chapter 7 multiple integral (chapter 14) double integral
TRANSCRIPT
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 π(π₯, π¦) is defined on the rectangular region given by
R: π β€ π₯ β€ π, π β€ π¦ β€ π
Then we can write
β¬ π(π₯, π¦)ππ΄
π
= β« β« π(π₯, π¦)ππ¦ππ₯
π
π
π
π
= β« β« π(π₯, π¦)ππ₯ππ¦
π
π
π
π
The double integral represents the volume under the surface π§ = π(π₯, π¦) within the region R
Ex: Evaluate the integral
β¬ π(π₯, π¦)ππ΄
π
Where π(π₯, π¦) = 1 β 6π₯2π¦ and R: 0 β€ π₯ β€ 2, β1 β€ π¦ β€ 1
Sol.
β¬ π(π₯, π¦)ππ΄
π
= β« β«(1 β 6π₯2π¦)ππ₯ππ¦
2
0
1
β1
= β« [π₯ β 2π₯3π¦]02ππ¦
β1
β1
= β«(2 β 16π¦)ππ¦
1
β1
= 2π¦ β 8π¦2|β11 = 2 β 8 β (β2 β 8) = 4
This integral can be evaluated in the order of integration reversed
β¬ π(π₯, π¦)ππ΄
π
= β« β«(1 β 6π₯2π¦)ππ¦ππ₯
1
β1
2
0
= β«[π¦ β 3π₯2π¦2]β11 ππ₯
2
0
= β«((1 β 3π₯2) β (β1 β 3π₯2))ππ₯
2
0
= β«(2)ππ₯
2
0
= 2π₯|02 = 4
Double Integral Over Bounded Nonrectangular Regions
β¬ π(π₯, π¦)ππ΄
π
= β« β« π(π₯, π¦)ππ¦ππ₯
π2(π₯)
π1(π₯)
π
π
= β« β« π(π₯, π¦)ππ₯ππ¦
β2(π¦)
β1(π¦)
π
π
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.
π = β« β«(3 β π₯ β π¦)ππ¦ππ₯
π₯
0
1
0
= β« [3π¦ β π₯π¦ βπ¦2
2]
0
π₯
ππ₯
1
0
= β« (3π₯ β π₯2 βπ₯2
2) ππ₯
1
0
= β« (3π₯ β3
2π₯2) ππ₯
1
0
= [3
2π₯ β
1
2π₯3]
0
1
=3
2β
1
2= 1
By reversing the order of integration
π = β« β«(3 β π₯ β π¦)ππ₯ππ¦
1
π¦
1
0
= β« [3π₯ βπ₯2
2β π₯π¦]
π¦
1
ππ¦
1
0
= β« ((3 β1
2β π¦) β (3π¦ β
π¦2
2β π¦2)) ππ¦
1
0
= β« (5
2β 4π¦ +
3
2π¦2) ππ¦
1
0
= [5
2π¦ β 2π¦2 +
1
2π¦3]
0
1
=5
2β 2 +
1
2= 1
Ex: Find
β¬sin π₯
π₯ππ΄
π
Where R is the triangular region in the xy-plane bounded by the lines y=x, x=1 and x-axis
Sol.:
β« β«sin π₯
π₯ππ¦ππ₯
π₯
0
= β« [π¦sin π₯
π₯]
0
π₯1
0
1
0
= β« sin π₯
1
0
= β cos π₯|01 = 1 β 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
β« β« (4π₯ + 2)ππ¦ππ₯
2π₯
π₯2
2
0
With the order of the integration reversed
Sol.:
First it is better to find the points of intersection of the curves.
