Math 243
Instructor: Longfei Li
Name:
Homework IV
Due on Feb 3, 2014
Suggested Practices (Not Collected):
Section 15.4: 1–6 7,9,11, 15, 21, 27, 39
Section 15.6: 1, 3, 5, 9
Section 15.7: 3, 7, 9, 11, 21, 27, 31, 39
Section 15.8: 1, 3, 5, 9, 11, 15, 19, 21, 23
Section 15.9: 1, 3, 5, 9, 11, 13, 17, 19, 21, 27
1. Use a double integral to find the area of the region inside the circle (x − 1)2 + y 2 = 1 and outside
the circle x2 + y 2 = 1
Solution: In polar coordinates, we have
x = r cos θ, y = r sin θ
So the equation for the circle
(x − 1)2 + y 2 = 1 ⇒ x2 + y 2 = 2x ⇒ r2 = 2r cos θ ⇒ r = 2 cos θ
And the equation for the circle x2 + y 2 = 1 is, in polar coordinates, r = 1. The intersection
happens when 2 cos θ = 1 ⇒ θ = ± π3 . So we have
−
π
π
≤ θ ≤ , 1 ≤ r ≤ 2 cos θ
3
3
So
Z
A(D) =
π
3
− π3
Z
2 cos θ
Z
r drdθ =
1
π
3
− π3
√
Z π
Z π
3
3 1
1 2 2 cos θ
1
π
3
2
r dθ =
2 cos θ− dθ =
+cos 2θ dθ = +
π
π 2
2 1
2
3
2
−
−
3
1
3
2. Use polar coordinates to find the volume of the solid above the cone z =
the sphere x2 + y 2 + z 2 = 1
p
x2 + y 2 and below
Solution:
The cone intersects the sphere at the curve
1
2
x2 + y 2 =
which is a circle with radius
√1 .
2
Use polar coordinates by letting
x = r cos θ, y = r sin θ
then we have 0 ≤ r ≤ √12 and 0 ≤ θ ≤ 2π. And the equation for the cone and sphere in polar
coordinates are, respectively,
p
z = r, z = 1 − r2
Therefore, the volume is
Z
V =
0
√1
2
Z
2π
Z
p
2
1 − r − r r dθdr = 2π
0
0
2
√1
2
r
p
√
π
1 − r2 − r2 dr = (2 − 2)
3
3. Evaluate the iterated integral by converting to polar coordinates.
√
3
Z
Z
9−x2
sin(x2 + y 2 )dydx
(a)
−3 0
Z 2 Z √2x−x2
(b)
0
p
x2 + y 2 dydx
0
Solution:
(a) From the integral, we know that
0≤y≤
p
9 − x2 ,
−3≤x≤3
⇒ x2 + y 2 ≤ 9, y ≥ 0 and − 3 ≤ x ≤ 3
Therefore, the region of the integral is an upper half circle with radius 3. Converting to polar
coordinates, we have x = r cos θ, y = r sin θ and
0 ≤ r ≤ 3, 0 ≤ θ ≤ π
Hence, the integral can be evaluated as
Z
3
√
Z
9−x2
2
Z
2
3Z π
sin(x + y )dydx =
−3
0
0
0
3 π(1 − cos 9)
1
sin(r2 )r dθdr = −π cos(r2 ) =
2
2
0
(b)
Z
0
2Z
0
√
2x−x2
Z
p
x2 + y 2 dydx =
0
π
2
Z
2 cos θ
r2 drdθ =
0
Z
0
3
π
2
π
8
8
1
16
2
cos3 θ dθ = (sin θ − sin3 θ) =
3
3
3
9
0
4. Find the surface area of the part of the hyperbolic paraboloid z = y 2 − x2 that lies between the
cylinders x2 + y 2 = 1 and x2 + y 2 = 4
Solution:
D = {(x, y)|1 ≤ x2 + y 2 ≤ 4}
And the surface area is
A(S) =
ZZ p
1+
4x2
+
4y 2 dA
=
D
= 2π
ZZZ
2 Z 2π
Z
1
1
(1 + 4r2 )3/2
12
2
1
p
1 + 4r2 r dθdr
0
√
√
π
= (17 17 − 5 5)
6
xydV , where E is bounded by the parabolic cylinders y = x2 and x = y 2 and the
5. Evaluate
E
planes z = 0 and z = x + y
Solution:
The region E can be represented as
0 ≤ z ≤ x + y, x2 ≤ y ≤
√
x, 0 ≤ x ≤ 1
So the integral is
ZZZ
Z
1Z
√
x Z x+y
Z
√
x
xy dzdydx =
x2
0
x2
0
√ 0 Z
Z 1
1
1 2 2 1 3 y= x
=
x y + xy 2 dx =
2
3
y=x
0
0
3
=
28
xydV =
E
1Z
4
