Chapter IV Vector and Tensor Analysis IV.3 Vector Analysis - Revisited September 19, 2018 269

IV.3 VECTOR ANALYSIS

Revisited and Enhanced

Chapter IV Vector and Tensor Analysis IV.3 Vector Analysis - Revisited September 19, 2018 270 IV.3.1. VECTOR FUNCTIONS Let V∈r be a vector in the Euclidian vector space. Then the vector function of

a scalar variable is a vector-valued function defined as a map from the set of real numbers to the space of vectors V :

( )t : I V⊂ →r (42)

The scalar variable t I∈ ⊂ can represent time, path, etc. The vector function can provide a convenient method for the definition of curves in space by tracing the points by the position vector ( ) [ ]t , t a,b∈r . The change of parameter

[ ]t a,b∈ also provides information about the position of the point on the curve for different moments of time. This definition of curves by vector functions is equivalent to the parametric definition of curves:

( )( )( )( )

x tt y t

z t

=

r or ( )( )( )

x x t y y t

z z t

= = =

[ ] t a,b∈ (43)

where ( ) ( ) ( ) [ ]x t , y t ,z t , t a,b∈ are real valued functions. LINE IN SPACE Let the point ( )0 0 0P x , y ,z= be defined by the position vector 0r and let u be

some vector ( )1 2 3u ,u ,u=u . Consider a line L which goes through the point P in the direction of vector u . Any vector in the line L is collinear to vector u and therefore is a multiple of the vector u with some coefficient tu . Then the position vector defining all points on the line L can be represented as a vector function of the variable t :

( ) 0t t= +r r u t∈ (44)

In the component form this function is given by the equation

( )( )( )( )

0 1

0 2

0 3

x t x ut y t y t u

z t z u

= = +

r t∈ (45)

which corresponds to a parametric definition of a line:

( )( )( )

0 1

0 2

0 3

x t x tuy t y tuz t z tu

= += += +

t∈ (46)

The other equation which determines the coordinates ( )x, y,z of the points on the line L can be obtained by solving each of the parametric equations for t :

0 0 0

1 2 3

x x y y z zt

u u u− − −

= = =

provided that all ku 0≠ . Then three equations specify the points of the line L :

0 0 0

1 2 3

x x y y z zu u u− − −

= = (47)

These are called the symmetric equations of a line.

0

( )( )( )( )

x tt y t

z t

=

r

( ) 0t t= +r r u

0r

u

0

L

P

Chapter IV Vector and Tensor Analysis IV.3 Vector Analysis - Revisited September 19, 2018 271 Let us derive now the equation of the line which goes through two given points

( )1 1 1 1P x , y ,z= and ( )2 2 2 2P x , y ,z= determined by the position vectors 1r and

2r . The direction vector for this line can be found as

( )2 1 2 1 2 1 2 1 2 1OP OP x x , y y ,z z→ →

= − = − = − − −u r r

According to Equation (44), the line is defined by the following vector function:

( ) ( )1 1 1 2 1

1 1 2 1 2 1 1 2 1

1 3 1 2 1

x u x x xt t y t u t y t y y

z u z z z

− = + = + = + − = + − −

r r u r r r t∈ (48)

The segment connecting the points ( )1 1 1 1P x , y ,z= and ( )2 2 2 2P x , y ,z= can be

defined by the equation:

( ) ( ) ( )1 2 1 2 1t t t 1 t== + − = + −r r r r r r 0 t 1≤ ≤ (49)

PLANE IN SPACE Geometrically a plane can be defined by several different ways – they yield different vector equations defining a plane:

1) The plane Π can be uniquely defined by the fixed point ( )0 0 0 0x , y ,z=r

and a vector ( )1 2 3n ,n ,n=n orthogonal to the plane Π :

Let vector ( )x, y,z=r define an arbitrary point on the plane Π , then the vector 0−r r belongs in the plane Π .

Hence all vectors on the plane Π are orthogonal to the vector n , the vectors 0−r r are orthogonal to n . Therefore,

( )0 0− ⋅ =r r n (50) Using the distributive law of dot products, one obtains the vector equation

of a plane

c⋅ =r n (50)

where 0 1 0 2 0 3 0c n x n y n z= ⋅ = + +n r . In the coordinate form, this equation can be written as

1 2 3 1 0 2 0 3 0n x n y n z n x n y n z+ + = + + (51) We can notice, that if the plane is defined by the equation

1 2 3a x a y a z c+ + = c∈ (52)

then the vector ( )1 2 3a ,a ,a=a is orthogonal to this plane. The planes defined by Equation (52) for different values of the coefficient c are parallel to each other. Equation (52) defines coordinates ( )x, y,z which belong to the plane.

( )tr

1r

u

0

1P

L

2P

2r

0

( )tr

1r

1P

2P

2r

0

Π

n

r0r

Chapter IV Vector and Tensor Analysis IV.3 Vector Analysis - Revisited September 19, 2018 272

2) The plane Π can be uniquely defined by the fixed point ( )0 0 0 0x , y ,z=r

and two vectors ( )1 2 3a ,a ,a=a and ( )1 2 3b ,b ,b=b lying on the plane Π . The vector obtained by cross product ×a b is orthogonal to the plane Π .

Then the vector equation defining the plane Π according to Equation (50) is

( ) ( )0 0− ⋅ × =r r a b

which yields

( ) ( )0⋅ × = ⋅ ×r a b r a b (53)

In the coordinate form this equation can be written as

( ) ( )0− ⋅ ×r r a b 0

0 1 2 3

0 1 2 3

x x y y a a a

z z b b b

− = − ⋅ −

i j k

( )( ) ( )( ) ( )( )0 2 3 3 2 0 1 3 3 1 0 1 2 2 1 x - x a b a b y - y a b a b z - z a b a b= − − − + −i j k

3) The plane Π can be uniquely defined by three fixed points ( )1 1 1 1x , y ,z=r

, ( )2 2 2 2x , y ,z=r , and ( )3 3 3 3x , y ,z=r :

Vectors ( )2 1 2 1 2 1 2 1x x , y y ,z z= − = − − −a r r and

( )3 1 3 1 3 1 3 1x x , y y ,z z= − = − − −b r r are lying in the plane Π . Then form Equation (53), it follows that

( ) ( ) ( )1 2 1 3 1 0 − ⋅ − × − = r r r r r r (54)

defines the plane Π . The left hand-side of the equation (54) is a triple scalar product, which according to (21) can be written as

1 1 1

2 1 2 1 2 1

3 1 3 1 3 1

x x y y z zx x y y z z 0x x y y z z

− − −− − − =− − −

(55)

SPHERE IN SPACE A spherical surface of radius a with the center at the origin 0 is a set of all points 3∈r the distance from which to the origin is equal to a . This condition can be

written in the vector form as

2 2 2a− = = ⋅ =r 0 r r r

Then the vector equation of the sphere of radius a with the center at the origin is 2a⋅ =r r (56) The equation of a sphere of radius a with the center at the point ( )0 0 0 0x , y ,z=r

can be derived in the following form ( ) ( ) 2

0 0 a− ⋅ − =r r r r (57)

0

Π

×a b

r0r

ab

0

Π

×a b

1r a

b

2r

3r

2a⋅ =r r

a

r

0

0

0r

0−r r

Chapter IV Vector and Tensor Analysis IV.3 Vector Analysis - Revisited September 19, 2018 273 EXAMPLES: 1. Find the distance from the point ( )0 0 0 0x , y ,z=r to the plane c⋅ =r n .

Because the plane is defined by the vector equation c⋅ =r n , vector n is orthogonal to the plane. Then a line going through the point ( )0 0 0 0x , y ,z=r

in a direction of the vector n is also orthogonal to the plane: 0 t= +r r n

This line intersects the plane at the moment of time 1t for which the equation of the line defines the point belonging to the plane c⋅ =r n :

( )0 1t c+ ⋅ =r n n 0 1t c⋅ + ⋅ =r n n n 0 1t c⋅ + =r n ⇒ 1 0t c= − ⋅r n

Then the point of intersection 1r is defined by

( )1 0 1 0 0t c= + = + − ⋅r r n r r n n The vector ( )0 1 0 c= − = ⋅ −d r r r n n

is a free vector connecting the points 0r and 1r and it is orthogonal to the plane. Therefore, the norm of d defines the distance from 0r to the plane:

( )0 0 0c c c= ⋅ − = ⋅ − = ⋅ −d r n n r n n r n (58)

2. Let the particle be traveling along the line with the constant velocity v .

If at t 0= the particle was at the point ( )0 0 0 0x , y ,z=r , determine at which moment of time it crosses the plane :Π c⋅ =r n .

The equation governing the motion of the point in the direction of the velocity vector v along the line through the point 0r is 0 t= +r r v Intersection of the plane c⋅ =r n yields ( )0 t c+ ⋅ =r v n Then, using the distributive property of dot product

0 t c⋅ + ⋅ =r n v n 0t c⋅ = − ⋅v n r n

0ct

− ⋅=

⋅r n

v n (59)

3. Derive the equation of the plane tangent to the sphere 2a⋅ =r r at 0r .

Because the point 0r belongs to the sphere, its norm is equal to a and it satisfies the equation 2 2

0 0 0r a⋅ = =r r ⇒ 0r a= The plane contains the point 0r and it is orthogonal to the vector 0r . Then the vector equation of the plane according to (50) is

( )0 0 0− ⋅ =r r r Using the distributive property of dot products, we can obtain 0 0 0 0⋅ − ⋅ =r r r r Then the vector equation of the tangent plane becomes: 2

0 a⋅ =r r (60) It can be rewritten in the following coordinate form

20 0 0x x y y z z a+ + =

0

0rd n

1r

c⋅ =r n

0

0r

0 t= +r r v

1r

c⋅ =r n

v

t 0=

1t t=

0

0rr

20 0 a⋅ =r r

Chapter IV Vector and Tensor Analysis IV.3 Vector Analysis - Revisited September 19, 2018 274 IV.3.2. DIFFERENTIATION AND INTEGRATION OF VECTOR FUNCTIONS Space Curve Consider a bounded space curve C in 3V with the end points A and B .

