Chapter 7 Multiple Integral (Chapter 14) Double Integral

University of Baghdad 2nd Class

College of Engineering Math II

Electronics and Communications Engineering Dpt. Chapter 7

Chapter 7 Multiple Integral (Chapter 14)

Double Integral (14.1)

If 𝑓(𝑥, 𝑦) is defined on the rectangular region given by

R: 𝑎 ≤ 𝑥 ≤ 𝑏, 𝑐 ≤ 𝑦 ≤ 𝑑

Then we can write

∬ 𝑓(𝑥, 𝑦)𝑑𝐴

𝑅

= ∫ ∫ 𝑓(𝑥, 𝑦)𝑑𝑦𝑑𝑥

𝑑

𝑐

𝑏

𝑎

= ∫ ∫ 𝑓(𝑥, 𝑦)𝑑𝑥𝑑𝑦

𝑏

𝑎

𝑑

𝑐

The double integral represents the volume under the surface 𝑧 = 𝑓(𝑥, 𝑦) within the region R

Ex: Evaluate the integral


𝑅

Where 𝑓(𝑥, 𝑦) = 1 − 6𝑥2𝑦 and R: 0 ≤ 𝑥 ≤ 2, −1 ≤ 𝑦 ≤ 1

Sol.


𝑅

= ∫ ∫(1 − 6𝑥2𝑦)𝑑𝑥𝑑𝑦

2

0

1

−1

= ∫ [𝑥 − 2𝑥3𝑦]02𝑑𝑦

−1

−1

= ∫(2 − 16𝑦)𝑑𝑦

1

−1

= 2𝑦 − 8𝑦2|−11 = 2 − 8 − (−2 − 8) = 4

This integral can be evaluated in the order of integration reversed


𝑅

= ∫ ∫(1 − 6𝑥2𝑦)𝑑𝑦𝑑𝑥

1

−1

2

0

= ∫[𝑦 − 3𝑥2𝑦2]−11 𝑑𝑥

2

0

= ∫((1 − 3𝑥2) − (−1 − 3𝑥2))𝑑𝑥

2

0

= ∫(2)𝑑𝑥

2

0

= 2𝑥|02 = 4

Double Integral Over Bounded Nonrectangular Regions


𝑅

= ∫ ∫ 𝑓(𝑥, 𝑦)𝑑𝑦𝑑𝑥

𝑔2(𝑥)

𝑔1(𝑥)

𝑏

𝑎

= ∫ ∫ 𝑓(𝑥, 𝑦)𝑑𝑥𝑑𝑦

ℎ2(𝑦)

ℎ1(𝑦)

𝑑

𝑐




Ex: Find the volume of the prism whose base is the triangle in the xy-plane bounded by the x-axis

and the lines y=x and x=1 and whose top lies in the plane z=3xy.

Sol.

𝑉 = ∫ ∫(3 − 𝑥 − 𝑦)𝑑𝑦𝑑𝑥

𝑥

0

1

0

= ∫ [3𝑦 − 𝑥𝑦 −𝑦2

2]

0

𝑥

𝑑𝑥

1

0

= ∫ (3𝑥 − 𝑥2 −𝑥2

2) 𝑑𝑥

1

0

= ∫ (3𝑥 −3

2𝑥2) 𝑑𝑥

1

0

= [3

2𝑥 −

1

2𝑥3]

0

1

=3

2−

1

2= 1

By reversing the order of integration

𝑉 = ∫ ∫(3 − 𝑥 − 𝑦)𝑑𝑥𝑑𝑦

1

𝑦

1

0

= ∫ [3𝑥 −𝑥2

2− 𝑥𝑦]

𝑦

1

𝑑𝑦

1

0

= ∫ ((3 −1

2− 𝑦) − (3𝑦 −

𝑦2

2− 𝑦2)) 𝑑𝑦

1

0

= ∫ (5

2− 4𝑦 +

3

2𝑦2) 𝑑𝑦

1

0

= [5

2𝑦 − 2𝑦2 +

1

2𝑦3]

0

1

=5

2− 2 +

1

2= 1

Ex: Find

∬sin 𝑥

𝑥𝑑𝐴

𝑅

Where R is the triangular region in the xy-plane bounded by the lines y=x, x=1 and x-axis

Sol.:

∫ ∫sin 𝑥

𝑥𝑑𝑦𝑑𝑥

𝑥

0

= ∫ [𝑦sin 𝑥

𝑥]

0

𝑥1

0

1

0

= ∫ sin 𝑥

1

0

= − cos 𝑥|01 = 1 − cos 1




Ex: Find the value of the integration

∫ ∫ (4𝑥 + 2)𝑑𝑦𝑑𝑥

2𝑥

𝑥2

2

0

With the order of the integration reversed

Sol.:

First it is better to find the points of intersection of the curves.