π₯2 = 2π₯ β π₯2 β 2π₯ = 0
π₯(π₯ β 2) = 0
π₯ = 0 β π¦ = 0
π₯ = 2 β π¦ = 4
π¦ = π₯2 β π₯ = βπ¦
π¦ = 2π₯ β π₯ = π¦/2
β« β« (4π₯ + 2)ππ¦ππ₯
2π₯
π₯2
2
0
= β« β« (4π₯ + 2)ππ₯ππ¦
βπ¦
π¦/2
4
0
= β«[2π₯2 + 2π₯]π¦/2βπ¦
4
0
= β« (2π¦ + 2βπ¦ βπ¦2
2β π¦) ππ¦
4
0
= β« (π¦ + 2βπ¦ βπ¦2
2) ππ¦
4
0
= [π¦2
2+
4
3π¦3/2 β
π¦3
6]
0
4
= 8 +32
3β
32
3β 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
π΄ = β¬ ππ΄
π
= β¬ ππ¦ππ₯
π
= β¬ ππ₯ππ¦
π
Ex: Find the area of the region R bounded by the parabola π¦ = π₯2 and the line y=x+2
Sol.:
First we find the points of intersections
π₯2 = π₯ + 2
π₯2 β π₯ β 2 = 0
(π₯ + 1)(π₯ β 2) = 0
University of Baghdad 2nd Class
College of Engineering Math II
Electronics and Communications Engineering Dpt. Chapter 7
π₯ = β1 β π¦ = 1
π₯ = 2 β π¦ = 4
π΄ = β« β« ππ¦ππ₯
π₯+2
π₯2
2
β1
= β«[π¦]π₯2π₯+2ππ₯
2
β1
= β«(π₯ + 2 β π₯2)ππ₯
2
β1
= [π₯2
2+ 2π₯ β
π₯3
3]
β1
2
= (2 + 4 β8
3) β (
1
2β 2 +
1
3) = 8 β 3 β
1
2=
9
2
First and Second Moments and Center of Mass
Density: πΏ(π₯, π¦)
Mass: π = β¬ πΏ(π₯, π¦)ππ΄
First Moments:
Moment about x-axis: ππ₯ = β¬ π¦πΏ(π₯, π¦)ππ΄
Moment about y-axis: ππ¦ = β¬ π₯πΏ(π₯, π¦)ππ΄
Center of Mass: οΏ½Μ οΏ½ =ππ¦
π, οΏ½Μ οΏ½ =
ππ₯
π
Moments of Inertia (Second Moments)
About x-axis: πΌπ₯ = β¬ π¦2πΏ(π₯, π¦)ππ΄
About y-axis: πΌπ¦ = β¬ π₯2πΏ(π₯, π¦)ππ΄
About origin: πΌπ = πΌπ₯ + πΌπ¦ = β¬(π₯2 + π¦2)πΏ(π₯, π¦)ππ΄
Centroids of Geometric Figures
When the density of an object is constant, it cancels out of the numerator and denominator of the
formulas for οΏ½Μ οΏ½ and οΏ½Μ οΏ½. So, we can take =1 in calculating οΏ½Μ οΏ½ and οΏ½Μ οΏ½.
Ex: Find the centroid of the region in the first quadrant that is bounded by the line y=x and the curve
π¦ = π₯2.
Sol.:
π = β« β«(1)ππ¦ππ₯
π₯
π₯2
1
0
= β«[π¦]π₯2π₯
1
0
ππ₯ = β«(π₯ β π₯2)
1
0
ππ₯ = [π₯2
2β
π₯3
3]
0
1
=1
2β
1
3=
1
6
University of Baghdad 2nd Class
College of Engineering Math II
Electronics and Communications Engineering Dpt. Chapter 7
ππ₯ = β« β«(1)π¦ππ¦ππ₯
π₯
π₯2
1
0
= β« [π¦2
2]
π₯2
π₯1
0
ππ₯ =1
2β«(π₯2 β π₯4)
1
0
ππ₯ =1
2[π₯3
3β
π₯5
5]
0
1
=1
2(
1
3β
1
5) =
1
15
ππ¦ = β« β«(1)π₯ππ¦ππ₯
π₯
π₯2
1
0
= β«[π₯π¦]π₯2π₯
1
0
ππ₯ = β«(π₯2 β π₯3)
1
0
ππ₯ = [π₯3
3β
π₯4
4]
0
1
= (1
3β
1
4) =
1
12
οΏ½Μ οΏ½ =ππ¦
π=
1/12
1/6=
1
2
οΏ½Μ οΏ½ =ππ₯
π=
1/15
1/6=
2
5
Ex: Evaluate the integral
β« β« cos(π¦3) ππ¦ππ₯
2
βπ₯
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
π¦ = βπ₯ (π₯ = π¦2) and y=2, the intersection point is (2,4), the integral in the reversed order will be
β« β« cos(π¦3) ππ₯ππ¦
π¦2
0
2
0
= β«[π₯ cos(π¦3)]0π¦2
ππ¦
2
0
= β« π¦2cos(π¦3) ππ¦
2
0
=1
3sin(π¦3)|
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 π₯ = π¦ β π¦2 and the line π₯ + π¦ = 0
Sol.
