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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 ๐(๐ฅ, ๐ฆ) is defined on the rectangular region given by
R: ๐ โค ๐ฅ โค ๐, ๐ โค ๐ฆ โค ๐
Then we can write
โฌ ๐(๐ฅ, ๐ฆ)๐๐ด
๐
= โซ โซ ๐(๐ฅ, ๐ฆ)๐๐ฆ๐๐ฅ
๐
๐
๐
๐
= โซ โซ ๐(๐ฅ, ๐ฆ)๐๐ฅ๐๐ฆ
๐
๐
๐
๐
The double integral represents the volume under the surface ๐ง = ๐(๐ฅ, ๐ฆ) within the region R
Ex: Evaluate the integral
โฌ ๐(๐ฅ, ๐ฆ)๐๐ด
๐
Where ๐(๐ฅ, ๐ฆ) = 1 โ 6๐ฅ2๐ฆ and R: 0 โค ๐ฅ โค 2, โ1 โค ๐ฆ โค 1
Sol.
โฌ ๐(๐ฅ, ๐ฆ)๐๐ด
๐
= โซ โซ(1 โ 6๐ฅ2๐ฆ)๐๐ฅ๐๐ฆ
2
0
1
โ1
= โซ [๐ฅ โ 2๐ฅ3๐ฆ]02๐๐ฆ
โ1
โ1
= โซ(2 โ 16๐ฆ)๐๐ฆ
1
โ1
= 2๐ฆ โ 8๐ฆ2|โ11 = 2 โ 8 โ (โ2 โ 8) = 4
This integral can be evaluated in the order of integration reversed
โฌ ๐(๐ฅ, ๐ฆ)๐๐ด
๐
= โซ โซ(1 โ 6๐ฅ2๐ฆ)๐๐ฆ๐๐ฅ
1
โ1
2
0
= โซ[๐ฆ โ 3๐ฅ2๐ฆ2]โ11 ๐๐ฅ
2
0
= โซ((1 โ 3๐ฅ2) โ (โ1 โ 3๐ฅ2))๐๐ฅ
2
0
= โซ(2)๐๐ฅ
2
0
= 2๐ฅ|02 = 4
Double Integral Over Bounded Nonrectangular Regions
โฌ ๐(๐ฅ, ๐ฆ)๐๐ด
๐
= โซ โซ ๐(๐ฅ, ๐ฆ)๐๐ฆ๐๐ฅ
๐2(๐ฅ)
๐1(๐ฅ)
๐
๐
= โซ โซ ๐(๐ฅ, ๐ฆ)๐๐ฅ๐๐ฆ
โ2(๐ฆ)
โ1(๐ฆ)
๐
๐
University of Baghdad 2nd Class
College of Engineering Math II
Electronics and Communications Engineering Dpt. Chapter 7
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.
๐ = โซ โซ(3 โ ๐ฅ โ ๐ฆ)๐๐ฆ๐๐ฅ
๐ฅ
0
1
0
= โซ [3๐ฆ โ ๐ฅ๐ฆ โ๐ฆ2
2]
0
๐ฅ
๐๐ฅ
1
0
= โซ (3๐ฅ โ ๐ฅ2 โ๐ฅ2
2) ๐๐ฅ
1
0
= โซ (3๐ฅ โ3
2๐ฅ2) ๐๐ฅ
1
0
= [3
2๐ฅ โ
1
2๐ฅ3]
0
1
=3
2โ
1
2= 1
By reversing the order of integration
๐ = โซ โซ(3 โ ๐ฅ โ ๐ฆ)๐๐ฅ๐๐ฆ
1
๐ฆ
1
0
= โซ [3๐ฅ โ๐ฅ2
2โ ๐ฅ๐ฆ]
๐ฆ
1
๐๐ฆ
1
0
= โซ ((3 โ1
2โ ๐ฆ) โ (3๐ฆ โ
๐ฆ2
2โ ๐ฆ2)) ๐๐ฆ
1
0
= โซ (5
2โ 4๐ฆ +
3
2๐ฆ2) ๐๐ฆ
1
0
= [5
2๐ฆ โ 2๐ฆ2 +
1
2๐ฆ3]
0
1
=5
2โ 2 +
1
2= 1
Ex: Find
โฌsin ๐ฅ
๐ฅ๐๐ด
๐
Where R is the triangular region in the xy-plane bounded by the lines y=x, x=1 and x-axis
Sol.:
โซ โซsin ๐ฅ
๐ฅ๐๐ฆ๐๐ฅ
๐ฅ
0
= โซ [๐ฆsin ๐ฅ
๐ฅ]
0
๐ฅ1
0
1
0
= โซ sin ๐ฅ
1
0
= โ cos ๐ฅ|01 = 1 โ cos 1
University of Baghdad 2nd Class
College of Engineering Math II
Electronics and Communications Engineering Dpt. Chapter 7
Ex: Find the value of the integration
โซ โซ (4๐ฅ + 2)๐๐ฆ๐๐ฅ
2๐ฅ
๐ฅ2
2
0
With the order of the integration reversed
Sol.:
First it is better to find the points of intersection of the curves.