The graph of the space curve C is traced by a vector function C : ( )tr [ ]t a,b∈ with ( )a =r A , ( )b =r B (61)

Parametric definition of the curve ( ) ( ) ( ) ( )( )t x t , y t ,z t=r :

C : ( ) ( )1x x t x t= =

( ) ( )2y y t x t= = [ ]t a,b∈ (62)

( ) ( )3z z t x t= =

Continuity Continuity of the vector function at the moment of time t :

( ) ( )t 0

lim t t t∆

∆→

+ =r r (63)

This condition is equivalent to the following condition:

( ) ( )t 0

lim t t t∆

∆→ + − = r r 0 (64)

Derivative The derivative of a vector function with respect to a scalar variable is defined as

( ) ( ) ( ) ( )t 0

t t tdt t limdt t∆

∆∆→

+ −′ = =

r rr r (65)

If the limit exists, then the derivative also is a vector function

( ) 3 3d t :dt

→r

The vector ddtr is on the line tangent to the curve C at the point ( )tr .

In parametric form, the differentiation of a vector function reduces to differentiation of its components:

( )( )( )( )

( )( )( )

x t x tt y t y t

z t z t

′ ′ ′ ′= = ′

r ( ) ( ) ( )x t y t z t′ ′ ′= + +i j k (66)

Repeatedly, the higher order derivatives can be defined:

( ) ( )dt tdt

′ =r r

( ) ( ) ( )2

2

d d dt t tdt dtdt ′′ = =

r r r

( ) ( ) ( ) ( )k k 1

kk k 1

d d dt t tdtdt dt

−

−

= =

r r r

0

( )t t∆+r

( )tr

( ) ( )t t t∆+ −r r

0

( )tr

( )t t∆+r

( )d tdt

r

Chapter IV Vector and Tensor Analysis IV.3 Vector Analysis - Revisited September 19, 2018 275 Mechanical sense of the vector derivatives: ( ) ( )t t t∆ ∆= + −r r r is displacement of the point in time t∆

t

∆∆

r is the average velocity over time t∆

( ) ( )t 0

lim t tt∆

∆∆→

′= ≡r r v is the instantaneous velocity at time t

( ) ( )t t t∆ ∆= + −v v v is the change of velocity in time t∆

( ) ( )t 0

lim t tt∆

∆∆→

′= ≡v v a is the acceleration at the time t , ( ) ( ) ( )t t t′ ′′= =a v r

If the motion of the point is defined by its components ( ) ( ) ( ) ( )( )t x t , y t ,z t=r , then

( ) ( ) ( ) ( )t x t y t z t= + +r i j k

( ) ( ) ( ) ( )t x t y t z t′ ′ ′= + +v i j k

( ) ( ) ( ) ( )t x t y t z t′′ ′′ ′′= + +a i j k

A differential element on the space curve is tangent to the curve: ( )1 2 3d dx ,dx ,dx=r

d d dsdt ds dt

=r r ds

dt= T d ds=r T 1=T (67)

PROPERTIES Let ( )tu and ( )tv be differentiable vector functions, then:

1. ( )c c′ ′=u u

2. ( ) ( ) ( )f t f t f t′ ′ ′ = + u u u

3. ( )′ ′ ′+ = +u v u v

4. ( )′ ′ ′⋅ = ⋅ + ⋅u v u v u v

5. ( )′ ′ ′× = × + ×u v u v u v

6. ( ) ( )d d d ddt dt dt dt

⋅ × = ⋅ × + ⋅ × + ⋅ ×

a u va u v u v a v a u

7. ( ) ( )d d d ddt dt dt dt

× × = × × + × × + × ×

a u va u v u v a v a u

8. ( )d d dss tdt ds dt

= rr (chain rule)

0

( )tr

( )tv

( )tv

( )t t∆+v( )ta

Chapter IV Vector and Tensor Analysis IV.3 Vector Analysis - Revisited September 19, 2018 276

Show how the property (2) can be derived:

( )f t ′ u ( ) ( ) ( ) ( ) ( )t 0

f t t t t f t td f t limdt t∆

∆ ∆∆→

+ + −′ = = u u

u

( ) ( ) ( ) ( )

t 0

f t f t f t tlim

t∆

∆ ∆

∆→

+ + − =u u u

t 0

f f f df dlim ft dt dt∆

∆ ∆ ∆ ∆∆→

+ += = +

u u u uu

In a similar manner the other properties are derived. TAYLOR SERIES If the components of vector function ( ) ( ) ( ) ( )( )t x t , y t ,z t=r are

expandable into a Taylor series over the point 0t t= then the vector

function ( )tr also can be expanded into a Taylor series:

Taylor Series ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )2 30 0 0

0 0 0 0

t t t t t tt t t t t ...

1! 2! 3!− − −

′ ′′ ′′′= + + +r r r r r (68)

INDEFINITE INTEGRAL The vector function ( )ta is the antiderivative of the vector function ( )tr if

( ) ( )d t tdt

=a r

Then the indefinite integral of a vector function is the operation of finding all antiderivatives, which is denoted in the traditional way:

Indefinite Integral ( ) ( ) ( ) ( )

( )

( )

( )

( )

x t dt

t dt x t dt y t dt z t dt y t dt t

z t dt

= + + = = +

∫

∫ ∫ ∫ ∫ ∫

∫

r i j k a c

where the constant vector c is a constant of integration. DEFINITE INTEGRAL Definite integral of the vector function ( )tr can be defined as a limit of

Riemann’s sum

( ) ( )( )v n

i i i 1n i 1u

t dt lim t t t −→∞ =

= = −∑∫b r r

where { }0 1 nu t ,t ,...,t v= = is a partition of the interval [ ]u,v . The Fundamental Theorem of Calculus for vector functions is formulated in a similar way: if ( )ta is the antiderivative of a vector function ( )tr then

Chapter IV Vector and Tensor Analysis IV.3 Vector Analysis - Revisited September 19, 2018 277

Definite Integral ( )

( )

( )

( )

( ) ( )

v

uv v

u uv

u

x t dt

t dt y t dt v u

z t dt

= = = −

∫

∫ ∫

∫

b r a a (69)

Derivation of these results can be performed for the component form of the vector function ( ) ( ) ( ) ( )( )t x t , y t ,z t=r based on the integration of the real-

valued functions of a single variable ( ) ( ) ( )x t , y t ,z t . LENGTH OF SPACE CURVE

If a space curve is defined parametrically by a vector function ( ) ( ) ( ) ( )( )t x t , y t ,z t=r , then the length of curve (arc length) is determined

by the integral

( ) ( ) ( ) ( )b b

2 2 2

a a

s t dt x t y t z t dt′ ′ ′ ′= = + + ∫ ∫r (70)

PARAMETRIZATION OF A SPACE CURVE IN TERMS OF ARC LENGTH

Natural parametrization: the pathlength s is used as a parameter for the vector function:

( )sr [ ]s a,b∈

( ) ( ) ( ) ( ) ( )t t

2 2 2

a a

s t t dt x t y t z t dt′ ′ ′ ′= = + + ∫ ∫r

Solve for t in terms of s , substitute into ( ) ( ) ( ) ( )( )t x t , y t ,z t=r .

( ) ( ) ( ) ( ) ( )2 2 2d s t t x t y t z tdt

′ ′ ′ ′= = + + r

The scalar variable in the vector function in terms of arc length:

( )s t r , ( )sr , [ ]0 1s s ,s∈ , ( )0 0s s t=

( ) ( )d d ds ds t tdt ds dt ds

′= = r rr r

( )s

a

s t dt′= ∫ r

( )s

a

d ds t dtds ds

′= ∫ r ⇒ ( )1 s′= r

Therefore, ( )s′r is always a unit tangent vector. It is denoted by

( ) ( )ds sds v

′′= = = =

′r vT r rr

( )ar

( )br

0

s

s 0=

s

( )sr

0

( )0r

( )s′=T r

Chapter IV Vector and Tensor Analysis IV.3 Vector Analysis - Revisited September 19, 2018 278 NORMAL PLANE TO THE CURVE

Normal plane to the curve

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

at the point ( )0 0t=r r .

The normal unit vector ( )( )

00

0

tt

′=

′r

Tr

is orthogonal to the plane.

The equation of the plane through the point ( )0tr normal to 0T (50):

( )0 0 0− ⋅ =r r T (71)

Definitions of some particular space curves: C is a closed curve if ( ) ( )a b=r r or ( ) ( )i ix a x b= C is a smooth curve if ( ) [ ]ix t C a,b′ ∈ (derivatives are continuous)

C is piece-wise smooth if kk

C C=

where kC are smooth curves

C is a simple curve if ( ) ( )i i 2x t x t≠ if 1 2t t≠ (w/o self- intersection)

EXAMPLE: The graph of the space curve traced by the vector function

( ) ( ) ( ) ( )t cos t ,sin t ,t sin t cos t t= = + +r i j k t∈

For discrete values of the parameter t , the coordinates of points traced by a vector function can be calculated from the given equation:

and the curve can be drawn through these points. The periodic functions defining the x and the y coordinates yield the circular projection of the space curve on the xy-plane: 2 2x y 1+ =

and the linear increase of the z-coordinate yields the spiral shape of the space curve.