π₯ + π¦ = 0 β π₯ = βπ¦
βπ¦ = π¦ β π¦2
2π¦ β π¦2 = 0 β π¦(2 β π¦) = 0
University of Baghdad 2nd Class
College of Engineering Math II
Electronics and Communications Engineering Dpt. Chapter 7
π¦ = 0 β π₯ = 0
π¦ = 2 β π₯ = β2
π΄ = β¬ ππ΄ = β« β« ππ₯ππ¦
π¦βπ¦2
βπ¦
2
0
π΄ = β«(π¦ β π¦2 β (βπ¦))ππ¦
2
0
= β«(2π¦ β π¦2)ππ¦
2
0
= π¦2 βπ¦3
3|
0
2
= 4 β8
3β 0 =
4
3
36- Find the centroid of the region in the xy-plane bounded by the curves =1
β1βπ₯2 , π¦ =
β1
β1βπ₯2 and
the lines x=0 and x=1.
Sol.:
From symmetry οΏ½Μ οΏ½ = 0
π = β¬(1)ππ΄
π = β« β« ππ¦ππ₯
1/β1βπ₯2
β1/β1βπ₯2
1
0
= β« (1
β1 β π₯2β
β1
β1 β π₯2)
1
0
ππ₯
= β« (2ππ₯
β1 β π₯2)
1
0
= 2sinβ1(π₯)|01
2 (π
2= 0) = π
ππ¦ = β« β« π₯ππ¦ππ₯
1/β1βπ₯2
β1/β1βπ₯2
1
0
= β«[π₯π¦]β1/β1βπ₯2
1/β1βπ₯2
1
0
ππ₯
= β« (2π₯
β1 β π₯2)
1
0
ππ₯ = ββ1 β π₯2
1/2|
0
1
= β2(0 β 1) = 2
οΏ½Μ οΏ½ =2
π
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 π(π, π) over the shaded region shown
β¬ π(π, π)ππ΄
π
= β« β« π(π, π)πππππ
1+cos π
1
π/2
βπ/2
Area in Polar Coordinates
π΄πππ ππ π = β¬ ππ΄
π
= β¬ πππππ
π
Ex: Find the area enclosed by the curve π2 = 4 cos 2π
Sol.:
π΄
4= β« β« πππππ
β4 cos 2π
0
π/4
0
π΄ = 4 β« [π2
2]
0
β4 cos 2π
ππ
π/4
0
= 4 β« 2 cos 2π ππ
π/4
0
= 4[sin(2π)]0π/4
= 4(1 β 0) = 4
Changing Cartesian Integration into Polar Integration
β¬ π(π₯, π¦)ππ₯ππ¦
π
= β¬ π(πcos(π), πsin(π))πππππ
πΊ
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 π₯2 + π¦2 = 1 in the first quadrant.