๐ฅ2 = 2๐ฅ โ ๐ฅ2 โ 2๐ฅ = 0
๐ฅ(๐ฅ โ 2) = 0
๐ฅ = 0 โ ๐ฆ = 0
๐ฅ = 2 โ ๐ฆ = 4
๐ฆ = ๐ฅ2 โ ๐ฅ = โ๐ฆ
๐ฆ = 2๐ฅ โ ๐ฅ = ๐ฆ/2
โซ โซ (4๐ฅ + 2)๐๐ฆ๐๐ฅ
2๐ฅ
๐ฅ2
2
0
= โซ โซ (4๐ฅ + 2)๐๐ฅ๐๐ฆ
โ๐ฆ
๐ฆ/2
4
0
= โซ[2๐ฅ2 + 2๐ฅ]๐ฆ/2โ๐ฆ
4
0
= โซ (2๐ฆ + 2โ๐ฆ โ๐ฆ2
2โ ๐ฆ) ๐๐ฆ
4
0
= โซ (๐ฆ + 2โ๐ฆ โ๐ฆ2
2) ๐๐ฆ
4
0
= [๐ฆ2
2+
4
3๐ฆ3/2 โ
๐ฆ3
6]
0
4
= 8 +32
3โ
32
3โ 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
๐ด = โฌ ๐๐ด
๐
= โฌ ๐๐ฆ๐๐ฅ
๐
= โฌ ๐๐ฅ๐๐ฆ
๐
Ex: Find the area of the region R bounded by the parabola ๐ฆ = ๐ฅ2 and the line y=x+2
Sol.:
First we find the points of intersections
๐ฅ2 = ๐ฅ + 2
๐ฅ2 โ ๐ฅ โ 2 = 0
(๐ฅ + 1)(๐ฅ โ 2) = 0
University of Baghdad 2nd Class
College of Engineering Math II
Electronics and Communications Engineering Dpt. Chapter 7
๐ฅ = โ1 โ ๐ฆ = 1
๐ฅ = 2 โ ๐ฆ = 4
๐ด = โซ โซ ๐๐ฆ๐๐ฅ
๐ฅ+2
๐ฅ2
2
โ1
= โซ[๐ฆ]๐ฅ2๐ฅ+2๐๐ฅ
2
โ1
= โซ(๐ฅ + 2 โ ๐ฅ2)๐๐ฅ
2
โ1
= [๐ฅ2
2+ 2๐ฅ โ
๐ฅ3
3]
โ1
2
= (2 + 4 โ8
3) โ (
1
2โ 2 +
1
3) = 8 โ 3 โ
1
2=
9
2
First and Second Moments and Center of Mass
Density: ๐ฟ(๐ฅ, ๐ฆ)
Mass: ๐ = โฌ ๐ฟ(๐ฅ, ๐ฆ)๐๐ด
First Moments:
Moment about x-axis: ๐๐ฅ = โฌ ๐ฆ๐ฟ(๐ฅ, ๐ฆ)๐๐ด
Moment about y-axis: ๐๐ฆ = โฌ ๐ฅ๐ฟ(๐ฅ, ๐ฆ)๐๐ด
Center of Mass: ๏ฟฝฬ ๏ฟฝ =๐๐ฆ
๐, ๏ฟฝฬ ๏ฟฝ =
๐๐ฅ
๐
Moments of Inertia (Second Moments)
About x-axis: ๐ผ๐ฅ = โฌ ๐ฆ2๐ฟ(๐ฅ, ๐ฆ)๐๐ด
About y-axis: ๐ผ๐ฆ = โฌ ๐ฅ2๐ฟ(๐ฅ, ๐ฆ)๐๐ด
About origin: ๐ผ๐ = ๐ผ๐ฅ + ๐ผ๐ฆ = โฌ(๐ฅ2 + ๐ฆ2)๐ฟ(๐ฅ, ๐ฆ)๐๐ด
Centroids of Geometric Figures
When the density of an object is constant, it cancels out of the numerator and denominator of the
formulas for ๏ฟฝฬ ๏ฟฝ and ๏ฟฝฬ ๏ฟฝ. So, we can take =1 in calculating ๏ฟฝฬ ๏ฟฝ and ๏ฟฝฬ ๏ฟฝ.