𝑥2 = 2𝑥 ⇒ 𝑥2 − 2𝑥 = 0

𝑥(𝑥 − 2) = 0

𝑥 = 0 ⇒ 𝑦 = 0

𝑥 = 2 ⇒ 𝑦 = 4

𝑦 = 𝑥2 ⇒ 𝑥 = √𝑦

𝑦 = 2𝑥 ⇒ 𝑥 = 𝑦/2

∫ ∫ (4𝑥 + 2)𝑑𝑦𝑑𝑥

2𝑥

𝑥2

2

0

= ∫ ∫ (4𝑥 + 2)𝑑𝑥𝑑𝑦

√𝑦

𝑦/2

4

0

= ∫[2𝑥2 + 2𝑥]𝑦/2√𝑦

4

0

= ∫ (2𝑦 + 2√𝑦 −𝑦2

2− 𝑦) 𝑑𝑦

4

0

= ∫ (𝑦 + 2√𝑦 −𝑦2

2) 𝑑𝑦

4

0

= [𝑦2

2+

4

3𝑦3/2 −

𝑦3

6]

0

4

= 8 +32

3−

32

3− 0 = 8

Areas, Moments and Center of Mass

Areas of Bounded Region in the Plane

The area of a bounded region R can be expressed in terms of double integral

𝐴 = ∬ 𝑑𝐴

𝑅

= ∬ 𝑑𝑦𝑑𝑥

𝑅

= ∬ 𝑑𝑥𝑑𝑦

𝑅

Ex: Find the area of the region R bounded by the parabola 𝑦 = 𝑥2 and the line y=x+2

Sol.:

First we find the points of intersections

𝑥2 = 𝑥 + 2

𝑥2 − 𝑥 − 2 = 0

(𝑥 + 1)(𝑥 − 2) = 0




𝑥 = −1 ⇒ 𝑦 = 1

𝑥 = 2 ⇒ 𝑦 = 4

𝐴 = ∫ ∫ 𝑑𝑦𝑑𝑥

𝑥+2

𝑥2

2

−1

= ∫[𝑦]𝑥2𝑥+2𝑑𝑥

2

−1

= ∫(𝑥 + 2 − 𝑥2)𝑑𝑥

2

−1

= [𝑥2

2+ 2𝑥 −

𝑥3

3]

−1

2

= (2 + 4 −8

3) − (

1

2− 2 +

1

3) = 8 − 3 −

1

2=

9

2

First and Second Moments and Center of Mass

Density: 𝛿(𝑥, 𝑦)

Mass: 𝑀 = ∬ 𝛿(𝑥, 𝑦)𝑑𝐴

First Moments:

Moment about x-axis: 𝑀𝑥 = ∬ 𝑦𝛿(𝑥, 𝑦)𝑑𝐴

Moment about y-axis: 𝑀𝑦 = ∬ 𝑥𝛿(𝑥, 𝑦)𝑑𝐴

Center of Mass: �̅� =𝑀𝑦

𝑀, �̅� =

𝑀𝑥

𝑀

Moments of Inertia (Second Moments)

About x-axis: 𝐼𝑥 = ∬ 𝑦2𝛿(𝑥, 𝑦)𝑑𝐴

About y-axis: 𝐼𝑦 = ∬ 𝑥2𝛿(𝑥, 𝑦)𝑑𝐴

About origin: 𝐼𝑜 = 𝐼𝑥 + 𝐼𝑦 = ∬(𝑥2 + 𝑦2)𝛿(𝑥, 𝑦)𝑑𝐴

Centroids of Geometric Figures

When the density of an object is constant, it cancels out of the numerator and denominator of the

formulas for �̅� and �̅�. So, we can take =1 in calculating �̅� and �̅�.

Ex: Find the centroid of the region in the first quadrant that is bounded by the line y=x and the curve

𝑦 = 𝑥2.