Sol.:
In Cartesian coordinates
πΌπ = β« β« (π₯2 + π¦2)ππ¦ππ₯
β1βπ₯2
0
1
0
University of Baghdad 2nd Class
College of Engineering Math II
Electronics and Communications Engineering Dpt. Chapter 7
β« [π₯2π¦ +π¦3
3]
0
β1βπ₯2
ππ₯ = β« (π₯2β1 β π₯2 +1
3(1 β π₯2)3/2) ππ₯
1
0
1
0
In polar coordinates
πΌπ = β« β« (π₯2 + π¦2)ππ¦ππ₯
β1βπ₯2
0
1
0
= β« β«(π2)πππππ
1
0
π/2
0
= β« β«(π3)ππππ
1
0
π/2
0
= β«π4
4|
0
1
ππ
π/2
0
= β«1
4ππ
π/2
0
=π
4|
0
π/2
=π
8
Ex: Evaluate the integral (the Gaussian integral)
β« πβπ₯2ππ₯
β
ββ
Sol.:
πΌ = β« πβπ₯2ππ₯
β
ββ
πΌ = β« πβπ¦2ππ¦
β
ββ
πΌ Β· πΌ = [ β« πβπ₯2ππ₯
β
ββ
] [ β« πβπ¦2ππ¦
β
ββ
]
πΌ2 = β« β« πβ(π₯2+π¦2)ππ₯ππ¦
β
ββ
β
ββ
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
πΌ2 = β« β« πβπ2πππππ
β
0
2π
0
= β« β1
2πβπ2
|0
β2π
0
ππ = β1
2β« (0 β 1)ππ
2π
0
=1
2(2π) = π
πΌ = βπ
β« πβπ₯2ππ₯
β
ββ
= βπ
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
ππππ’ππ ππ π· = β ππ
π·
= β ππ§ππ¦ππ₯
π·
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 π§ = 4 β π₯2 β π¦
Sol.:
In the xy-plane z=0 π¦ = 4 β π₯2
0 β€ π§ β€ 4 β π₯2 β π¦
0 β€ π¦ β€ 4 β π₯2
0 β€ π₯ β€ 2
π = β« β« β« ππ§ππ¦ππ₯
4βπ₯2βπ¦
0
4βπ₯2
0
2
0
π = β« β« π§|04βπ₯2βπ¦
ππ¦ππ₯
4βπ₯2
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 π§ = 4 β
π₯2 β π¦, 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
π = β« β« (4 β π₯2 β π¦)ππ¦ππ₯
4βπ₯2
0
2
0
π = β« [(4 β π₯2)π¦ βπ¦2
2]
0
4βπ₯2
ππ₯
2
0
= β« ((4 β π₯2)2 β(4 β π₯2)2
2) ππ₯
2
0
=1
2β«((4 β π₯2)2)ππ₯
2
0
π =1
2β«(16 β 8π₯2 + π₯4)ππ₯
2
0
=1
2β« [16π₯ β
8
3π₯3 +
1
5π₯5] ππ₯
2
0
=1
2(32 β
64
3+
32
5) =
128
15
Masses and Moments in Three Dimensions
Mass:
π = β πΏππ
π
(πΏ = ππππ ππ‘π¦)
First Moments:
ππ¦π§ = β π₯πΏππ
π
ππ₯π§ = β π¦πΏππ
π
ππ₯π¦ = β π§πΏππ
π
Center of Mass
οΏ½Μ οΏ½ =ππ¦π§
π, οΏ½Μ οΏ½ =
ππ₯π§
π, π§Μ =
ππ₯π¦
π
Moments of Inertia
πΌπ₯ = β(π¦2 + π§2)πΏππ
π
πΌπ¦ = β(π₯2 + π§2)πΏππ
π
πΌπ§ = β(π₯2 + π¦2)πΏππ
π
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
π§1(π, π) β€ π§ β€ π§2(π, π)
π1(π) β€ π β€ π2(π)
π1 β€ π β€ π2
β« β« β« π(π, π, π§)ππ§πππππ
π§2(π,π)
π§1(π,π)
π2(π)
π1(π)
π2
π1
Ex: Find the volume of the solid bounded by the cylinder π₯2 + π¦2 = 4 and the planes y+z=4 and
z=0.
Sol.:
π = β« β« β« πππ§ππππ
4βπ sin π
0
2
0
2π
0
π = β« β«ππ§|04βπ sin ππππππ
2
0
2π
0
π = β« β«(4π β π2 sin π)ππππ
2
0
2π
0
π = β« [2π2 β1
3π3 sin π]
0
2
ππ
2π
0
π = β« (8 β8
3sin π) ππ
2π
0
= 8π +8
3cos π|
0
2π
= 8(2π β 1) +8
3(1 β 1) = 16π
Ex: Find the centroid of the solid enclosed by the cylinder π₯2 + π¦2 = 4, bounded above by the
paraboloid π§ = π₯2 + π¦2 and below by the xy-plane.