Ex: Find the centroid of the region in the first quadrant that is bounded by the line y=x and the curve
๐ฆ = ๐ฅ2.
Sol.:
๐ = โซ โซ(1)๐๐ฆ๐๐ฅ
๐ฅ
๐ฅ2
1
0
= โซ[๐ฆ]๐ฅ2๐ฅ
1
0
๐๐ฅ = โซ(๐ฅ โ ๐ฅ2)
1
0
๐๐ฅ = [๐ฅ2
2โ
๐ฅ3
3]
0
1
=1
2โ
1
3=
1
6
University of Baghdad 2nd Class
College of Engineering Math II
Electronics and Communications Engineering Dpt. Chapter 7
๐๐ฅ = โซ โซ(1)๐ฆ๐๐ฆ๐๐ฅ
๐ฅ
๐ฅ2
1
0
= โซ [๐ฆ2
2]
๐ฅ2
๐ฅ1
0
๐๐ฅ =1
2โซ(๐ฅ2 โ ๐ฅ4)
1
0
๐๐ฅ =1
2[๐ฅ3
3โ
๐ฅ5
5]
0
1
=1
2(
1
3โ
1
5) =
1
15
๐๐ฆ = โซ โซ(1)๐ฅ๐๐ฆ๐๐ฅ
๐ฅ
๐ฅ2
1
0
= โซ[๐ฅ๐ฆ]๐ฅ2๐ฅ
1
0
๐๐ฅ = โซ(๐ฅ2 โ ๐ฅ3)
1
0
๐๐ฅ = [๐ฅ3
3โ
๐ฅ4
4]
0
1
= (1
3โ
1
4) =
1
12
๏ฟฝฬ ๏ฟฝ =๐๐ฆ
๐=
1/12
1/6=
1
2
๏ฟฝฬ ๏ฟฝ =๐๐ฅ
๐=
1/15
1/6=
2
5
Ex: Evaluate the integral
โซ โซ cos(๐ฆ3) ๐๐ฆ๐๐ฅ
2
โ๐ฅ
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
๐ฆ = โ๐ฅ (๐ฅ = ๐ฆ2) and y=2, the intersection point is (2,4), the integral in the reversed order will be
โซ โซ cos(๐ฆ3) ๐๐ฅ๐๐ฆ
๐ฆ2
0
2
0
= โซ[๐ฅ cos(๐ฆ3)]0๐ฆ2
๐๐ฆ
2
0
= โซ ๐ฆ2cos(๐ฆ3) ๐๐ฆ
2
0
=1
3sin(๐ฆ3)|
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 ๐ฅ = ๐ฆ โ ๐ฆ2 and the line ๐ฅ + ๐ฆ = 0
Sol.
๐ฅ + ๐ฆ = 0 โ ๐ฅ = โ๐ฆ
โ๐ฆ = ๐ฆ โ ๐ฆ2
2๐ฆ โ ๐ฆ2 = 0 โ ๐ฆ(2 โ ๐ฆ) = 0
University of Baghdad 2nd Class
College of Engineering Math II
Electronics and Communications Engineering Dpt. Chapter 7
๐ฆ = 0 โ ๐ฅ = 0
๐ฆ = 2 โ ๐ฅ = โ2
๐ด = โฌ ๐๐ด = โซ โซ ๐๐ฅ๐๐ฆ
๐ฆโ๐ฆ2
โ๐ฆ
2
0
๐ด = โซ(๐ฆ โ ๐ฆ2 โ (โ๐ฆ))๐๐ฆ
2
0
= โซ(2๐ฆ โ ๐ฆ2)๐๐ฆ
2
0
= ๐ฆ2 โ๐ฆ3
3|
0
2
= 4 โ8
3โ 0 =
4
3
36- Find the centroid of the region in the xy-plane bounded by the curves =1
โ1โ๐ฅ2 , ๐ฆ =
โ1
โ1โ๐ฅ2 and
the lines x=0 and x=1.