0r

0

( )tr

0T

t x y z0 1 0 0

4 2 2 2 2 42 0 1 2

1 0

π ππ ππ π−

( )tr

x

y

z

t 0=t 4π=

t 2π=

t π=

0

Chapter IV Vector and Tensor Analysis IV.3 Vector Analysis - Revisited September 19, 2018 279 IV.3.3. SCALAR AND VECTOR FIELDS

We considered vector functions of the scalar variable (time). Now, we introduce functions of the vector variable (point in space). Let 3∈r be a position vector which specifies a location in Euclidian space.

A scalar field is defined as a real valued function of a vector variable:

( ) 3:ϕ →r

( ) 3x, y,z :ϕ →

By this function, a scalar value is specified for any point of space r .

A scalar field can describe distribution in space of temperature, density, concentration, etc.

Function ( ) ( ) 2x, y :ϕ ϕ≡ →r defines a scalar field on a plane.

A vector field is defined as a vector valued function of a vector variable:

( ) 3 3: →v r

( ) ( ) ( ) ( )P x, y,z Q x, y,z R x, y,z= + +v r i j k

By this function, a vector value is specified for any point of space r .

A vector field can be described by a distribution in space of velocity, acceleration, force, etc.

A non-stationary scalar or vector field are defined as time-dependent maps

( ) 3,t :ϕ × →r

( ) 3 3,t : × →v r All considered vector functions are assumed to be continuous:

( ) ( )0

0limϕ ϕ→

=r r

r r

( ) ( )0

0lim→

=r r

v r v r ⇒ ( ) ( )0

0lim→

− = r rv r v r 0

All operations defined for vectors can be applied for vector fields point-wise.

( )ϕ r

r

r

( )v r

( ),tv r

r

r

0r

( )v r

( )0v r

( ) ( )0−v r v r

Chapter IV Vector and Tensor Analysis IV.3 Vector Analysis - Revisited September 19, 2018 280 LEVEL CURVES AND SURFACES Let ( ) 2f : →r be a scalar field, then equation

( )f c=r or

describes the curves in the plane 1 2x x called the level curves:

( ) ( ){ }1 2 1 2x ,x f x ,x c, c= ∈

Level curves are described by the functions of two variables ( )1 2f x ,x . Let ( ) 3:ϕ →r be a scalar field. Then equation

( ) cϕ =r or ( )1 2 3x ,x ,x cϕ =

describes the surface in 3 called the level surface:

( ) ( ){ }1 2 3 1 2 3x ,x ,x x ,x ,x c, cϕ = ∈

For different values of the constant c∈ , we obtain the families of un-intersected level curves and surfaces (why do level curves not intersect?). Level surfaces are described by the functions of three variables ( )1 2 3x ,x ,xϕ . OPERATOR “NABLA”, GRADIENT, DIRECTIONAL DERIVATIVE nabla ( )1

3 3: C V V∇ → is a differential vector operator called nabla and defined as

, ,x y z

∂ ∂ ∂∇ = ∂ ∂ ∂

(72)

Sometimes it is also called the Hamilton operator. gradient Operator nabla applied to a scalar-valued vector function yields a vector called the gradient of the scalar field

gradϕ ≡ ϕ∇ 1 2 3

, ,x x xϕ ϕ ϕ ∂ ∂ ∂

= ∂ ∂ ∂

1 2 31 2 3

+ +x x xϕ ϕ ϕ∂ ∂ ∂

=∂ ∂ ∂

i i i (73)

(73) which is a vector orthogonal to the level surface of the scalar field or to the level curve in the 2-D case.

( )1 2f x ,x c=

( )ϕ r

( ) ( ){ }0x, y,z T x, y,z T=

Isothermal Surface:

( ) ( ){ }x, y T x, y c=

Isotherms:

Chapter IV Vector and Tensor Analysis IV.3 Vector Analysis - Revisited September 19, 2018 281 Example: Let the position vector ( )x, y=r define points on the plane.

And let a scalar field be defined by ( ) ( ) 2 2x, y x yϕ ϕ= = +r . This scalar field can be visualized by the level curves 2 2x y c+ = which are concentric circles around the origin. The gradient of this scalar field is

( ) ( ) ( ) ( )2 2 2 2x, y x y x y 2x 2y 2 x, yx y

ϕ ∂ ∂∇ = + + + = + =

∂ ∂i j i j

or in vector form it can be written as ( ) 2ϕ∇ =r r

The gradient at the point ( )x, y=r is normal to the level curve passing through this point.

The point ( )P 1,2= belongs to the level curve 2 2x y 5+ = . The value of the gradient at this point is ( ) ( )1,2 2 1 2 2 2 4 2,4ϕ∇ = ⋅ + ⋅ = + =i j i j

Properties of gradient: 1. ( ) ϕ ψ ϕ ψ∇ + = ∇ +∇

2. ( )c cϕ ϕ∇ = ∇ , c∈

3. ( ) ϕψ ψ ϕ ϕ ψ∇ = ∇ + ∇ Directional derivative of the scalar field Let the function define a scalar field, and let be a unit vector,

1=s . Vector hs with a small h 0> is an increment in the direction .

Then a derivative of the function in the direction s at the point in space r is defined as

D ϕs ( ) ( )h 0

hlim

hϕ ϕ

→

+ −=

r s r (74)

It determines the rate of change of the scalar field at the point r in the direction s . Using the differentiation rules for the multivariable functions and the definition of the operator nabla, let us write the other representations of the directional derivative:

sϕ∂∂

31 2

1 2 3

xx x + +x s x s x sϕ ϕ ϕ ∂∂ ∂∂ ∂ ∂

=∂ ∂ ∂ ∂ ∂ ∂

( ) ( ) ( )1 1 11 2 3

cos , + cos , + cos ,x x xϕ ϕ ϕ∂ ∂ ∂

=∂ ∂ ∂

s i s i s i

1 2 31 2 3

s + s + sx x xϕ ϕ ϕ∂ ∂ ∂

=∂ ∂ ∂

i

i

sxφ∂

=∂

ϕ= ∇ ⋅s (75)

( ) 3:Vϕ →r s

s

( )ϕ r

sϕ∂

=∂

( )ϕ r

( ) 2ϕ∇ =r r

2 2x y c+ =

Chapter IV Vector and Tensor Analysis IV.3 Vector Analysis - Revisited September 19, 2018 282 Therefore, the derivative of ϕ in any direction is equal to the projection of the

gradient ϕ∇ onto this direction:

sϕ∂∂

ϕ= ∇ ⋅s ( )cos ,ϕ ϕ= ∇ ∇ s (76)

It follows from this equation that the maximum value of the directional derivative

is achieved in the direction of the gradient of the scalar field at this point. So we can conclude that the gradient of the scalar field is a vector which has a direction of the greatest increase and its magnitude is equal to the directional derivative in this direction. The opposite direction ϕ−∇ corresponds to the direction of the greatest decrease.

From equation (76) also yields that

d dϕ ϕ= ∇ ⋅ r If n is the unit normal vector to the level surface of ϕ , then

nϕϕ ∂

∇ =∂

n (77)

Directional derivative of the vector field Let the function ( ) 3 3:V V→a r define a vector field, and let be a unit vector,

1=s . Vector hs with a small h 0> is an increment in the direction .

Then the derivative of the function ( )a r in a direction s at the point of space r is defined as

Dsa s∂

=∂a ( ) ( )

h 0

hlim

h→

+ −=

a r s a r (78)

provided that the limit exists. It determines the rate of change of the vector field ( )a r at the point r in the direction s :

s∂∂a 31 2

1 2 3

xx x + +x s x s x s

∂∂ ∂∂ ∂ ∂=

∂ ∂ ∂ ∂ ∂ ∂a a a

( ) ( ) ( )1 1 11 2 3

cos , + cos , + cos ,x x x∂ ∂ ∂

=∂ ∂ ∂

a a as i s i s i

( ) = ∇ ⋅s a (79)

s

s

1 2 31 2 3

s + s + sx x x∂ ∂ ∂

=∂ ∂ ∂

a a a

Chapter IV Vector and Tensor Analysis IV.3 Vector Analysis - Revisited September 19, 2018 283 IV.3.4. DOUBLE INTEGRAL Consider the function of two variables ( )y f= r defined in 2A∈ ⊂r . Definition of the double integral:

( )A

f dA∫∫ r ( )k k0lim f Aσ

∆→

= ∑ r (80)

Values of scalar function ( )y f= r define the surface:

( ) ( ) ( ){ }S x, y,z x, y A,z f x, y= ∈ = ,

then A is a projection of S on xy-plane. Geometrical sense: if ( )f x, y 1≡ Area of A

A

dA= ∫∫ (81)

if ( )f x, y 0≥ Volume between S and A

A

f dA= ∫∫ (82)

Cartesian coordinates: ( )x, y A= ∈r : ( )A

f dA∫∫ r ( )A

f x, y dA= ∫∫ ( )k kf A

k k k k0lim f x , y x y

∆

σ∆ ∆

→= ∑

( )A

f x, y dxdy= ∫∫

I ( ) ( ) ( ){ }1 2A x, y a x b, g x y g x = ≤ ≤ ≤ ≤ Iterated integral:

( )A

f x, y dA∫∫ ( )( )

( )2

1

g xb

a g x

f x, y dy dx

= ∫ ∫ (83)

II ( ) ( ) ( ){ }1 2A x, y h y x h y , c y d = ≤ ≤ ≤ ≤ Iterated integral:

( )A

f x, y dA∫∫ ( )( )