Sol.:

𝑀 = ∫ ∫(1)𝑑𝑦𝑑𝑥

𝑥

𝑥2

1

0

= ∫[𝑦]𝑥2𝑥

1

0

𝑑𝑥 = ∫(𝑥 − 𝑥2)

1

0

𝑑𝑥 = [𝑥2

2−

𝑥3

3]

0

1

=1

2−

1

3=

1

6




𝑀𝑥 = ∫ ∫(1)𝑦𝑑𝑦𝑑𝑥

𝑥

𝑥2

1

0

= ∫ [𝑦2

2]

𝑥2

𝑥1

0

𝑑𝑥 =1

2∫(𝑥2 − 𝑥4)

1

0

𝑑𝑥 =1

2[𝑥3

3−

𝑥5

5]

0

1

=1

2(

1

3−

1

5) =

1

15

𝑀𝑦 = ∫ ∫(1)𝑥𝑑𝑦𝑑𝑥

𝑥

𝑥2

1

0

= ∫[𝑥𝑦]𝑥2𝑥

1

0

𝑑𝑥 = ∫(𝑥2 − 𝑥3)

1

0

𝑑𝑥 = [𝑥3

3−

𝑥4

4]

0

1

= (1

3−

1

4) =

1

12

�̅� =𝑀𝑦

𝑀=

1/12

1/6=

1

2

�̅� =𝑀𝑥

𝑀=

1/15

1/6=

2

5

Ex: Evaluate the integral

∫ ∫ cos(𝑦3) 𝑑𝑦𝑑𝑥

2

√𝑥

4

0

Sol.:

The integral cannot be solved in the given order, so, we have

to reverse the order of integration. The limits of the y variable are

𝑦 = √𝑥 (𝑥 = 𝑦2) and y=2, the intersection point is (2,4), the integral in the reversed order will be

∫ ∫ cos(𝑦3) 𝑑𝑥𝑑𝑦

𝑦2

0

2

0

= ∫[𝑥 cos(𝑦3)]0𝑦2

𝑑𝑦

2

0

= ∫ 𝑦2cos(𝑦3) 𝑑𝑦

2

0

=1

3sin(𝑦3)|

0

2

=1

3sin(8)

Exercises 14.2

Sketch the region bounded by the given lines and curves, then find the area by double integration.

6- The parabola 𝑥 = 𝑦 − 𝑦2 and the line 𝑥 + 𝑦 = 0

Sol.

𝑥 + 𝑦 = 0 ⇒ 𝑥 = −𝑦

−𝑦 = 𝑦 − 𝑦2

2𝑦 − 𝑦2 = 0 ⇒ 𝑦(2 − 𝑦) = 0




𝑦 = 0 ⇒ 𝑥 = 0

𝑦 = 2 ⇒ 𝑥 = −2

𝐴 = ∬ 𝑑𝐴 = ∫ ∫ 𝑑𝑥𝑑𝑦

𝑦−𝑦2

−𝑦

2

0

𝐴 = ∫(𝑦 − 𝑦2 − (−𝑦))𝑑𝑦

2

0

= ∫(2𝑦 − 𝑦2)𝑑𝑦

2

0

= 𝑦2 −𝑦3

3|

0

2

= 4 −8

3− 0 =

4

3

36- Find the centroid of the region in the xy-plane bounded by the curves =1

√1−𝑥2 , 𝑦 =

−1

√1−𝑥2 and

the lines x=0 and x=1.

Sol.:

From symmetry �̅� = 0

𝑀 = ∬(1)𝑑𝐴

𝑀 = ∫ ∫ 𝑑𝑦𝑑𝑥

1/√1−𝑥2

−1/√1−𝑥2

1

0

= ∫ (1

√1 − 𝑥2−

−1

√1 − 𝑥2)

1

0

𝑑𝑥

= ∫ (2𝑑𝑥

√1 − 𝑥2)

1

0

= 2sin−1(𝑥)|01

2 (𝜋

2= 0) = 𝜋

𝑀𝑦 = ∫ ∫ 𝑥𝑑𝑦𝑑𝑥

1/√1−𝑥2

−1/√1−𝑥2

1

0

= ∫[𝑥𝑦]−1/√1−𝑥2

1/√1−𝑥2

1

0

𝑑𝑥

= ∫ (2𝑥

√1 − 𝑥2)

1

0

𝑑𝑥 = −√1 − 𝑥2

1/2|

0

1

= −2(0 − 1) = 2

�̅� =2

𝜋




Double Integral in Polar Form

To integrate a function in polar form 𝑓(𝑟, 𝜃) over the shaded region shown

∬ 𝑓(𝑟, 𝜃)𝑑𝐴

𝑅

= ∫ ∫ 𝑓(𝑟, 𝜃)𝑟𝑑𝑟𝑑𝜃

1+cos 𝜃

1

𝜋/2

−𝜋/2

Area in Polar Coordinates

𝐴𝑟𝑒𝑎 𝑜𝑓 𝑅 = ∬ 𝑑𝐴

𝑅

= ∬ 𝑟𝑑𝑟𝑑𝜃

𝑅

Ex: Find the area enclosed by the curve 𝑟2 = 4 cos 2𝜃

Sol.:

𝐴

4= ∫ ∫ 𝑟𝑑𝑟𝑑𝜃

√4 cos 2𝜃

0

𝜋/4

0

𝐴 = 4 ∫ [𝑟2

2]

0

√4 cos 2𝜃

𝑑𝜃

𝜋/4

0

= 4 ∫ 2 cos 2𝜃 𝑑𝜃

𝜋/4

0

= 4[sin(2𝜃)]0𝜋/4

= 4(1 − 0) = 4

Changing Cartesian Integration into Polar Integration

∬ 𝑓(𝑥, 𝑦)𝑑𝑥𝑑𝑦

𝑅

= ∬ 𝑓(𝑟cos(𝜃), 𝑟sin(𝜃))𝑟𝑑𝑟𝑑𝜃

𝐺

R is the region of integration described in Cartesian coordinates and G is the same region described

in polar coordinates.

Ex: Find the moment of inertia about the origin of a thin plate of density (x,y)=1 bounded by the

quarter circle 𝑥2 + 𝑦2 = 1 in the first quadrant.

Sol.:

In Cartesian coordinates

𝐼𝑜 = ∫ ∫ (𝑥2 + 𝑦2)𝑑𝑦𝑑𝑥

√1−𝑥2

0

1

0




∫ [𝑥2𝑦 +𝑦3

3]

0

√1−𝑥2

𝑑𝑥 = ∫ (𝑥2√1 − 𝑥2 +1

3(1 − 𝑥2)3/2) 𝑑𝑥

1

0

1

0

In polar coordinates

𝐼𝑜 = ∫ ∫ (𝑥2 + 𝑦2)𝑑𝑦𝑑𝑥

√1−𝑥2

0

1

0

= ∫ ∫(𝑟2)𝑟𝑑𝑟𝑑𝜃

1

0

𝜋/2

0

= ∫ ∫(𝑟3)𝑑𝑟𝑑𝜃

1

0

𝜋/2

0

= ∫𝑟4

4|

0

1

𝑑𝜃

𝜋/2

0

= ∫1

4𝑑𝜃

𝜋/2

0

=𝜃

4|

0

𝜋/2

=𝜋

8

Ex: Evaluate the integral (the Gaussian integral)

∫ 𝑒−𝑥2𝑑𝑥

∞

−∞

Sol.:

𝐼 = ∫ 𝑒−𝑥2𝑑𝑥

∞

−∞

𝐼 = ∫ 𝑒−𝑦2𝑑𝑦

∞

−∞

𝐼 · 𝐼 = [ ∫ 𝑒−𝑥2𝑑𝑥

∞

−∞

] [ ∫ 𝑒−𝑦2𝑑𝑦

∞

−∞

]

𝐼2 = ∫ ∫ 𝑒−(𝑥2+𝑦2)𝑑𝑥𝑑𝑦

∞

−∞

∞

−∞

The above integral can be converted to polar coordinates. We note that the region of integration is

the entire xy-plane that can be described as below

𝐼2 = ∫ ∫ 𝑒−𝑟2𝑟𝑑𝑟𝑑𝜃

∞

0

2𝜋

0

= ∫ −1

2𝑒−𝑟2

|0

∞2𝜋

0

𝑑𝜃 = −1

2∫ (0 − 1)𝑑𝜃

2𝜋

0

=1

2(2𝜋) = 𝜋

𝐼 = √𝜋

∫ 𝑒−𝑥2𝑑𝑥

∞

−∞

= √𝜋




Triple Integral in Rectangular Coordinates

The volume of a closed bounded region D in space is

𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝐷 = ∭ 𝑑𝑉

𝐷

= ∭ 𝑑𝑧𝑑𝑦𝑑𝑥

𝐷

In triple integral we always make the order of integration starts with dz and the limits of the integral

with respect to the variable z should always be clear from the problem statement. Next, the problem

reduces to double integral and we follow the same rules described in earlier sections. Usually we

only need to draw the region of integration with respect to x and y variable in 2-D only and no need

to draw a 3-dimesional drawing for the overall region.