Sol.:
From symmetry ๏ฟฝฬ ๏ฟฝ = 0
๐ = โฌ(1)๐๐ด
๐ = โซ โซ ๐๐ฆ๐๐ฅ
1/โ1โ๐ฅ2
โ1/โ1โ๐ฅ2
1
0
= โซ (1
โ1 โ ๐ฅ2โ
โ1
โ1 โ ๐ฅ2)
1
0
๐๐ฅ
= โซ (2๐๐ฅ
โ1 โ ๐ฅ2)
1
0
= 2sinโ1(๐ฅ)|01
2 (๐
2= 0) = ๐
๐๐ฆ = โซ โซ ๐ฅ๐๐ฆ๐๐ฅ
1/โ1โ๐ฅ2
โ1/โ1โ๐ฅ2
1
0
= โซ[๐ฅ๐ฆ]โ1/โ1โ๐ฅ2
1/โ1โ๐ฅ2
1
0
๐๐ฅ
= โซ (2๐ฅ
โ1 โ ๐ฅ2)
1
0
๐๐ฅ = โโ1 โ ๐ฅ2
1/2|
0
1
= โ2(0 โ 1) = 2
๏ฟฝฬ ๏ฟฝ =2
๐
University of Baghdad 2nd Class
College of Engineering Math II
Electronics and Communications Engineering Dpt. Chapter 7
Double Integral in Polar Form
To integrate a function in polar form ๐(๐, ๐) over the shaded region shown
โฌ ๐(๐, ๐)๐๐ด
๐
= โซ โซ ๐(๐, ๐)๐๐๐๐๐
1+cos ๐
1
๐/2
โ๐/2
Area in Polar Coordinates
๐ด๐๐๐ ๐๐ ๐ = โฌ ๐๐ด
๐
= โฌ ๐๐๐๐๐
๐
Ex: Find the area enclosed by the curve ๐2 = 4 cos 2๐
Sol.:
๐ด
4= โซ โซ ๐๐๐๐๐
โ4 cos 2๐
0
๐/4
0
๐ด = 4 โซ [๐2
2]
0
โ4 cos 2๐
๐๐
๐/4
0
= 4 โซ 2 cos 2๐ ๐๐
๐/4
0
= 4[sin(2๐)]0๐/4
= 4(1 โ 0) = 4
Changing Cartesian Integration into Polar Integration
โฌ ๐(๐ฅ, ๐ฆ)๐๐ฅ๐๐ฆ
๐
= โฌ ๐(๐cos(๐), ๐sin(๐))๐๐๐๐๐
๐บ
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 ๐ฅ2 + ๐ฆ2 = 1 in the first quadrant.