( )2

1

h xd

c h y

f x, y dx dy

= ∫ ∫ (84)

Polar coordinates: ( )r, Aθ= ∈r : ( )A

f dA∫∫ r ( )A

f r , dAθ= ∫∫ ( )k

kA

f

k k 1k k k k0

r rlim f r , r

2

∆

σθ ∆ ∆θ+

→

+= ∑

( )

A

f r , rdrdθ θ= ∫∫

I ( ) ( ) ( ){ }1 2A r , a r b, g r g r θ θ= ≤ ≤ ≤ ≤ Iterated integral:

( )A

f r , dAθ∫∫ ( )( )

( )2

1

g rb

a g r

r f x, y d drθ

=

∫ ∫ (85)

II ( ) ( ) ( ){ }1 2A r , h r h , θ θ θ α θ β= ≤ ≤ ≤ ≤ Iterated integral:

( )A

f x, y dA∫∫ ( )( )

( )2

1

h

h

f r , rdr dθβ

α θ

θ θ

= ∫ ∫ (86)

y

x

kx∆ky∆

( )k k kf f x , y=

kA∆

A

y

x

kr∆k∆θ

A

kA∆

kA∆ k∆θ

k k 1 kr r r∆ += −

2 2k 1 k

kr r

2∆θ+ −

=

k 1 kk k

r rr

2∆ ∆θ+ +

=

( )k k kf f r ,θ=

kr

kA∆

( )kf r ( )f r

A

2

r

S

Chapter IV Vector and Tensor Analysis IV.3 Vector Analysis - Revisited September 19, 2018 284 RATE OF HEAT TRANSFER Rate of Heat Transfer through the Surface A :

(87) q ( )A

q x, y dA′′= ∫ ∫ where ( )q x, y′′ is the local heat flux on the surface

A in a direction of axis z (normal heat flux) AVERAGING Averaging of Scalar Field ( )z f x, y= on the Surface A :

(88) avT ( )

A

T x, y dA

A=∫ ∫

averaged temperature on A

(89) h ( )

A

h x, y dA

A=∫ ∫

( )h x, y is the local convective coefficient on A

Example: Given the heat flux in z direction through the area

( ){ }A x, y 2 x 3, 5 y 2x 1= ≤ ≤ ≤ ≤ +

(normal distribution of heat flux over area A ):

( )q x, y′′ 2 26 x y= 2

Wm

What is the rate of heat transfer q through the area A ? What is the averaged value of the heat flux on A?

The rate of heat transfer: q ( )A

q x, y dA′′= ∫ ∫

( )3 2 x 1

2 5

q x, y dy dx+

′′= ∫ ∫

3 2 x 1

2 2

2 5

6 x y dy dx+

= ∫ ∫

3 2 x 1

2 2

2 5

6 x y dy dx+

=

∫ ∫

2 x 13 3

2

2 5

y6 x dx3

+

=

∫

( )3

32

2

2 x 2x 1 125 dx = + − ∫

( )3

32

2

2 x 2x 1 125 dx = + − ∫ 2993.8 W=

Area: A 3 2 x 1

2 5

dy dx+

= ∫ ∫ [ ]

32

2

2 x 2 dx 1 m= − =∫

Averaged value: avq′′ ( )

A

q x, y dA

A

′′=∫ ∫

2

W2993.8 m

=

( ) 2

Wq x, y ,m

′′

y 2x 1= +

q

Chapter IV Vector and Tensor Analysis IV.3 Vector Analysis - Revisited September 19, 2018 285 Example: Consider a cross-section of a circular pipe of radius R 2= .

Let the temperature in the cross-section be defined by

( ) 2 2

10T x, yx y 4

=+ +

Find the mean temperature mT of the cross-section.

mT ( )

A

T dA

A=∫ ∫ r

2A R 4π π= =

Conversion to polar coordinates: ( ) 2

10T r,r 4

θ =+

x r cosθ= y r sinθ= ( ){ }A r, 0 r 2, 0 2 θ θ π= ≤ ≤ ≤ ≤

2 2 2x y r+ =

mT ( )

2 2

0 0

T r, rdrd

A

π

θ θ=∫ ∫

2 2

20 0

1 10 rdrd4 r 4

π

θπ

=+∫ ∫

2

20

10 rdr24 r 4

ππ

=+∫

2

20

5 2rdr2 r 4

=+∫

( )22

20

d r 452 r 4

+=

+∫

( ) 22

0

5 ln r 42 = +

5 ln 22

= 1.73≈

2 2x y 4+ =

( )T x, y

( ) 2

10T r,r 4

θ =+

( ) 2 2

10T x, yx y 4

=+ +

Chapter IV Vector and Tensor Analysis IV.3 Vector Analysis - Revisited September 19, 2018 286 SURFACE AREA Consider a surface S defined by the equation

( )z f x, y= , ( )x, y A∈ Consider a partition of area A onto small area increments k k kA x y∆ ∆ ∆=

Area increments in the xy-plane define the area increments of the surface S k kf : A S∆ ∆→

Consider a point ( )( )k k k k kx , y ,z f x , y= on the surface S . Construct a tangent

plane to the surface S at this point, and then the projection kT∆ of the area

kA∆ and correspondingly of kS∆ onto the tangent plane. First find the vectors u and v

kk k k k k x ,k k

fx z x x x f x

x∆ ∆ ∆ ∆ ∆ ∆

∂= + = + = +

∂u i k i k i k

kk k k k k y ,k k

fy z y y y f y

y∆ ∆ ∆ ∆ ∆ ∆

∂= + = + = +

∂v j k j k j k

Area kT∆ = ×u v

k x ,k k

k y ,k k

det x 0 f x0 y f y∆ ∆

∆ ∆

=

i j k

x ,k k k y ,k k k k kf x y f x y x y∆ ∆ ∆ ∆ ∆ ∆= − − +i j k

k k x ,k y ,kx y f f∆ ∆= − − +i j k

2 2k k x ,k y ,kx y f f 1∆ ∆= + +

Then the area of the surface S can be approximated as:

S kk

S∆= ∑ kk

T∆≈ ∑ 2 2x,k y ,k k k

kf f 1 x y∆ ∆= + +∑

In a limit, with the norm of partition approaching zero, the area is defined by the double integral:

S 2 2x,k y ,k k k0 k

lim f f 1 x yσ

∆ ∆→

= + +∑ 2 2x y

A

f f 1 dxdy= + +∫∫

S 2 2x y

A

f f 1 dA= + +∫∫ (90)

kS∆

( )z f x, y=S

kT∆

kx∆ kx

( )k k kz f x , y=

x

z

u

i

k

{k kfz xx

∆ ∆∂=∂

k k k kfx z x xx

∆ ∆ ∆ ∆∂= + = +

∂u i k i k

Chapter IV Vector and Tensor Analysis IV.3 Vector Analysis - Revisited September 19, 2018 287 : Representation of surface integral of scalar function with different choice of projection of the surface S

S 2 2x,k y ,k k k0 k

lim f f 1 x yσ

∆ ∆→

= + +∑ xy

2 2x y

A

f f 1 dxdy= + +∫∫ (91)

( )z f x, y= , ( ) xyx, y A∈

S 2 2x,k z ,k k k0 k

lim g g 1 x zσ

∆ ∆→

= + +∑ xz

2 2x z

A

g g 1 dxdz= + +∫∫ (92)

( )y g x,z= , ( ) xzx,z A∈

S 2 2y ,k z ,k k k0 k

lim h h 1 y zσ

∆ ∆→

= + +∑ yz

2 2y z

A

h h 1 dydz= + +∫∫ (93)

( )x h y,z= , ( ) yzy,z A∈

x

y

( )z f x, y=

xyA

z

x

y ( )x h y,z=yzA

z

S

x

y

( )y g x,z=

xzA

z

S

Chapter IV Vector and Tensor Analysis IV.3 Vector Analysis - Revisited September 19, 2018 288 IV.3.5. SURFACE INTEGRAL OF THE SCALAR FUNCTION Consider the scalar function defined on the surface S : ( ) ( ) 3G G x, y,z : S= ⊂ →r Let the surface S be defined by the equation: ( )g x, y,z 0= Then the normal vector to S any point on the surface is defined by

gg

∇=

∇n

Assume that the surface S allows also representations: ( )z f x, y= , ( ) xyx, y A∈

( )y g x,z= , ( ) xzx,z A∈

( )x h y,z= , ( ) yzy,z A∈ Construct a surface G.