Sol.:
π§ = π₯2 + π¦2 = π2
π₯2 + π¦2 = 4 = π2 β π = 2
0 β€ π§ β€ π2
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 β€ π β€ 2
0 β€ π β€ 2π
The centroid lies on the axis of symmetry in this case the z-axis
οΏ½Μ οΏ½ = 0, οΏ½Μ οΏ½ = 0
π = β« β« β« (1)πππ§ππππ
π2
0
2
0
2π
0
= β« β«[ππ§]0π2
ππππ
2
0
2π
0
= β« β« π3ππππ
2
0
2π
0
= [π4
4]
0
2
[π]02π = (4)(2π) = 8π
ππ₯π¦ = β« β« β« π§(1)πππ§ππππ
π2
0
2
0
2π
0
= β« β« [ππ§2
2]
0
π2
ππππ
2
0
2π
0
= β« β«π5
2ππππ
2
0
2π
0
=1
2[π6
6]
0
2
[π]02π =
1
2(
32
3) (2π) =
32π
3
π§Μ =ππ₯π¦
π=
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 π₯2 + π¦2 = 4.
Sol.:
π = β πΏππ
= β« β« β« π§πππ§ππππ
πsinπ
0
2
0
π
0
= β« β«π§2π
2|
0
πsinπ
ππππ
2
0
π
0
= β« β«1
2π3 sin2 π ππππ
2
0
π
0
= β« β«1
4π3(1 β cos2π)ππππ
2
0
π
0
=1
4[π4
4]
0
2
[π β1
2sin2π]
0
π
=1
4(
16
4β 0) (π β 0) = π
ππ₯π¦ = β« β« β« π§πΏπππ§ππππ
πsinπ
0
2
0
π
0
= β« β« β« π§2πππ§ππππ
πsinπ
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
University of Baghdad 2nd Class
College of Engineering Math II
Electronics and Communications Engineering Dpt. Chapter 7
= β« β«π§3
3π|
0
πsinπ
ππππ
2
0
π
0
= β« β«1
3π4 sin3 π ππππ
2
0
π
0
= β« β«1
3π4 sin π (1 β cos2 π) ππππ
2
0
π
0
=1
3[π5
5]
0
2
[βcos π +cos3 π
3]
0
π
=1
3(
32
5) (1 β
1
3+ 1 β
1
3)
=32
15(
4
3) =
128
45
ππ₯π§ = β« β« β« π¦πΏπππ§ππππ
πsinπ
0
2
0
π
0
= β« β« β« π§π2 sin π ππ§ππππ
πsinπ
0
2
0
π
0
= β« β«π§2
2|
0
πsinπ
π2 sin π ππππ
2
0
π
0
= β« β«1
2π4 sin3 π ππππ
2
0
π
0
=64
15
ππ¦π§ = β« β« β« π₯πΏπππ§ππππ
πsinπ
0
2
0
π
0
= β« β« β« π§π2 cos π ππ§ππππ
πsinπ
0
2
0
π
0
= β« β«π§2
2|
0
πsinπ
π2 cos π ππππ
2
0
π
0
= β« β«1
2π4 sin2 π cos π ππππ
2
0
π
0
=1
2[π5
5]
0
2
[sin3 π
3]
0
π
=1
2(
32
5) (0 β 0) = 0
οΏ½Μ οΏ½ =ππ¦π§
π= 0
οΏ½Μ οΏ½ =ππ₯π§
π=
64/15
π=
64
15π
π§Μ =ππ₯π¦
π=
128/45
π=
128
45π
Spherical Coordinates
To integrate a continuous function π(π, π, π) over a region given by
π1(π, π) β€ π β€ π2(π, π)
π1(π) β€ π β€ π2(π)
University of Baghdad 2nd Class
College of Engineering Math II
Electronics and Communications Engineering Dpt. Chapter 7
π1 β€ π β€ π2
The integral will be
β« β« β« π(π, π, π)π2 sin π ππππππ
π2(π,π)
π1(π,π)
π2(π)
π1(π)
π2
π1
Ex: Find the volume of the upper region cut from the solid sphere 1 by the cone =/3.