Sol.:
In Cartesian coordinates
๐ผ๐ = โซ โซ (๐ฅ2 + ๐ฆ2)๐๐ฆ๐๐ฅ
โ1โ๐ฅ2
0
1
0
University of Baghdad 2nd Class
College of Engineering Math II
Electronics and Communications Engineering Dpt. Chapter 7
โซ [๐ฅ2๐ฆ +๐ฆ3
3]
0
โ1โ๐ฅ2
๐๐ฅ = โซ (๐ฅ2โ1 โ ๐ฅ2 +1
3(1 โ ๐ฅ2)3/2) ๐๐ฅ
1
0
1
0
In polar coordinates
๐ผ๐ = โซ โซ (๐ฅ2 + ๐ฆ2)๐๐ฆ๐๐ฅ
โ1โ๐ฅ2
0
1
0
= โซ โซ(๐2)๐๐๐๐๐
1
0
๐/2
0
= โซ โซ(๐3)๐๐๐๐
1
0
๐/2
0
= โซ๐4
4|
0
1
๐๐
๐/2
0
= โซ1
4๐๐
๐/2
0
=๐
4|
0
๐/2
=๐
8
Ex: Evaluate the integral (the Gaussian integral)
โซ ๐โ๐ฅ2๐๐ฅ
โ
โโ
Sol.:
๐ผ = โซ ๐โ๐ฅ2๐๐ฅ
โ
โโ
๐ผ = โซ ๐โ๐ฆ2๐๐ฆ
โ
โโ
๐ผ ยท ๐ผ = [ โซ ๐โ๐ฅ2๐๐ฅ
โ
โโ
] [ โซ ๐โ๐ฆ2๐๐ฆ
โ
โโ
]
๐ผ2 = โซ โซ ๐โ(๐ฅ2+๐ฆ2)๐๐ฅ๐๐ฆ
โ
โโ
โ
โโ
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
๐ผ2 = โซ โซ ๐โ๐2๐๐๐๐๐
โ
0
2๐
0
= โซ โ1
2๐โ๐2
|0
โ2๐
0
๐๐ = โ1
2โซ (0 โ 1)๐๐
2๐
0
=1
2(2๐) = ๐
๐ผ = โ๐
โซ ๐โ๐ฅ2๐๐ฅ
โ
โโ
= โ๐
University of Baghdad 2nd Class
College of Engineering Math II
Electronics and Communications Engineering Dpt. Chapter 7
Triple Integral in Rectangular Coordinates
The volume of a closed bounded region D in space is
๐๐๐๐ข๐๐ ๐๐ ๐ท = โญ ๐๐
๐ท
= โญ ๐๐ง๐๐ฆ๐๐ฅ
๐ท
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 ๐ง = 4 โ ๐ฅ2 โ ๐ฆ
Sol.:
In the xy-plane z=0 ๐ฆ = 4 โ ๐ฅ2
0 โค ๐ง โค 4 โ ๐ฅ2 โ ๐ฆ
0 โค ๐ฆ โค 4 โ ๐ฅ2
0 โค ๐ฅ โค 2
๐ = โซ โซ โซ ๐๐ง๐๐ฆ๐๐ฅ
4โ๐ฅ2โ๐ฆ
0
4โ๐ฅ2
0
2
0
๐ = โซ โซ ๐ง|04โ๐ฅ2โ๐ฆ
๐๐ฆ๐๐ฅ
4โ๐ฅ2
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 ๐ง = 4 โ
๐ฅ2 โ ๐ฆ, 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
University of Baghdad 2nd Class
College of Engineering Math II
Electronics and Communications Engineering Dpt. Chapter 7
๐ = โซ โซ (4 โ ๐ฅ2 โ ๐ฆ)๐๐ฆ๐๐ฅ
4โ๐ฅ2
0
2
0
๐ = โซ [(4 โ ๐ฅ2)๐ฆ โ๐ฆ2
2]
0
4โ๐ฅ2
๐๐ฅ
2
0
= โซ ((4 โ ๐ฅ2)2 โ(4 โ ๐ฅ2)2
2) ๐๐ฅ
2
0
=1
2โซ((4 โ ๐ฅ2)2)๐๐ฅ
2
0
๐ =1
2โซ(16 โ 8๐ฅ2 + ๐ฅ4)๐๐ฅ
2
0
=1
2โซ [16๐ฅ โ
8
3๐ฅ3 +
1
5๐ฅ5] ๐๐ฅ
2
0
=1
2(32 โ
64
3+
32
5) =
128
15
Masses and Moments in Three Dimensions
Mass:
๐ = โญ ๐ฟ๐๐
๐
(๐ฟ = ๐๐๐๐ ๐๐ก๐ฆ)
First Moments:
๐๐ฆ๐ง = โญ ๐ฅ๐ฟ๐๐
๐
๐๐ฅ๐ง = โญ ๐ฆ๐ฟ๐๐
๐
๐๐ฅ๐ฆ = โญ ๐ง๐ฟ๐๐
๐
Center of Mass
๏ฟฝฬ ๏ฟฝ =๐๐ฆ๐ง
๐, ๏ฟฝฬ ๏ฟฝ =
๐๐ฅ๐ง
๐, ๐งฬ =
๐๐ฅ๐ฆ
๐
Moments of Inertia
๐ผ๐ฅ = โญ(๐ฆ2 + ๐ง2)๐ฟ๐๐
๐
๐ผ๐ฆ = โญ(๐ฅ2 + ๐ง2)๐ฟ๐๐
๐
๐ผ๐ง = โญ(๐ฅ2 + ๐ฆ2)๐ฟ๐๐
๐
University of Baghdad 2nd Class
College of Engineering Math II
Electronics and Communications Engineering Dpt. Chapter 7
Triple Integral in Cylindrical and Spherical Coordinates
Cylindrical Coordinates
To integrate a continuous function f(r,,z) over a region given by
๐ง1(๐, ๐) โค ๐ง โค ๐ง2(๐, ๐)
๐1(๐) โค ๐ โค ๐2(๐)
๐1 โค ๐ โค ๐2
โซ โซ โซ ๐(๐, ๐, ๐ง)๐๐ง๐๐๐๐๐
๐ง2(๐,๐)
๐ง1(๐,๐)
๐2(๐)
๐1(๐)
๐2
๐1
Ex: Find the volume of the solid bounded by the cylinder ๐ฅ2 + ๐ฆ2 = 4 and the planes y+z=4 and
z=0.