What we are going to obtain with the surface integration of scalar function G is the area between surfaces S and G (with positive sign if G is to direction of n , and minus sign if it is in the opposite direction):

( )S

G x, y,z dS∫∫ k k0 klim G Sσ

∆→

= ∑ 2 2k x ,k y ,k k k0 k

lim G f f 1 x yσ

∆ ∆→

= + +∑

( )

S

G x, y,z dS∫∫ ( )xy

2 2x y

A

G x, y, f x, y f f 1 dxdy= + + ∫∫ (94)

, ( )

S

G x, y,z dS∫∫ ( )xz

2 2x z

A

G x,g x,z ,z g g 1 dxdz= + + ∫∫ (95)

( )

S

G x, y,z dS∫∫ ( )yz

2 2y z

A

G h y,z , y,z h h 1 dydz= + + ∫∫ (96)

Chapter IV Vector and Tensor Analysis IV.3 Vector Analysis - Revisited September 19, 2018 289

( )T R,R,0 2=

T 1.85=

T 1.62=

T 2.66=

Example: Find lateral surface area of a cone with h and R :

( ) 2 2hz f x, y x y hR

= = − + + equation of the surface

( )x 2 2

h xf x, yR x y

= −+

( )2 2

2x 2 2 2

h xf x, yR x y

=+

( )y 2 2

h yf x, yR x y

= −+

( )2 2

2y 2 2 2

h yf x, yR x y

=+

S

xy

2 2x y

A

f f 1 dxdy= + +∫∫

2 2R R x

2 2x y

0 0

4 f f 1 dxdy−

= + +∫ ∫ consider one quarter of a cone

2 2R R x 2

20 0

h4 1 dxdyR

−

= +∫ ∫ integrate with Maple

2 2R R hπ= + Rsπ=

Let h 20= , R 2= Temperature distribution: ( ) 2 2 2T x, y,z x y z 1= + + +

Surface average temperature: averT ?=

S 2 2R R h 31.57π= + ≈ surface area

averT ( )S

1 T x, y,z dSS

= ∫∫

( )xy

2 2x y

A

1 T x, y, f x, y f f 1 dxdyS

= + + ∫∫

( )2 2R R x

2 2x y

0 0

4 T x, y, f x, y f f 1 dxdyS

−

= + + ∫ ∫

1.78≈ integrated with Maple Remark: integration can be simpler in cylindrical coordinates

S

h

R

s

( )T x, y,0

z 0=

z 0=

Chapter IV Vector and Tensor Analysis IV.3 Vector Analysis - Revisited September 19, 2018 290 IV.3.6. SURFACE INTEGRAL OF THE VECTOR FUNCTION – FLUX OF A VECTOR FIELD

Let the surface S be defined by the equation: ( )g x, y,z 0= The unit normal vector to surface S is defined by

gg

∇=

∇n

(for the direction of the normal vector, we will agree to take the exterior direction for closed surfaces, and one of two directions for non-closed surfaces and stick to this direction when changing position on the surface).

Assume that the surface S allows representations: ( )z f x, y= , ( ) xyx, y A∈

( )y g x,z= , ( ) xzx,z A∈

( )x h y,z= , ( ) yzy,z A∈

Let the vector field ( ) 3, S∈ ⊂a r r be defined by

( )a r ( ) ( ) ( )1 2 3x x xa a a= + +r i r j r k

( ) ( ) ( )x y za a a= + +r i r j r k

( ) ( ) ( )x y za x, y,z a x, y,z a x, y,z= + +i j k The dot product na⋅ =a n ( ) ( ) ( )

1 2 3x 1 x 2 x 3a cos , a cos , a cos ,= + +n i n i n i

x x y y z za n a n a n= + +

is the magnitude of the projection of vector in the normal direction n . Subdivide surface S onto subsurfaces with areas S∆ which can be assumed to

be flat and be characterized by the normal vector ∆S with the magnitude equal to the area S∆ : S∆ ∆=S n . Then the surface integral can be defined as a limit of the sums

Using the differential relation d dS=S n , we can express d⋅a S ( )dS= ⋅a n

na dS=

( ) ( ) ( )1 2 3x 1 x 2 x 3a cos , a cos , a cos , dS = + + n i n i n i

( ) ( ) ( )1 2 3x 1 x 2 x 3a cos , dS a cos , dS a cos , dS = + + n i n i n i

( ) ( ) ( )1 2 3x 2 3 x 1 3 x 1 2a dx dx a dx dx a dx dx= + +r r r (97)

( )a r

kS 0 kS

d = lim∆

∆→

⋅ ⋅∑∫a S a S

( )g x, y,z 0=

Chapter IV Vector and Tensor Analysis IV.3 Vector Analysis - Revisited September 19, 2018 291 FLUX Let us define a flux of a vector field ( )a r through the surface S as a surface

integral of the vector function ( )a r :

{ }S

flux of through the surface S = d⋅∫a a S (98)

Using Equation (81), the flux of the vector field through the surface S can be written in the traditional form of the surface integral:

S S

d = dS ⋅ ⋅∫ ∫a S a n (99)

Evaluation of the surface integral can be performed with the help of any form of

equation (97). If we denote scalar product by the scalar function

( ) ( )G G x, y,z= = ⋅r a n

then the flux of the vector field can be calculated using one of the equations (94-96), if the corresponding representation of the surface S (p.43) is allowed:

S

d ⋅∫a S ( )S

G x, y,z dS= ∫∫ ( )xy

2 2x y

A

G x, y, f x, y f f 1 dxdy= + + ∫∫ (100)

S

d ⋅∫a S ( )S

G x, y,z dS= ∫∫ ( )xz

2 2y z

A

G x,g x,z ,z g g 1 dxdz= + + ∫∫ (101)

S

d ⋅∫a S ( )S

G x, y,z dS= ∫∫ ( )yz

2 2y z

A

G h y,z , y,z h h 1 dydz= + + ∫∫ (102)

All three forms of the surface integral with the help of components of scalar

product of vector field with the surface normal vector ⋅a n x x y y z za n a n a n= + +

( ) ( ) ( )x y zG x, y,z G x, y,z G x, y,z= + + are traditionally combined in a single equation [Korn&Korn, p.156]:

S

d ⋅∫a S ( )xy

2 2x x y

A

G x, y, f x, y f f 1 dxdy= + + ∫∫

( )xz

2 2y y z

A

G x,g x,z ,z g g 1 dxdz+ + + ∫∫ (103)

( )yz

2 2z y z

A

G h y,z , y,z h h 1 dydz+ + + ∫∫

The flux through the closed surface S is denoted by

(104)

Useful fact: the flux through any closed surface in the constant vector field

( ) 0=a r a is zero (conservation law): 0S

d = 0⋅∫ a S

[ ]

2

In the Heat Transfer theory, Wflux of the vector field m

(which is defined as a heat flux itself) through the surface Sdefines the rate of heat transfer q Wthrough the surface S (if it i

′′ q

s positive, then the heat transferoccurs from the surface in the directionof the normal vector ; and if it is negative, then it occurs in the opposite to normal vector direction).Therefore,

flux of

n

[ ]

through the surface S is the rate of heat transfer through S

q q A W

′′

′′= ⋅

q

S S

d = dS ⋅ ⋅∫ ∫a S a n

Chapter IV Vector and Tensor Analysis IV.3 Vector Analysis - Revisited September 19, 2018 292 Example: (Zill, p.529, my solution) Find the flux of the vector field

( )x, y,z z z= +a j k (vector field)

through the surface S defined by the plane 3x 2y z 6+ + = in the first octant oriented upward. Surface S is defined alternatively by the following equations: ( )g x, y,z 3x 2y z 6 0≡ + + − = ( )z f x, y 3x 2y 6= ≡ − − + ⇒ xf 3= − yf 2= −

( ) 3 1y g x,z x z 32 2

= ≡ − − + ⇒ x3g2

= − z1g2

= −

( ) 2 1x h y,z y z 23 3

= ≡ − − + ⇒ y2h3

= − z1h3

= −

Unit normal vector to surface S :

gg

∇=

∇n ( )3,2,1

9 4 1=

+ + 3 2 1

14 14 14= + +i j k

Scalar field ( )G x, y,z is defined by a scalar product of the vector field with the normal vector:

( )G x, y,z = ⋅a n

yx zGG G

3 2 10 z z14 14 14

= ⋅ + +

G

3 z14

=

Calculate the flux using equation (100):

S

d ⋅∫a SS

= dS ⋅∫a n ( )S

G x, y,z dS= ∫∫ ( )xy

2 2x y

A

G x, y, f x, y f f 1 dxdy= + + ∫∫

( ) ( ) ( )3- x 32 2

2 2

0 0

3 3x 2y 6 3 2 1dydx 14

+

= − − + − + − +

∫ ∫

( )3- x 32 2

0 0

3 3x 2y 6 14dydx 14

+

= − − +

∫ ∫

[ ]3- x 32 2

0 0

3 3x 2y 6 dydx +

= − − +∫ ∫

32 - x 32 2

00

3 3xy 6 y y dx +

= − + − ∫

22

0

3 3 33 3x x 3 6 x 3 x 3 dx 2 2 2

= − − + + − + − − +

∫

2

2 2

0

9 93 x 9x 9x 18 x +9x 9 dx 2 4 = − + − + − − ∫

2

2

0

27 x 4x 4 dx 4

= − + ∫

23

2

0

27 x 2x 4x 4 3

= − +

27 8 8 8 4 3 = − +

18 =

x

y

z

xyA

Sn

yzAxzA

2

3

6

F

3y x 32

= − +

1y z 32

= − +

1x z 23

= − +

Chapter IV Vector and Tensor Analysis IV.3 Vector Analysis - Revisited September 19, 2018 293 Other solution: Calculate the flux using equation (103):

S

d ⋅∫a S ( )xy

2 2x x y

A

G x, y, f x, y f f 1 dxdy= + + ∫∫

( )xz

2 2y y z

A

G x,g x,z ,z g g 1 dxdz+ + + ∫∫

( )yz

2 2z y z

A

G h y,z , y,z h h 1 dydz+ + + ∫∫

0=

xz

2 2x z

A

2z g g 1 dxdz14

+ + +∫∫

yz

2 2y z

A

z h h 1 dydz14

+ + +∫∫

xzA

2z 9 1 1dxdz 4 414

= + +∫ ∫

yzA

z 4 1 1dydz9 914

+ + +∫ ∫

z 2

6 3

0

zdxdz − +

= ∫ ∫

z 36 2

0 0

1 zdydz3

− +

+ ∫ ∫

6

0

zz 2 dz 3

= − + ∫ 6

0

1 zz 3 dz3 2

+ − + ∫

6 2

0

z 2z dz 3

= − +

∫

6 2

0

1 z 3z dz3 2

+ − +

∫

632

0

z z9

= − +

63

2

0

1 z 3 z3 6 2

+ − +

216 369

= − + 1 216 3 363 6 2 + − +

24 36= − + 12 18− +

18=

Answer: Rate of heat transfer through the surface S : nS

q d = ⋅∫a S 18=

Chapter IV Vector and Tensor Analysis IV.3 Vector Analysis - Revisited September 19, 2018 294

( ) 2 2

25T x, y,zx y 4

=+ +

Example: Consider the long rectangular column with the known temperature distribution

( ) ( ) 3T T x, y,z :V= ⊂ →r scalar field ( ) 2 2

25T x, y,zx y 4

=+ +

L 4.0= M 3.0= K 1.0= k 2= thermal conductivity

Describe and visualize the heat transfer in the column, and determine the rate of heat transfer through the surface ( ){ }S L, y,z 0 y M ,0 z K= ≤ ≤ ≤ ≤ .