Exercises 14.4

24- Find the volume of the region in the 1st octant bounded by the surface 𝑧 = 4 − 𝑥2 − 𝑦

Sol.:

In the xy-plane z=0 𝑦 = 4 − 𝑥2

0 ≤ 𝑧 ≤ 4 − 𝑥2 − 𝑦

0 ≤ 𝑦 ≤ 4 − 𝑥2

0 ≤ 𝑥 ≤ 2

𝑉 = ∫ ∫ ∫ 𝑑𝑧𝑑𝑦𝑑𝑥

4−𝑥2−𝑦

0

4−𝑥2

0

2

0

𝑉 = ∫ ∫ 𝑧|04−𝑥2−𝑦

𝑑𝑦𝑑𝑥

4−𝑥2

0

2

0

The first step is to find the limits of the variable

z. Since the region is in the first octant this

means that x 0, y 0 and z 0. Combining this

and the fact that the upper bound is 𝑧 = 4 −

𝑥2 − 𝑦, then we have found the limits of z

variable.

The second step is to draw the region in the xy-

plane and finding the limits of x and y variables.

The limits of the x, y, z variables are




𝑉 = ∫ ∫ (4 − 𝑥2 − 𝑦)𝑑𝑦𝑑𝑥

4−𝑥2

0

2

0

𝑉 = ∫ [(4 − 𝑥2)𝑦 −𝑦2

2]

0

4−𝑥2

𝑑𝑥

2

0

= ∫ ((4 − 𝑥2)2 −(4 − 𝑥2)2

2) 𝑑𝑥

2

0

=1

2∫((4 − 𝑥2)2)𝑑𝑥

2

0

𝑉 =1

2∫(16 − 8𝑥2 + 𝑥4)𝑑𝑥

2

0

=1

2∫ [16𝑥 −

8

3𝑥3 +

1

5𝑥5] 𝑑𝑥

2

0

=1

2(32 −

64

3+

32

5) =

128

15

Masses and Moments in Three Dimensions

Mass:

𝑀 = ∭ 𝛿𝑑𝑉

𝑅

(𝛿 = 𝑑𝑒𝑛𝑠𝑖𝑡𝑦)

First Moments:

𝑀𝑦𝑧 = ∭ 𝑥𝛿𝑑𝑉

𝑅

𝑀𝑥𝑧 = ∭ 𝑦𝛿𝑑𝑉

𝑅

𝑀𝑥𝑦 = ∭ 𝑧𝛿𝑑𝑉

𝑅

Center of Mass

�̅� =𝑀𝑦𝑧

𝑀, �̅� =

𝑀𝑥𝑧

𝑀, 𝑧̅ =

𝑀𝑥𝑦

𝑀

Moments of Inertia

𝐼𝑥 = ∭(𝑦2 + 𝑧2)𝛿𝑑𝑉

𝑅

𝐼𝑦 = ∭(𝑥2 + 𝑧2)𝛿𝑑𝑉

𝑅

𝐼𝑧 = ∭(𝑥2 + 𝑦2)𝛿𝑑𝑉

𝑅




Triple Integral in Cylindrical and Spherical Coordinates

Cylindrical Coordinates

To integrate a continuous function f(r,,z) over a region given by

𝑧1(𝑟, 𝜃) ≤ 𝑧 ≤ 𝑧2(𝑟, 𝜃)

𝑟1(𝜃) ≤ 𝑟 ≤ 𝑟2(𝜃)

𝜃1 ≤ 𝜃 ≤ 𝜃2

∫ ∫ ∫ 𝑓(𝑟, 𝜃, 𝑧)𝑑𝑧𝑟𝑑𝑟𝑑𝜃

𝑧2(𝑟,𝜃)

𝑧1(𝑟,𝜃)

𝑟2(𝜃)

𝑟1(𝜃)

𝜃2

𝜃1

Ex: Find the volume of the solid bounded by the cylinder 𝑥2 + 𝑦2 = 4 and the planes y+z=4 and

z=0.

Sol.:

𝑉 = ∫ ∫ ∫ 𝑟𝑑𝑧𝑑𝑟𝑑𝜃

4−𝑟 sin 𝜃

0

2

0

2𝜋

0

𝑉 = ∫ ∫𝑟𝑧|04−𝑟 sin 𝜃𝑟𝑑𝑟𝑑𝜃

2

0

2𝜋

0

𝑉 = ∫ ∫(4𝑟 − 𝑟2 sin 𝜃)𝑑𝑟𝑑𝜃

2

0

2𝜋

0

𝑉 = ∫ [2𝑟2 −1

3𝑟3 sin 𝜃]

0

2

𝑑𝜃

2𝜋

0

𝑉 = ∫ (8 −8

3sin 𝜃) 𝑑𝜃

2𝜋

0

= 8𝜃 +8

3cos 𝜃|

0

2𝜋

= 8(2𝜋 − 1) +8

3(1 − 1) = 16𝜋

Ex: Find the centroid of the solid enclosed by the cylinder 𝑥2 + 𝑦2 = 4, bounded above by the

paraboloid 𝑧 = 𝑥2 + 𝑦2 and below by the xy-plane.