Sol.:
๐ = โซ โซ โซ ๐๐๐ง๐๐๐๐
4โ๐ sin ๐
0
2
0
2๐
0
๐ = โซ โซ๐๐ง|04โ๐ sin ๐๐๐๐๐๐
2
0
2๐
0
๐ = โซ โซ(4๐ โ ๐2 sin ๐)๐๐๐๐
2
0
2๐
0
๐ = โซ [2๐2 โ1
3๐3 sin ๐]
0
2
๐๐
2๐
0
๐ = โซ (8 โ8
3sin ๐) ๐๐
2๐
0
= 8๐ +8
3cos ๐|
0
2๐
= 8(2๐ โ 1) +8
3(1 โ 1) = 16๐
Ex: Find the centroid of the solid enclosed by the cylinder ๐ฅ2 + ๐ฆ2 = 4, bounded above by the
paraboloid ๐ง = ๐ฅ2 + ๐ฆ2 and below by the xy-plane.
Sol.:
๐ง = ๐ฅ2 + ๐ฆ2 = ๐2
๐ฅ2 + ๐ฆ2 = 4 = ๐2 โ ๐ = 2
0 โค ๐ง โค ๐2
2 2
y
z
x2+y2=4
y=2
x=0
z=4y
x2+y2=4
r=2
x
y
University of Baghdad 2nd Class
College of Engineering Math II
Electronics and Communications Engineering Dpt. Chapter 7
0 โค ๐ โค 2
0 โค ๐ โค 2๐
The centroid lies on the axis of symmetry in this case the z-axis
๏ฟฝฬ ๏ฟฝ = 0, ๏ฟฝฬ ๏ฟฝ = 0
๐ = โซ โซ โซ (1)๐๐๐ง๐๐๐๐
๐2
0
2
0
2๐
0
= โซ โซ[๐๐ง]0๐2
๐๐๐๐
2
0
2๐
0
= โซ โซ ๐3๐๐๐๐
2
0
2๐
0
= [๐4
4]
0
2
[๐]02๐ = (4)(2๐) = 8๐
๐๐ฅ๐ฆ = โซ โซ โซ ๐ง(1)๐๐๐ง๐๐๐๐
๐2
0
2
0
2๐
0
= โซ โซ [๐๐ง2
2]
0
๐2
๐๐๐๐
2
0
2๐
0
= โซ โซ๐5
2๐๐๐๐
2
0
2๐
0
=1
2[๐6
6]
0
2
[๐]02๐ =
1
2(
32
3) (2๐) =
32๐
3
๐งฬ =๐๐ฅ๐ฆ
๐=
32๐/3
8๐=
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 ๐ฅ2 + ๐ฆ2 = 4.