Temperature scalar field defines the gradient vector field

( )T x, y,z∇ T T Tx x x

∂ ∂ ∂= + +∂ ∂ ∂

i j k

( ) ( )2 22 2 2 2

50x 50 y

x y 4 x y 4= − −

+ + + +i j

Gradient vectors are orthogonal to isotherms, they are in the direction of greatest incease of temperature. Gradient vectors are tangent to the lines of heat flow.

Gradient vector field defines the heat flux vector field by Fourier’s Law:

( )x, y,z′′q ( )k T x, y,z= − ∇ ( ) ( )2 22 2 2 2

100x 100 y

x y 4 x y 4= +

+ + + +i j

Heat flux vectors are in a direction of greatest decrease of temperature.

isotherms

x

x

x L=

y M=

0

( )x, y,z′′q

( )T x, y,z∇

gradient vector field

heat flux vector field

n

qrate of heat transferthrough the surface Sin a direction of n

isotherms

x

y

z

n

K S

L

M

( )T x, y,z

0

q ?=

Chapter IV Vector and Tensor Analysis IV.3 Vector Analysis - Revisited September 19, 2018 295 Surface S is in the plane defined by equation ( )g x, y,z x L 0≡ − = Equation for S defined by projection on yz − plane (p.43): ( )x L h y,z= ≡ , 0 y M≤ ≤ , 0 z K≤ ≤ ⇒ y zh h 0= = Unit normal vector to surface S: 0 0= + + =n i j k i

The scalar field defined by the projection of the heat flux on normal direction:

( )nG x, y,z ( )x, y,z′′= ⋅q n ( )22 2

100x

x y 4=

+ +

Rate of heat transfer through the surface S:

nS

q = dS ′′ ⋅∫q n ( )S

G x, y,z dS= ∫∫ ( )yz

2 2y z

A

G h y,z , y,z g g 1 dydz= + + ∫∫

( )

K M

22 20 0

100L dydzL y 4

=+ +

∫ ∫

( )

1 3

220 0

1400 dydzy 20

=+

∫ ∫

( )

3

220

1400 dyy 20

=+

∫ 3.79≈

Answer: Rate of heat transfer through the surface S : nS

q d = ⋅∫a S 3.79≈

Chapter IV Vector and Tensor Analysis IV.3 Vector Analysis - Revisited September 19, 2018 296 IV.3.7. DIVERGENCE Divergence of the vector field ( )a r at the point of space r is defined as a limit of the averaged flux through the surface of the arbitrary volume

containing point r :

diva (105)

Use a parallelepiped for the arbitrary volume V with one corner located at the point ( )1 2 3x ,x ,x=r , sides ix∆ and faces perpendicular to the coordinate axes , and 1 2 3V x x x∆ ∆ ∆= with the surfaces 1 2 3S x x∆ ∆ ∆= ,… (see picture). Then

diva

k

k

V 0= lim

V

∆

→

⋅∑a S

k k

k

V 0

S= lim

V

∆

→

⋅∑a n

( ) ( )1 1 2 3 1 2 3 1 1

V 0

x x ,x ,x x ,x ,x S ...= lim

V

∆ ∆→

+ − ⋅ + a a i

( ) ( )

1 2 3

1 1 2 3 1 2 3 1 2 3

x x x 01 2 3

x x ,x ,x x ,x ,x x x ...= lim

x x x∆ ∆ ∆

∆ ∆ ∆

∆ ∆ ∆→

+ − ⋅ + a a i

( ) ( )

1 2 3

1 1 2 3 1 1 2 3 1

x x x 01

x x ,x ,x x ,x ,x ...= lim

x∆ ∆ ∆

∆

∆→

+ ⋅ − ⋅ + a i a i

( ) ( )

1 1

1 2 3

x 1 1 2 3 x 1 2 3

x x x 01

a x x ,x ,x a x ,x ,x ...= lim

x∆ ∆ ∆

∆

∆→

+ − +

( ) ( )

1 1

1 2 3

x 1 1 2 3 x 1 2 3

x x x 01

a x x ,x ,x a x ,x ,x ...= lim

x∆ ∆ ∆

∆

∆→

+ − +

31 2 xx x

1 2 3

aa a=

x x x∂∂ ∂

+ +∂ ∂ ∂

= ∇ ⋅a (106)

For incompressible fluid, the divergence of the velocity vector field is zero: div =v 0

S

V 0

d lim

V→

⋅=

∫ a S

S

V 0

d lim

V→

⋅=

∫ a S

S

V 0

dS = lim

V→

⋅∫ a n

Chapter IV Vector and Tensor Analysis IV.3 Vector Analysis - Revisited September 19, 2018 297 IV.3.8. CURL The curl of the vector field ( )a r at the point of space r is defined as a limit of the averaged flux through the surface of the arbitrary volume

containing point r :

S

V 0

dS lim

V→

×=

∫ n a

(107)

Compare with divergence (equation (106)):

diva S

V 0

dS = lim

V→

⋅∫ n a

= ∇ ⋅a (108)

Application of the operator nabla yields a similar representation of the curl

= ∇×a (109)

Useful formulas:

= ∇×a

1 2 3

1 2 3

1 2 3

x x x

x x x

a a a

∂ ∂ ∂=

∂ ∂ ∂

i i i

(110)

The components can be determined by

( )icurla jk

j k

aa

x x∂∂

= −∂ ∂

where indices i, j ,k are a cyclic

permutation of the numbers . For irrotational fluid, the curl of the velocity vector field is zero, curl =v 0

Laplacian Operator ( )2 2 2

22 2 2div grad

x y zϕ ϕ ϕ∆ϕ ϕ ϕ ϕ ∂ ∂ ∂

= ∇ = ∇ ⋅∇ = = + +∂ ∂ ∂

( ) 3:ϕ →r

IDENTITIES: Let ( ) ( ) 3 3, : →a r b r be vector fields, ( ) 3:ϕ →r be scalar field, c∈

1. ( )curl gradϕ = 0 ϕ∇×∇ = 0

2. ( )div curl =a 0 ( )∇ ⋅ ∇× =a 0

3. ( )div div div+ = +a b a b ( )∇ ⋅ + = ∇ ⋅ +∇ ⋅a b a b

4. ( )div div gradϕ ϕ ϕ= + ⋅a a a ( )ϕ ϕ ϕ∇ ⋅ = ∇ ⋅ + ⋅∇a a a

5. ( )div curl curl× = ⋅ − ⋅a b b a a b ( ) ( ) ( )∇ ⋅ × = ⋅ ∇× − ⋅ ∇×a b b a a b

6. ( )div c cdiv=a a ( )c c∇⋅ = ∇ ⋅a a

7. ( )curl curl curl+ = +a b a b ( )∇× + = ∇× +∇×a b a b

8. ( )curl curl gradϕ ϕ ϕ= + ×a a a ( ) ( )ϕ ϕ ϕ∇× = ∇× +∇ ×a a a

9. ( ) ( )curl curl grad curl curlϕ+ =a a ( ) ( )ϕ∇× ∇× +∇ = ∇× ∇×a a

10. ( ) ( ) 2curl curl grad div= −∇a a a ( ) ( ) 2∇× ∇× = ∇ ∇⋅ −∇a a a

curla

curla

curla

1,2,3

Chapter IV Vector and Tensor Analysis IV.3 Vector Analysis - Revisited September 19, 2018 298 IV.3.9. LINE INTEGRAL Consider a vector-valued function , where vectors belong to the space curve

C : ( )tr [ ]t a,b∈ with ( )a =r A , ( )b =r B (111) We will consider a line integral which symbolically is written as:

(112)

Let us see how this integral is defined in its physical sense. Set up a partition nP of the curve C into a discrete set of n points:

( ) ( ){ }n 0 1 2 nP a , , ,..., b= = =r r r r r r and define the increment Define the norm of partition as the biggest increment in the partition: n kk

P max ∆= r

Denote the values of function at the points of partition

Form a dot product k k∆⋅a r which has a physical sense of the work performed by the force ka over path

k∆r . Then the line integral is defined similarly to the definition of the definite integral as a limit of the Riemann sum:

(113)

which physically expresses the work done by the force along the space curve C .