Sol.:

𝑧 = 𝑥2 + 𝑦2 = 𝑟2

𝑥2 + 𝑦2 = 4 = 𝑟2 ⇒ 𝑟 = 2

0 ≤ 𝑧 ≤ 𝑟2

2 2

y

z

x2+y2=4

y=2

x=0

z=4y

x2+y2=4

r=2

x

y




0 ≤ 𝑟 ≤ 2

0 ≤ 𝜃 ≤ 2𝜋

The centroid lies on the axis of symmetry in this case the z-axis

�̅� = 0, �̅� = 0

𝑀 = ∫ ∫ ∫ (1)𝑟𝑑𝑧𝑑𝑟𝑑𝜃

𝑟2

0

2

0

2𝜋

0

= ∫ ∫[𝑟𝑧]0𝑟2

𝑑𝑟𝑑𝜃

2

0

2𝜋

0

= ∫ ∫ 𝑟3𝑑𝑟𝑑𝜃

2

0

2𝜋

0

= [𝑟4

4]

0

2

[𝜃]02𝜋 = (4)(2𝜋) = 8𝜋

𝑀𝑥𝑦 = ∫ ∫ ∫ 𝑧(1)𝑟𝑑𝑧𝑑𝑟𝑑𝜃

𝑟2

0

2

0

2𝜋

0

= ∫ ∫ [𝑟𝑧2

2]

0

𝑟2

𝑑𝑟𝑑𝜃

2

0

2𝜋

0

= ∫ ∫𝑟5

2𝑑𝑟𝑑𝜃

2

0

2𝜋

0

=1

2[𝑟6

6]

0

2

[𝜃]02𝜋 =

1

2(

32

3) (2𝜋) =

32𝜋

3

𝑧̅ =𝑀𝑥𝑦

𝑀=

32𝜋/3

8𝜋=

4

3

Ex: Find the center of mass of the solid whose density =z and bounded above by the plane z=y,

below by the xy-plane and laterally by the cylinder 𝑥2 + 𝑦2 = 4.

Sol.:

𝑀 = ∭ 𝛿𝑑𝑉

= ∫ ∫ ∫ 𝑧𝑟𝑑𝑧𝑑𝑟𝑑𝜃

𝑟sin𝜃

0

2

0

𝜋

0

= ∫ ∫𝑧2𝑟

2|

0

𝑟sin𝜃

𝑑𝑟𝑑𝜃

2

0

𝜋

0

= ∫ ∫1

2𝑟3 sin2 𝜃 𝑑𝑟𝑑𝜃

2

0

𝜋

0

= ∫ ∫1

4𝑟3(1 − cos2𝜃)𝑑𝑟𝑑𝜃

2

0

𝜋

0

=1

4[𝑟4

4]

0

2

[𝜃 −1

2sin2𝜃]

0

𝜋

=1

4(

16

4− 0) (𝜋 − 0) = 𝜋

𝑀𝑥𝑦 = ∫ ∫ ∫ 𝑧𝛿𝑟𝑑𝑧𝑑𝑟𝑑𝜃

𝑟sin𝜃

0

2

0

𝜋

0

= ∫ ∫ ∫ 𝑧2𝑟𝑑𝑧𝑑𝑟𝑑𝜃

𝑟sin𝜃

0

2

0

𝜋

0

2 2

x,y,r

z

z=r2

x2+y2=4

r=2

z=4

x

y

2 2

y

z x2+y2=4

y=2

x=0

z=y

r=2 y

x

r=2 x

y




= ∫ ∫𝑧3

3𝑟|

0

𝑟sin𝜃

𝑑𝑟𝑑𝜃

2

0

𝜋

0

= ∫ ∫1


2

0

𝜋

0

= ∫ ∫1

3𝑟4 sin 𝜃 (1 − cos2 𝜃) 𝑑𝑟𝑑𝜃

2

0

𝜋

0

=1

3[𝑟5

5]

0

2

[−cos 𝜃 +cos3 𝜃

3]

0

𝜋

=1

3(

32

5) (1 −

1

3+ 1 −

1

3)