Sol.:
Limits:
0 β€ π β€ 1
0 β€ π β€ π/3
0 β€ π β€ 2π
π = β« β« β« π2 sin π ππππππ
1
0
π/3
0
2π
0
= [π3
3]
0
1
[β cos π]0π/3[π]0
2π = (1
3) (β
1
2+ 1) (2π) =
π
3
Ex: find the moment of inertia about the z-axis of the region in example above.
Sol.:
πΌπ§ = β(π₯2 + π¦2)ππ = β π2ππ = β(π sin π)2ππ
πΌπ§ = β« β« β«(π sin π)2π2 sin π ππππππ
1
0
π/3
0
2π
0
πΌπ§ = β« β« β« π4 sin3 π ππππππ
1
0
π/3
0
2π
0
= β« β« β« π4(1 β cos2 π) sin π ππππππ
1
0
π/3
0
2π
0
= [π4
5]
0
1
[β cos π +cos3 π
3]
0
π/3
[π]02π
y
z /3
=1
University of Baghdad 2nd Class
College of Engineering Math II
Electronics and Communications Engineering Dpt. Chapter 7
= (1
5) (β
1
2+
1/8
3+ 1 β
1
3) (2π) = (
2π
5) (
1
2β
7
24) = (
2π
5) (
5
24) =
π
12
Exercises: (14.6)
16- Convert the integral
β« β« β«(π₯2 + π¦2)ππ§ππ₯ππ¦
π₯
0
β1βπ¦2
0
1
β1
into cylindrical coordinates and evaluate the result.
Sol.:
= β« β« β« (π2)πππ§ππππ
πcosπ
0
1
0
π/2
βπ/2
= β« β« β« π3ππ§ππππ
πcosπ
0
1
0
π/2
βπ/2
= β« β«[π§]0πcosππ3ππππ
1
0
π/2
βπ/2
= β« β« π4cosπππππ
1
0
π/2
βπ/2
= [π5
5]
0
1
[sin π]βπ/2π/2
= (1
5) (1 β (β1)) =
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
π§ = 1 β π cos π = 1 β π = 1/ cos π
Find the limit of as the intersection of the plane z=1 and the sphere
=2
π§ = 1 β π cos π = 1 β 2 cos π = 1 β cos π =1
2β π = π/3
π = β« β« β« π2 sin π ππππππ
2
1/ cos π
π/3
0
2π
0
x
y π₯ = β1 β π¦2
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
= β« β« [π3
3]
1/ cos π
2
sin π ππππ
π/3
0
2π
0
=1
3β« β« (8 sin π β
sin π
cos3 π) ππππ
π/3
0
2π
0
=1
3[β8 cos π β
1
2cosβ2 π]
0
π/3
[π]02π =
1
3(β8 (
1
2β 1) β
1
2((
1
2)
β2
β 1)) (2π)
=1
3(4 β
1
2(4 β 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 οΏ½Μ οΏ½ = 0, οΏ½Μ οΏ½ = 0
π = β πΏππ = β« β« β«(1)π2 sin π ππππππ
2
0
π/2
π/3
2π
0
= [π3
3]
0
2
[β cos π]π/3π/2[π]0
2π
= (8
3) (0 +
1
2) (2π) =
8π
3
ππ₯π¦ = β π§πΏππ = β« β« β« (π cos π)(1)π2 sin π ππππππ
2π
0
π/2
π/3
2π
0
= β« β« β« π3 sin π cos π ππππππ
2π
0
π/2
π/3
2π
0
= [π4
4]
0
2
[sin2 π
2]
π/3
π/2
[π]02π = (
16
4) (
1 β 3/4
2) (2π) = π
π§Μ =ππ₯π¦
π=
π
8π/3=
3
8
Center of mass at (0, 0, 3/8)
y
z /3