Sol.:
๐ = โญ ๐ฟ๐๐
= โซ โซ โซ ๐ง๐๐๐ง๐๐๐๐
๐sin๐
0
2
0
๐
0
= โซ โซ๐ง2๐
2|
0
๐sin๐
๐๐๐๐
2
0
๐
0
= โซ โซ1
2๐3 sin2 ๐ ๐๐๐๐
2
0
๐
0
= โซ โซ1
4๐3(1 โ cos2๐)๐๐๐๐
2
0
๐
0
=1
4[๐4
4]
0
2
[๐ โ1
2sin2๐]
0
๐
=1
4(
16
4โ 0) (๐ โ 0) = ๐
๐๐ฅ๐ฆ = โซ โซ โซ ๐ง๐ฟ๐๐๐ง๐๐๐๐
๐sin๐
0
2
0
๐
0
= โซ โซ โซ ๐ง2๐๐๐ง๐๐๐๐
๐sin๐
0
2
0
๐
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
University of Baghdad 2nd Class
College of Engineering Math II
Electronics and Communications Engineering Dpt. Chapter 7
= โซ โซ๐ง3
3๐|
0
๐sin๐
๐๐๐๐
2
0
๐
0
= โซ โซ1
3๐4 sin3 ๐ ๐๐๐๐
2
0
๐
0
= โซ โซ1
3๐4 sin ๐ (1 โ cos2 ๐) ๐๐๐๐
2
0
๐
0
=1
3[๐5
5]
0
2
[โcos ๐ +cos3 ๐
3]
0
๐
=1
3(
32
5) (1 โ
1
3+ 1 โ
1
3)
=32
15(
4
3) =
128
45
๐๐ฅ๐ง = โซ โซ โซ ๐ฆ๐ฟ๐๐๐ง๐๐๐๐
๐sin๐
0
2
0
๐
0
= โซ โซ โซ ๐ง๐2 sin ๐ ๐๐ง๐๐๐๐
๐sin๐
0
2
0
๐
0
= โซ โซ๐ง2
2|
0
๐sin๐
๐2 sin ๐ ๐๐๐๐
2
0
๐
0
= โซ โซ1
2๐4 sin3 ๐ ๐๐๐๐
2
0
๐
0
=64
15
๐๐ฆ๐ง = โซ โซ โซ ๐ฅ๐ฟ๐๐๐ง๐๐๐๐
๐sin๐
0
2
0
๐
0
= โซ โซ โซ ๐ง๐2 cos ๐ ๐๐ง๐๐๐๐
๐sin๐
0
2
0
๐
0
= โซ โซ๐ง2
2|
0
๐sin๐
๐2 cos ๐ ๐๐๐๐
2
0
๐
0
= โซ โซ1
2๐4 sin2 ๐ cos ๐ ๐๐๐๐
2
0
๐
0
=1
2[๐5
5]
0
2
[sin3 ๐
3]
0
๐
=1
2(
32
5) (0 โ 0) = 0
๏ฟฝฬ ๏ฟฝ =๐๐ฆ๐ง
๐= 0
๏ฟฝฬ ๏ฟฝ =๐๐ฅ๐ง
๐=
64/15
๐=
64
15๐
๐งฬ =๐๐ฅ๐ฆ
๐=
128/45
๐=
128
45๐
Spherical Coordinates
To integrate a continuous function ๐(๐, ๐, ๐) over a region given by
๐1(๐, ๐) โค ๐ โค ๐2(๐, ๐)
๐1(๐) โค ๐ โค ๐2(๐)
University of Baghdad 2nd Class
College of Engineering Math II
Electronics and Communications Engineering Dpt. Chapter 7
๐1 โค ๐ โค ๐2
The integral will be
โซ โซ โซ ๐(๐, ๐, ๐)๐2 sin ๐ ๐๐๐๐๐๐
๐2(๐,๐)
๐1(๐,๐)
๐2(๐)
๐1(๐)
๐2
๐1
Ex: Find the volume of the upper region cut from the solid sphere 1 by the cone =/3.
Sol.:
Limits:
0 โค ๐ โค 1
0 โค ๐ โค ๐/3
0 โค ๐ โค 2๐
๐ = โซ โซ โซ ๐2 sin ๐ ๐๐๐๐๐๐
1
0
๐/3
0
2๐
0
= [๐3
3]
0
1
[โ cos ๐]0๐/3[๐]0
2๐ = (1
3) (โ
1
2+ 1) (2๐) =
๐
3
Ex: find the moment of inertia about the z-axis of the region in example above.