=32

15(

4

3) =

128

45

𝑀𝑥𝑧 = ∫ ∫ ∫ 𝑦𝛿𝑟𝑑𝑧𝑑𝑟𝑑𝜃

𝑟sin𝜃

0

2

0

𝜋

0

= ∫ ∫ ∫ 𝑧𝑟2 sin 𝜃 𝑑𝑧𝑑𝑟𝑑𝜃

𝑟sin𝜃

0

2

0

𝜋

0

= ∫ ∫𝑧2

2|

0

𝑟sin𝜃

𝑟2 sin 𝜃 𝑑𝑟𝑑𝜃

2

0

𝜋

0

= ∫ ∫1


2

0

𝜋

0

=64

15

𝑀𝑦𝑧 = ∫ ∫ ∫ 𝑥𝛿𝑟𝑑𝑧𝑑𝑟𝑑𝜃

𝑟sin𝜃

0

2

0

𝜋

0

= ∫ ∫ ∫ 𝑧𝑟2 cos 𝜃 𝑑𝑧𝑑𝑟𝑑𝜃

𝑟sin𝜃

0

2

0

𝜋

0

= ∫ ∫𝑧2

2|

0

𝑟sin𝜃

𝑟2 cos 𝜃 𝑑𝑟𝑑𝜃

2

0

𝜋

0

= ∫ ∫1

2𝑟4 sin2 𝜃 cos 𝜃 𝑑𝑟𝑑𝜃

2

0

𝜋

0

=1

2[𝑟5

5]

0

2

[sin3 𝜃

3]

0

𝜋

=1

2(

32

5) (0 − 0) = 0

�̅� =𝑀𝑦𝑧

𝑀= 0

�̅� =𝑀𝑥𝑧

𝑀=

64/15

𝜋=

64

15𝜋


𝑀=

128/45

𝜋=

128

45𝜋

Spherical Coordinates

To integrate a continuous function 𝑓(𝜌, 𝜙, 𝜃) over a region given by

𝜌1(𝜙, 𝜃) ≤ 𝜌 ≤ 𝜌2(𝜙, 𝜃)

𝜙1(𝜃) ≤ 𝜙 ≤ 𝜙2(𝜃)




𝜃1 ≤ 𝜃 ≤ 𝜃2

The integral will be

∫ ∫ ∫ 𝑓(𝜌, 𝜙, 𝜃)𝜌2 sin 𝜙 𝑑𝜌𝑑𝜙𝑑𝜃

𝜌2(𝜙,𝜃)

𝜌1(𝜙,𝜃)

𝜙2(𝜃)

𝜙1(𝜃)

𝜃2

𝜃1

Ex: Find the volume of the upper region cut from the solid sphere 1 by the cone =/3.

Sol.:

Limits:

0 ≤ 𝜌 ≤ 1

0 ≤ 𝜙 ≤ 𝜋/3

0 ≤ 𝜃 ≤ 2𝜋

𝑉 = ∫ ∫ ∫ 𝜌2 sin 𝜙 𝑑𝜌𝑑𝜙𝑑𝜃

1

0

𝜋/3

0

2𝜋

0

= [𝜌3

3]

0

1

[− cos 𝜙]0𝜋/3[𝜃]0

2𝜋 = (1

3) (−

1

2+ 1) (2𝜋) =

𝜋

3

Ex: find the moment of inertia about the z-axis of the region in example above.

Sol.:

𝐼𝑧 = ∭(𝑥2 + 𝑦2)𝑑𝑉 = ∭ 𝑟2𝑑𝑉 = ∭(𝜌 sin 𝜙)2𝑑𝑉

𝐼𝑧 = ∫ ∫ ∫(𝜌 sin 𝜙)2𝜌2 sin 𝜙 𝑑𝜌𝑑𝜙𝑑𝜃

1

0

𝜋/3

0

2𝜋

0

𝐼𝑧 = ∫ ∫ ∫ 𝜌4 sin3 𝜙 𝑑𝜌𝑑𝜙𝑑𝜃

1

0

𝜋/3

0

2𝜋

0

= ∫ ∫ ∫ 𝜌4(1 − cos2 𝜙) sin 𝜙 𝑑𝜌𝑑𝜙𝑑𝜃

1

0

𝜋/3

0

2𝜋

0

= [𝜌4

5]

0

1

[− cos 𝜙 +cos3 𝜙

3]

0

𝜋/3

[𝜃]02𝜋

y

z /3

=1




= (1

5) (−

1

2+

1/8

3+ 1 −

1

3) (2𝜋) = (

2𝜋

5) (

1

2−

7

24) = (

2𝜋

5) (

5

24) =

𝜋

12

Exercises: (14.6)

16- Convert the integral

∫ ∫ ∫(𝑥2 + 𝑦2)𝑑𝑧𝑑𝑥𝑑𝑦

𝑥

0

√1−𝑦2

0

1

−1

into cylindrical coordinates and evaluate the result.