Calculation of the line integral: Let the vector function have the following specification:

( ) ( ) ( )1 2 3 1 2 3 1 2 3P x ,x ,x Q x ,x ,x R x ,x ,x= + +i j k

(114)

Let the parameterization of the curve C be defined by

( )tr ( ) ( ) ( )1 1 2 2 3 3x t x t x t= + +i i i (115)

differentiation of this equation yields:

( )d tdt

r ( ) ( ) ( )1 1 2 2 3 3x t x t x t′ ′ ′= + +i i i or

( ) ( ) ( )1 1 2 2 3 3x t x t x t dt′ ′ ′ = + + i i i 1 1 2 2 3 3dx dx dx= + +i i i (116) from which follows that

k d⋅i r kdx= k 1,2,3=

Then the line integral (112) can be transformed to

( )a r

C

d⋅∫a r

k k k 1∆ −= −r r r

( )a r

( )k k=a a r

P 0n

n

k kn k 1C

d lim ∆→

→∞ =

⋅ = ⋅∑∫a r a r

( )a r

( )a r

( )a r

( ) ( ) ( )1 1 2 3 1 2 1 2 3 2 3 1 2 3 3P x ,x ,x P x ,x ,x P x ,x ,x= + +i i i

( )d tr

Chapter IV Vector and Tensor Analysis IV.3 Vector Analysis - Revisited September 19, 2018 299

C

d⋅∫a r

( ) ( ) ( )1 1 2 3 1 2 1 2 3 2 3 1 2 3 3C

P x ,x ,x dx P x ,x ,x dx P x ,x ,x dx = + + ∫

C

d⋅∫a r

this is the traditional form without parentheses. Then applying i idx x dt′= ,

C

d⋅∫a r ( ) ( ) ( )b

1 1 2 3 1 2 1 2 3 2 3 1 2 3 3a

P x ,x ,x x P x ,x ,x x P x ,x ,x x dt′ ′ ′ = + + ∫ (117)

Example: Find for the vector function

3p,q,x=a p,q∈ along the space curve C : ( )tr ( ) ( )1 2 3cos t sin t t= + +i i i t from 0 to π Identify: ( )1x t cos t= ( )1x t sin t′ = −

( )2x t sin t= ( )2x t cos t′ =

( )3x t t= ( )3x t 1′ = Then, using equation (117), one ends up with

[ ]0

p sin t q cos t t dtπ

= − + +∫

2

0

tp cos t q sin t2

π

= + +

2

p p2π

= − + −

2

2 p2π

= − ■

If the curve is defined with the natural parameterization , then

( )d d ds dssdt ds dt dt

= =rr T where T is a unit vector tangent to the curve C .

Then d ds=r T . Recall also

ds 2 2 21 2 3dx dx dx= + + 2 2 2

1 2 3x x x dt′ ′ ′= + +

The line integral then is calculated according to

( ) ( ) ( )1 1 2 3 1 2 1 2 3 2 3 1 2 3 3C

P x ,x ,x P x ,x ,x P x ,x ,x d = + + ⋅ ∫ i i i r

( ) ( ) ( )1 1 2 3 1 2 1 2 3 2 3 1 2 3 3C

P x ,x ,x d P x ,x ,x d P x ,x ,x d = ⋅ + ⋅ + ⋅ ∫ i r i r i r

( ) ( ) ( )1 1 2 3 1 2 1 2 3 2 3 1 2 3 3C

P x ,x ,x dx P x ,x ,x dx P x ,x ,x dx= + +∫

C

d⋅∫a r

C

d⋅∫a r

( )sr

Chapter IV Vector and Tensor Analysis IV.3 Vector Analysis - Revisited September 19, 2018 300

C

d⋅∫a r

( ) ( ) ( )b

2 2 21 1 2 3 1 2 1 2 3 2 3 1 2 3 3 1 2 3

a

P x ,x ,x P x ,x ,x P x ,x ,x x x x dt′ ′ ′ = ⋅ + ⋅ + ⋅ + + ∫ i T i T i T

b2 2 2

1 2 3a

x x x dt′ ′ ′= ⋅ + +∫a T (118)

Therefore, the work is performed only by the tangential component of the force. Conservative vector fields If ( )a r is a gradient field of some scalar field

( ) ϕ= ∇a r (119)

In this case function ( )ϕ r is called a potential function for the gradient field

( )a r . Then a linear integral along the curve connecting two points 1r and 2r is equal to the difference between values of the scalar function at these end points:

( ) ( )1 2C C C

d d dϕ ϕ ϕ ϕ⋅ = ∇ ⋅ = = −∫ ∫ ∫a r r r r (120)

It means that the same result will occur for any curve connecting points 1r and

2r , and the line integral is said to be independent of path. We have for a gradient field that d dϕ ϕ= ∇ ⋅ r d= ⋅a r

( )1 1 2 2 3 3P P P d= + + ⋅i i i r

( )1 1 2 2 3 3P d P d P d= ⋅ + ⋅ + ⋅i r i r i r 1 1 2 2 3 3P dx P dx P dx= + + (121)

Therefore, 1 1 2 2 3 3P dx P dx P dx+ + is an exact differential. So, the line integral

1 1 2 2 3 3C

P dx P dx P dx+ +∫

is independent of path, if 1 1 2 2 3 3P dx P dx P dx dϕ+ + = is an exact differential. Test for path independence Recall from calculus that the differential form 1 1 2 2 3 3P dx P dx P dx+ + is an exact differential if and only if

i k

k i

P Px x∂ ∂

=∂ ∂

(122)

It is called the test for path independence of a linear integral in space. Of particular interest are the linear integrals along the closed curves denoted by

Circulation C

d⋅∫ a r

(123)

It is called the circulation of the vector ( )a r around the contour C . If ( )a r is a force field, then the circulation is the work done by the force around C . It is obvious that for the gradient field

C C

d d 0ϕ⋅ = ∇ ⋅ =∫ ∫a r r

If a vector field is a gradient field of some scalar field it is said to be conservative.

( ) ( ) ( )1 1 2 3 1 2 1 2 3 2 3 1 2 3 3C

P x ,x ,x P x ,x ,x P x ,x ,x d = + + ⋅ ∫ i i i r

( ) ( )1 2C C C

d d dϕ ϕ ϕ ϕ⋅ = ∇ ⋅ = = −∫ ∫ ∫a r r r r

Chapter IV Vector and Tensor Analysis IV.3 Vector Analysis - Revisited September 19, 2018 301 IV.3.10. VOLUME INTEGRAL Consider a scalar field ( )ϕ r and let V be a volume. Subdivide the volume V

into subvolumes kk

V V∆= ∑ and define an integral of the function ( )ϕ r over

the volume V as a limit

( )V

dVϕ∫ r ( )k

k kV 0 k lim V∆

ϕ ∆→

= ∑ r (124)

where kr is an arbitrary point in the subvolume kV∆ . TRIPLE INTEGRAL OF THE SCALAR FIELD

One of the possible types of region V in which the volume integral can be reduced to iterated triple integral:

V : ( ) 3x, y,z ∈ 1 2a x a≤ ≤

( ) ( )1 2f x y f x≤ ≤

( ) ( )1 2g x, y z g x, y≤ ≤

There are totally six standard types of volume region which allow reduction to iterated integration:

V : ( ) 3x, y,z ∈ 1 i 2a x a≤ ≤

( ) ( )1 i j 2 if x x f x≤ ≤

( ) ( )1 i j l 2 i jg x ,x x g x ,x≤ ≤ i, j ,l 1,2,3= i j l≠ ≠

The arbitrary region V should be subdivided into combination of standard regions.

Construction of the iterated triple integral of the scalar field ( )F x, y,z , ( )x, y,z V∈ :

( )V

F dV∫ r ( )( )

( )

( )

( )2 22

1 1 1

f x g x ,ya

a f x g x ,y

F x, y,z dzdydx= ∫ ∫ ∫ (125)

If F 1≡ , then the triple integral yields the volume of the region:

V V

dV= ∫ ( )

( )

( )

( )2 22

1 1 1

f x g x ,ya

a f x g x ,y

dzdydx= ∫ ∫ ∫ (126)

Other coordinate systems (see Table 14): cylindrical: dV rdrd dzθ= , spherical: 2dV r sin drd dθ φ θ= .

VkV∆

( )2g x, y

( )1g x, y

ky

kr

y

x

z

kz

kx ( )1f x

( )2f x( )k kx , y

( )k k kx , y ,z

2a

1a0

Chapter IV Vector and Tensor Analysis IV.3 Vector Analysis - Revisited September 19, 2018 302 Example (Zill, p.541, modified) Consider the solid in the first octant bounded by the

the surfaces: x 3= , y 2x= and 2z 1 y= − .

Let the temperature distribution is given by

( ) 2 2 2

10T x, y,zx y z 4

=+ + +

Find the average temperature of the solid avT :

avV

1T TdVV

= ∫∫∫

First, find the volume of the solid: V

V

dV= ∫∫∫

21 y1 3

y0 02

dzdxdy−

= ∫ ∫ ∫

( )1 3

2

y02

1 y dxdy= −∫ ∫

( )1

2

0

11 y 3 y dy2

= − − ∫

15 1.8758

= =

Second, integrate temperature field over the volume:

V

TdV∫∫∫ 21 y1 3

2 2 2y0 02

10 dzdxdyx y z 4

−

=+ + +∫ ∫ ∫

2.77≈ integrated with Maple Then the average temperature is:

avT V

1 TdVV

= ∫∫∫ 1.48≈

Remark: To find the volume, the problem could be formulated as a double integral

V A

FdA= ∫∫ ( )1 3

2

y02

1 y dxdy= −∫ ∫

x

y

z

y 2x=

x 3=

y 1=

z 1=

2z 1 y= −

( ) 2 2

10T x, y,0x y 4

=+ +

( ) 2 2

10T x, y,0x y 4

=+ +

Temperature distribution on the bottom of the solid

Chapter IV Vector and Tensor Analysis IV.3 Vector Analysis - Revisited September 19, 2018 303 IV.3.11. INTEGRAL THEOREMS

GREEN’S THEOREM (Green’s Theorem – Plane case) Let S be a plane region bounded by a closed contour C .