Sol.:
๐ผ๐ง = โญ(๐ฅ2 + ๐ฆ2)๐๐ = โญ ๐2๐๐ = โญ(๐ sin ๐)2๐๐
๐ผ๐ง = โซ โซ โซ(๐ sin ๐)2๐2 sin ๐ ๐๐๐๐๐๐
1
0
๐/3
0
2๐
0
๐ผ๐ง = โซ โซ โซ ๐4 sin3 ๐ ๐๐๐๐๐๐
1
0
๐/3
0
2๐
0
= โซ โซ โซ ๐4(1 โ cos2 ๐) sin ๐ ๐๐๐๐๐๐
1
0
๐/3
0
2๐
0
= [๐4
5]
0
1
[โ cos ๐ +cos3 ๐
3]
0
๐/3
[๐]02๐
y
z /3
=1
University of Baghdad 2nd Class
College of Engineering Math II
Electronics and Communications Engineering Dpt. Chapter 7
= (1
5) (โ
1
2+
1/8
3+ 1 โ
1
3) (2๐) = (
2๐
5) (
1
2โ
7
24) = (
2๐
5) (
5
24) =
๐
12
Exercises: (14.6)
16- Convert the integral
โซ โซ โซ(๐ฅ2 + ๐ฆ2)๐๐ง๐๐ฅ๐๐ฆ
๐ฅ
0
โ1โ๐ฆ2
0
1
โ1
into cylindrical coordinates and evaluate the result.
Sol.:
= โซ โซ โซ (๐2)๐๐๐ง๐๐๐๐
๐cos๐
0
1
0
๐/2
โ๐/2
= โซ โซ โซ ๐3๐๐ง๐๐๐๐
๐cos๐
0
1
0
๐/2
โ๐/2
= โซ โซ[๐ง]0๐cos๐๐3๐๐๐๐
1
0
๐/2
โ๐/2
= โซ โซ ๐4cos๐๐๐๐๐
1
0
๐/2
โ๐/2
= [๐5
5]
0
1
[sin ๐]โ๐/2๐/2
= (1
5) (1 โ (โ1)) =
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
๐ง = 1 โ ๐ cos ๐ = 1 โ ๐ = 1/ cos ๐
Find the limit of as the intersection of the plane z=1 and the sphere
=2
๐ง = 1 โ ๐ cos ๐ = 1 โ 2 cos ๐ = 1 โ cos ๐ =1
2โ ๐ = ๐/3
๐ = โซ โซ โซ ๐2 sin ๐ ๐๐๐๐๐๐
2
1/ cos ๐
๐/3
0
2๐
0
x
y ๐ฅ = โ1 โ ๐ฆ2
r=1
y
z /3
=2
z=1
University of Baghdad 2nd Class
College of Engineering Math II
Electronics and Communications Engineering Dpt. Chapter 7
= โซ โซ [๐3
3]
1/ cos ๐
2
sin ๐ ๐๐๐๐
๐/3
0
2๐
0
=1
3โซ โซ (8 sin ๐ โ
sin ๐
cos3 ๐) ๐๐๐๐
๐/3
0
2๐
0
=1
3[โ8 cos ๐ โ
1
2cosโ2 ๐]
0
๐/3
[๐]02๐ =
1
3(โ8 (
1
2โ 1) โ
1
2((
1
2)
โ2
โ 1)) (2๐)
=1
3(4 โ
1
2(4 โ 1)) (2๐) =
2๐
3(
5
2) =
5๐
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 ๏ฟฝฬ ๏ฟฝ = 0, ๏ฟฝฬ ๏ฟฝ = 0
๐ = โญ ๐ฟ๐๐ = โซ โซ โซ(1)๐2 sin ๐ ๐๐๐๐๐๐
2
0
๐/2
๐/3
2๐
0
= [๐3
3]
0
2
[โ cos ๐]๐/3๐/2[๐]0
2๐
= (8
3) (0 +
1
2) (2๐) =
8๐
3
๐๐ฅ๐ฆ = โญ ๐ง๐ฟ๐๐ = โซ โซ โซ (๐ cos ๐)(1)๐2 sin ๐ ๐๐๐๐๐๐
2๐
0
๐/2
๐/3
2๐
0
= โซ โซ โซ ๐3 sin ๐ cos ๐ ๐๐๐๐๐๐
2๐
0
๐/2
๐/3
2๐
0
= [๐4
4]
0
2
[sin2 ๐
2]
๐/3
๐/2
[๐]02๐ = (
16
4) (
1 โ 3/4
2) (2๐) = ๐
๐งฬ =๐๐ฅ๐ฆ
๐=
๐
8๐/3=
3
8
Center of mass at (0, 0, 3/8)
y
z /3