Sol.:

= ∫ ∫ ∫ (𝑟2)𝑟𝑑𝑧𝑑𝑟𝑑𝜃

𝑟cos𝜃

0

1

0

𝜋/2

−𝜋/2

= ∫ ∫ ∫ 𝑟3𝑑𝑧𝑑𝑟𝑑𝜃

𝑟cos𝜃

0

1

0

𝜋/2

−𝜋/2

= ∫ ∫[𝑧]0𝑟cos𝜃𝑟3𝑑𝑟𝑑𝜃

1

0

𝜋/2

−𝜋/2

= ∫ ∫ 𝑟4cos𝜃𝑑𝑟𝑑𝜃

1

0

𝜋/2

−𝜋/2

= [𝑟5

5]

0

1

[sin 𝜃]−𝜋/2𝜋/2

= (1

5) (1 − (−1)) =

2

5

37- Find the volume of the smaller region cut from the sphere =2 by the plane z=1.

Sol.:

Convert the plane equation z=1 to spherical coordinates

𝑧 = 1 ⇒ 𝜌 cos 𝜙 = 1 ⇒ 𝜌 = 1/ cos 𝜙

Find the limit of as the intersection of the plane z=1 and the sphere

=2

𝑧 = 1 ⇒ 𝜌 cos 𝜙 = 1 ⇒ 2 cos 𝜙 = 1 ⇒ cos 𝜙 =1

2⇒ 𝜙 = 𝜋/3

𝑉 = ∫ ∫ ∫ 𝜌2 sin 𝜙 𝑑𝜌𝑑𝜙𝑑𝜃

2

1/ cos 𝜙

𝜋/3

0

2𝜋

0

x

y 𝑥 = √1 − 𝑦2

r=1

y

z /3

=2

z=1




= ∫ ∫ [𝜌3

3]

1/ cos 𝜙

2

sin 𝜙 𝑑𝜙𝑑𝜃

𝜋/3

0

2𝜋

0

=1

3∫ ∫ (8 sin 𝜙 −

sin 𝜙

cos3 𝜙) 𝑑𝜙𝑑𝜃

𝜋/3

0

2𝜋

0

=1

3[−8 cos 𝜙 −

1

2cos−2 𝜙]

0

𝜋/3

[𝜃]02𝜋 =

1

3(−8 (

1

2− 1) −

1

2((

1

2)

−2

− 1)) (2𝜋)

=1

3(4 −

1

2(4 − 1)) (2𝜋) =

2𝜋

3(

5

2) =

5𝜋

3

40- A conical hole is drilled inside the solid hemisphere, the cone equation is =/3. Find the center

of mass, 2, z0

Sol.:

From symmetry �̅� = 0, �̅� = 0

𝑀 = ∭ 𝛿𝑑𝑉 = ∫ ∫ ∫(1)𝜌2 sin 𝜙 𝑑𝜌𝑑𝜙𝑑𝜃

2

0

𝜋/2

𝜋/3

2𝜋

0

= [𝜌3

3]

0

2

[− cos 𝜙]𝜋/3𝜋/2[𝜃]0

2𝜋

= (8

3) (0 +

1

2) (2𝜋) =

8𝜋

3

𝑀𝑥𝑦 = ∭ 𝑧𝛿𝑑𝑉 = ∫ ∫ ∫ (𝜌 cos 𝜙)(1)𝜌2 sin 𝜙 𝑑𝜌𝑑𝜙𝑑𝜃

2𝜋

0

𝜋/2

𝜋/3

2𝜋

0

= ∫ ∫ ∫ 𝜌3 sin 𝜙 cos 𝜙 𝑑𝜌𝑑𝜙𝑑𝜃

2𝜋

0

𝜋/2

𝜋/3

2𝜋

0

= [𝜌4

4]

0

2

[sin2 𝜙

2]

𝜋/3

𝜋/2

[𝜃]02𝜋 = (

16

4) (

1 − 3/4

2) (2𝜋) = 𝜋


𝑀=

𝜋

8𝜋/3=

3

8

Center of mass at (0, 0, 3/8)

y

z /3

Documents

Chapter 7 Multiple Integral (Chapter 14) Double Integral