Let ( ) ( ) ( )P x, y Q x, y= +a r i j be a two-dimensional vector field

where ( ) ( ) ( ) ( )P x, y Q x, yP x, y ,Q x, y , ,

y x∂ ∂

∂ ∂ are continuous in

S C∪ . Then

C S

Q PPdx Qdy dSx y

∂ ∂+ = − ∂ ∂

∫ ∫∫

(127)

where C is traced in the direction such that region S appears to the left of an observer moving along contour C . Equation (124) can be written also in the following vector form: ( )

C S

d curl dS⋅ = ⋅∫ ∫∫a r a k

(128)

STOKES’ THEOREM (Stokes’ Theorem – Green’s Theorem in Space) Let S be a surface bounded by a closed contour C .

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

where P P Q Q R RP,Q,R, , , , , ,y z x z x y

∂ ∂ ∂ ∂ ∂ ∂∂ ∂ ∂ ∂ ∂ ∂

are continuous in S C∪ .

Then

C

Pdx Qdy Rdz+ +∫ ( ) ( ) ( )S

R Q P R Q Pcos ,x cos , y cos ,z dSy z z x x y

∂ ∂ ∂ ∂ ∂ ∂ = − + − + − ∂ ∂ ∂ ∂ ∂ ∂ ∫∫ n n n (129)

Where orientation of the surface S is defined by the exterior unit normal vector n , and the contour C is traced in the direction such that surface S appears to the left of an observer moving along contour C with the vector n at points near C pointing from the observer’s feet to his head. Equation (129) can be written also in the following vector form: ( )

C S

d curl dS⋅ = ⋅∫ ∫∫a r a n

(130)

claming that circulation of a vector field around the boundary is equal to the flux of the curl through the surface.

( )n r

n

r

0

C

S

S

Csimple closed curve

Chapter IV Vector and Tensor Analysis IV.3 Vector Analysis - Revisited September 19, 2018 304 DIVERGENCE THEOREM (the Gauss-Ostrogradsky Theorem or the Divergence Theorem) Let V be a volume bounded by a closed surface S .

Then flux of the vector field ( )a r through the surface S is equal to the integral of the divergence of the vector field over the volume

S V

d = div dV⋅∫ ∫a S a

(131)

Proof: We will show that equation (131) is approximately valid with any

degree of accuracy, i.e. that

S V

d - div dV ε⋅ <∫ ∫a S a

for any 0ε > .

Subdivide volume V into kk

V V∆= ∑ such that

( )k

kk S

1 d - divV

δ⋅ <∫ a S a r

This is possible according to the definition of the divergence as a limit (105). Multiply this inequality by kV

( )k

k k kS

d - V div Vδ⋅ <∫ a S a r

then summation over all k yields

( )k kkS

d - V div Vδ⋅ <∑∫ a S a r

In this result, the surface integral only over the exterior surface S is left. All

interior surfaces kS have to be the boundaries of some adjacent volumes mV and kV . Having the opposite normal vectors, k m= −n n , the surface integrals over kS cancel each other in the summation:

k mS S

d d ⋅ + ⋅∫ ∫a S a S

k m

k mS S

dS dS = ⋅ + ⋅∫ ∫a n a n

k k

k kS S

dS dS = ⋅ − ⋅∫ ∫a n a n

0=

Let k →∞ with kV 0→ , then according to the definition of the volume integral

( )k

k kV 0 klim V div∆ →

∑ a rV

div dV= ∫ a

and

S V

d - div dV Vδ⋅ <∫ ∫a S a

Choose Vεδ = , then

S V

d - div dV ε⋅ <∫ ∫a S a

for any specified 0ε > . ■

Chapter IV Vector and Tensor Analysis IV.3 Vector Analysis - Revisited September 19, 2018 305 Recall Equations (85) and (12):

diva 31 2 xx x

1 2 3

aa a=

x x x∂∂ ∂

+ +∂ ∂ ∂

( ) ( ) ( )1 2 3x 1 x 2 x 3 na cos , a cos , a cos , a⋅ = + + =a n n i n i n i (normal projection)

The Gauss-Ostrogradsky Theorem or the Divergence Theorem

can be written also in the following forms:

S V

d = div dV⋅∫ ∫a S a

S V

dS = div dV⋅∫ ∫a n a

(132)

nS V

a dS = div dV∫ ∫ a

( ) ( ) ( ) 31 2

1 2 3

xx xx 1 x 2 x 3

1 2 3S V

aa aa cos , a cos , a cos , dS = dV

x x x∂∂ ∂

+ + + + ∂ ∂ ∂ ∫ ∫n i n i n i

The Divergence Theorem has a great importance in mathematical modeling in

engineering. In derivation of the governing equations for physical processes in the continuous media, the conservation laws are applied to the control volume yielding an equation which contains both surface integrals and volume integrals. Application of the Divergence Theorem reduces all integrals to the volume integral which allows combination of all terms in one volume integral and concludes with the partial differential equation which governs the physical process under consideration.

Chapter IV Vector and Tensor Analysis IV.3 Vector Analysis - Revisited September 19, 2018 306 EXAMPLE (Application of the Divergence Theorem - Equation of Continuity)

Consider fluid flow with the velocity field ( )v r .

The flux of the velocity field through some surface S determines the volume of fluid flowing through the surface S per unit time:

S

q dS= ⋅∫ v n

Then for incompressible fluid

m qρ=S

dS= ⋅∫ v n

is the mass of fluid flowing through the surface S per unit time. For compressible fluid, density varies in space ( )ρ ρ= r , so

( )S

m dSρ= ⋅∫ r v n

This equation represents the net amount of fluid flowing through the surface S with normal vector n representing the positive direction.

Let now S be the closed surface of some finite control volume V containing the point of space r . In the stationary case, without any sources or sinks,

( )S

m dS 0ρ= ⋅ =∫ r v n

i.e. the mass of fluid flowing into V is equal to the mass flowing out of V.

The mass of fluid in the control volume is defined by the volume integral

V

dVρ∫

In the non-stationary fluid flow, density depends on time

( ),tρ ρ= r

Then the conservation of mass in the control volume yields that the change of mass in the control volume V is equal to the mass flowing through the surface of the volume:

(the sign is minus because the negative direction is toward the control volume). For fixed boundaries of control volume, differentiation can be moved inside of the integral

V S

dV dS 0tρ ρ∂

+ ⋅ =∂∫ ∫ v n

Apply the Divergence Theorem to replace the surface integral by the volume integral

( )V V

dV div dV 0tρ ρ∂

+ =∂∫ ∫ v

Now both integrals are over the volume. Adding the integrals, we obtain

V S

dV dSt

ρ ρ∂= − ⋅

∂ ∫ ∫ v n

S

V

r

0

n

n

v

v

0⋅ >v n

0⋅ <v n

Chapter IV Vector and Tensor Analysis IV.3 Vector Analysis - Revisited September 19, 2018 307

( )V

div dV 0tρ ρ∂ + = ∂ ∫ v

Because this equation is valid for any control volume containing the point r and the velocity and density fields are continuous, the integrand should be equal to zero provided that it also is continuous:

equation of continuity ( )div 0tρ ρ∂+ =

∂v (133)

This equation is called the equation of continuity. Using operator nabla, we can rewrite it in the form:

( ) 0tρ ρ∂+∇ ⋅ =

∂v

For incompressible fluid, density does not depend on location, and 0tρ∂≈

∂

is negligible. Then the equation of continuity reduces to

incompressible fluid 0∇⋅ =v

for both stationary and non-stationary flow.

Exercize Convert equation of continuity to Cartesian, cylindrical and spherical

coordinates.

Chapter IV Vector and Tensor Analysis IV.3 Vector Analysis - Revisited September 19, 2018 308 IV.4 EXERCISES AND EXAMPLES REVIEW QUESTIONS: 1) What is a vector function of a scalar variable? How it can be defined? What

is the graph of a vector function? 2) How can lines in space be defined? How can a segment in space be defined? 3) What is the vector equation defining a plane in space? What is the meaning

of the constant term in this equation? 4) What is the vector equation of the plane containing three fixed points?

5) How is a sphere defined by a vector equation? 6) What is the continuity of a vector function? 7) How are the vector functions differentiated?

8) What is the geometrical sense of the derivative of a vector function? 9) What is the mechanical sense of the first and second derivatives of a vector

function? 10) Recall the main rules of differentiation of vector functions. 11) How is the Taylor series defined for a vector function? 12) How are the vector functions integrated? 13) Give the geometrical interpretation of the indefinite integral of a vector

function. 14) How is the definite integral defined for a vector function? 15) How can the length of a space curve be calculated? 16) What is the parameterization of a curve in terms of arc length? 17) What property is possessed by the derivative of the space curve

parameterized in terms of arc length? 18) What is a scalar field? 19) What is a vector field? 20) What are level curves and level surfaces? 21) What is the operator “nabla”? How is it applied to a scalar field? 22) What is a gradient of a scalar field? Give a geometrical interpretation of a

gradient. 23) What is a directional derivative of a scalar field? How it is defined with the

help of operator nabla? 24) What is the directional derivative of a vector field? How is it defined with

the help of the operator nabla? 25) How the surface integral of a vector field calculated? 26) What is the flux of a vector field through the surface? How is it defined? 27) What is the divergence of a vector field? 28) What is the curl of the vector field? 29) What is a line integral of a vector field? 30) What is a gradient field? 31) What is it called when the line integral is independent of path? 32) What does it mean that some gradient field is conservative? 33) What is the circulation of a vector field? 34) Recall the Integral Theorems. 35) What is the sense of application of the Divergence Theorem for derivation of

the Equation of continuity and The Heat Equation?