force field 2 2 2, ,F x y zx z y x z y= + + + when a particle moves

under its influence around the close edge of the part of the sphere 2 2 2 4x y z+ + = that lies in thefirst octant, in a counterclockwise direction as viewed from above.

6.4 Use the Divergence Theorem to evaluate F SS

d⋅∫∫a. 2 3 2 21

3, sin ,F z x y z x z y= + +

S is the upward oriented top half of the sphere 2 2 2 1x y z+ + = .

b. 2 4 2 3 2 2, 4 sin , 2z y x y z x z y= + +F and S is the outward oriented surface of the cube3[ 1, 1]− .


87

c. 2 2 3 2, 7 , 3z x z y z y= − +F and S is the outward oriented surface of the sphere2 2 2 2x y z a+ + = .

d. 2, 3 , 3xy y zy= −F and S is the outward oriented surface of the tetrahedron with

vertices (0, 0, 0), (1, 0, 0), (0, 1, 0), and (0, 0, 1) .

e. 2 2 2 36 , 3 , 2zxy x e z=F and S is the outward oriented surface of the solid bounded by

the cylinder 2 2 4x y+ = and the planes 1,x = − and 2x = .

f. 5 2 3 4103, , 5x x y zy=F and S is the outward oriented surface of the solid bounded by

the paraboloid 2 21z x y= − − and the plane 0z = .

6.5 Use the Divergence Theorem to evaluate

a. 2 2 2(2 3 4 )x y z dS+ +∫∫S

, S is the sphere 2 2 2 2x y z a+ + = .

b.2 2 2

2 2 2

2 4

16 81 625

x y zdS

x y z

− + +

+ +∫∫S

, S is the ellipsoid2 2 2

14 9 25

x y z+ + = .

c. 3 2 2 2, 2 , , 6zdS x y e y z⋅ = +∫∫F n FS

,

isS the outward oriented surface of the solid bounded by 2 2 4, 1, 2.x y z z+ = = − =

6.6 Verify that the Stokes’ Theorem is true for the given vector field F and surface S .

a. 2 2 2 23 , 4 , 6 , is the part of 9, 9, oriented downward.F Sy z x z x y x y= − = + − + ≤

b. , , , is the part of 1, 0, 0, 0,oriented downward.F Sy z x x y z x y z= + + = − ≤ ≤ ≤

6.7 Verify that the Divergence Theorem is true for the given vector field F on the domain D .

a. 33 , 4 , 6 , is the cube [0,1] .F xy y xz D= −

b. 2 2 2, , 3 , is the region bounded by and 9.F x xy xz D z x y z= = + =

c. 2 2, , , is the region bounded by 4, 1 and 4.xy zy xz D x y z z= + = = − =F

d. 2 2 2 2, , , 0, and is the ball .F ax ay az a D x y z a= > + + ≤

6.8 Use Stockes’ Theorem to evaluate curl dS⋅∫∫ F nS

a. 2, , 2xe yz y z z=F , S is the part of the hemisphere 2 2 2 9, 0,x y z x+ + = ≥ that lies

inside the cylinder 2 2 4y z+ = , oriented in the direction of the positive x-axis.


88

b. 2, , ,xyz xy x yz=F S consists of the top and the four sides but not the bottom of the

cube 3[0, 1] , oriented outward.

ANSWERS

6.1 a. 2 1c + ; b. 3 2 ln 2+ ; c. 62 / 3

6.2 a. 16I = ; b.23 6 16 69 6 128

280 105 840I

−= − = ; c. 44

3a ; d. ( ) 33

2 2 a+

6.3 a. 0; b. 16;

6.4 a. 13 / 20 (Hint: Add the disk oriented downward)

b. 16. c. 54 a ; d. 1/3; e. 144 ; f. 5 /12

6.5 a. 412 a ; b. 4560 ; c. 144

6.8 a. 4− ; b. 1/2 ;

REFERENCES

[1] James Stewart, CALCULUS, Fourth edition, Brooks/Cole, 1999.

[2] George B. Thomas, Jr., CALCULUS, Pearson Education, Inc., 2006.

A COURSE IN CALCULUS 2 NGUYỄN VĂN HỘ

1

CONTENTS

CHAPTER 1 VECTORS AND GEOMETRY OF SPACE 3

1.1 Vectors 3A. Vectors 3B. Dot Product 5C. Cross Product 7D. Scalar Triple Product 8

1.2 Equations of Lines and Planes 10

A. Equations of Lines 10B. Equations of Planes 11

1.3 Cylinders and Quadratic Equations 12

1.3 Cylindrical and Spherical Coordinates 13A. Cylindrical Coordinates 13B. Spherical Coordinates 14

Exercises 15

CHAPTER 2 VECTOR FUNCTIONS 18

2.1 Vector Functions and Space Curves 18

2.2 Derivatives of Vector Functions 19

2.3 Integrals of Vector Functions 21

2.4 Arc Length of Space Curves 22

2.5 Curvature 23

2.6 Normal and Binormal Vectors 25

2.7 Motions in Space. Velocity and Acceleration 26

Exercises 27

CHAPTER 3 DOUBLE INTEGRALS 29

3.1 Definitions and Properties 29

3.2 Iterated Integrals 31

3.3 Change of Variables in Double Integrals 32

3.4 Double Integrals in Polar Coordinates 34

3.5 Application of Double Integrals 35

1. Area and Volume 35

2. Density and Mass 36

3. Moment and Center of Mass 37

A COURSE IN CALCULUS 2 NGUYỄN VĂN HỘ

2

4. Moment of Inertia 38

5. Surface Area 38

Exercises 40

CHAPTER 4 TRIPLE INTEGRALS 42

4.1 Definitions and Properties 42

4.2 Iterated Integrals 43

4.3 Change of Variables in Triple Integrals 44

4.4 Triple Integrals in Cylindrical Coordinates 45

4.5 Application of Triple Integrals 46

4.5.1. Volume 46

4.5.2. Density and Mass 47

4.5.3. Moments, Center of Mass, and Moment of Inertia 47

Exercises 49

CHAPTER 5 LINE INTEGRALS 50

5.1 Line Integrals 50

5.2 Line Integrals of Vector Fields 53

5.3 The Fundamental Theorem 56

5.4 Green’s Theorem 59

5.5 Curl 62

5.6 Divergence 65

5.7 Vector Form of Green’s Theorem 66

Exercises 68

CHAPTER 6 SURFACE INTEGRALS 70

6.1 Surface Integrals 70

6.2 Surface Integrals of Vector Fields 75

6.3 Stokes’ Theorem 81

6.4 The Divergence Theorem 84

Exercises 86

REFERENCES 88

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