xy(x + y) dydx
1 3 1 5/2 1 6 1 7
x + x − x − x dx
2
3
2
3
6. Use triple integral to find the volume of the solid enclosed by the cylinder x2 + z 2 = 4 and the
planes y = −1 and y + z = 4
Solution:
The solid E can be represented as:
E : −1 ≤ y ≤ 4 − z, (x, y) ∈ D
where D is the projection of E onto the xz−plane, which is a circle of radius 2. So in polar
coordinates:
x = r cos θ, z = r sin θ
So
D : 0 ≤ θ ≤ 2π, 0 ≤ r ≤ 2
and
−1 ≤ y ≤ 4 − r sin θ
So
ZZZ
Z
V =
2 Z 2π
Z
4−r sin θ
1dV =
V
Z
rdydθdr
0
0
−1
2 Z 2π
r(5 − r sin θ)dθdr
=
0
Z
0
2
=
10πrdr
0
= 20π
5
7. Find the mass and the center of the mass of the solid S bounded by the paraboloid z = 4x2 + 4y 2
and the plane z = a(a > 0) if S has a constant density K. (Use cylindrical coordinates)
Solution:
The intersection of the parabola z = 4x2 + 4y 2 and the plane z = a(a > 0) is
a = 4x2 + 4y 2 ⇒ x2 + y 2 =
a
4
The solid S can be represented as
x2 + y 2 ≤
a
, 4x2 + 4y 2 ≤ z ≤ a
4
So, in cylindrical coordinates, x = r cos θ, y = r sin θ, z = z and
√
a
0≤r≤
, 0 ≤ θ ≤ 2π, 4r2 ≤ z ≤ a
2
Therefore, the mass is found
ZZZ
Z
M=
K dV =
E
√
a
2
0
√
Z
2π
Z
a
Z
Kr dzdθdr = 2π
0
4r2
a
2
Kr(a − 4r2 ) dr =
0
By Symmetry and constant density, we have
Myz = Mxz = 0
For the moment on the xy−plane, we have
ZZZ
Z
Mxy =
Kz dV =
E
√
a
2
0
Z
2π
Z
a
Krz dzdθdr =
0
4r2
Hence, the center of mass is
(x̄, ȳ, z̄) = (0, 0,
6
2a
)
3
πa3 K
12
πa2 K
8
8. Evaluate
ZZZ p
x2 + y 2 dV , where E is the region that lies inside the cylinder x2 + y 2 = 16 and
E
between the planes z = −5 and z = 4. (Use cylindrical coordinates)
Solution:
For cylindrical coordinates:
x = r cos θ, y = r sin θ, z = z
and the region E for this problem can be represented as
−5 ≤ z ≤ 4, 0 ≤ r ≤ 4, 0 ≤ θ ≤ 2π
So
4
ZZZ p
Z
2
2
x + y dV =
xex
9. Evaluate
2 +y 2 +z 2
4 Z 2π
r · r dθdrdz = 384π
−5
E
ZZZ
Z
0
0
dV , where E is the portion of the unit ball x2 + y 2 + z 2 ≤ 1 that lies
E
in the first octant.
Solution:
The region is the part of the ball x2 + y 2 + z 2 ≤ 1 in the first octant. If we use polar coordinates,
then
x = ρ sin φ cos θ, y = ρ sin φ sin θ, z = ρ cos φ
and
0 ≤ ρ ≤ 1, 0 ≤ θ ≤
π
π
, 0≤φ≤
2
2
Hence, the integral can be evaluated as
ZZZ
xe
x2 +y 2 +z 2
Z
π
2
1Z
dV =
E
0
0
Z
=
1
Z
0
ρ2
2
ρ sin φ cos θeρ ρ2 sin φ dφdθdρ
! Z π
Z π
2
ρ3 e dρ
0
0
=
π
2
π
8
7
2
cos θ dθ
0
!
sin2 φ dφ
10. Evaluate the integral by changing to spherical coordinates
Z Z √
Z √
1−x2
1
2−x2 −y 2
√
0
0
xydzdydx
x2 +y 2
Solution:
p
The region E for this integral is the region above the cone z = x2 + y 2 and below the sphere
x2 + y 2 + z 2 = 2 in the first octant. So in spherical coordinates, we have
0≤ρ≤
√
And the integral becomes
Z Z √
Z √
1−x2
1
0
√
2−x2 −y 2
√
0
π
π
, 0≤φ≤
2
4
2, 0 ≤ θ ≤
Z
π
2
2Z
xydzdydx =
x2 +y 2
0
√
Z
0
Z
0
2
√
4 2−5
=
15
8
ρ sin φ cos θρ sin φ sin θρ2 sin φ dφdθdρ
0
ρ4 dρ
=
π
4
! Z
0
π
2
! Z
cos θ sin θ dθ
0
π
4
!
sin3 φ dφ