Calculus Homework
Assignment 9
1. Find a parametrization of the portion of the sphere x2 + y 2 +
z 2 =p8 in the first octant between the xy-plane and the cone
z = x2 + y 2 .
[like §16.6 #6]
Sol.
Consider in cylindrical coordinate, then the sphere √has parametric equation as p
r(r, θ) = (r cos θ)i + (r sin θ)j ± 8 − r2 k.
Substituting z =
x2 + y 2 into x2 + y 2 + z 2 = 8, we get
x2 + y 2 = r2 = 4 ⇒ r = 2. Thus, the spherical cap in first
octant can be parametrized as
√
√
π
r(r, θ) = (r cos θ)i+(r sin θ)j+ 8 − r2 k, 2 ≤ r ≤ 2 2, 0 ≤ θ ≤
2
2. Use a parametrization to express the area of the portion of the
paraboloid z = 2(x2 + y 2 ) between the planes z = 2 and z = 8
as a double integral. Then evaluate the integral.
[like §16.6 #24]
Sol. Let x = r cos θ, y = r sin θ, then z = 2(x2 + y 2 ) = 2r2 .
Note that z = 2r2 = 2 ⇒ r = 1 and z = 2r2 = 8 ⇒ r = 2. Thus
we can parametrize the portion of the paraboloid z = 2(x2 +y 2 )
between the planes z = 2 and z = 8 as
r(r, θ) = (r cos θ)i + (r sin θ)j + 2r2 k, 1 ≤ r ≤ 2, 0 ≤ θ ≤ 2π
Further, we have that
rr = (cos θ)i + (sin θ)j + 4rk, rθ = (−r sin θ)i + (r cos θ)j
so
i
j
k sin θ 2r = (−4r2 cos θ)i+(−4r2 sin θ)j+rk
rr ×rθ = cos θ
−r sin θ r cos θ 0 Hence, the area A of the surface is
ˆ 2π
ˆ 2
ˆ 2π
ˆ 2p
A=
|rr × rθ |drdθ =
16r4 cos2 θ + 16r4 sin2 θ + r2 drdθ
0 1
0 1
ˆ 2√
ˆ 2π
ˆ 2π ˆ 2 √
2
dθ
=
16r + 1 rdrdθ =
16r2 + 1 rdr
0 1
0
1
ˆ 65
65
√
√
√
π
π
π 3
=
zdz = z 2 = (65 65 − 17 17)
16 17
24 17 24
3. Integrate G(x, y, z) = y 2 over the sphere x2 + y 2 + z 2 = 4.
[like §16.6 #29)]
Sol. Consider the spherical coordinate, then the sphere S :
x2 + y 2 + z 2 = 4 can be parametrized as
S : r(φ, θ) = (2 sin φ cos θ)i+(2 sin φ sin θ)j+(2 cos φ)k, 0 ≤ φ ≤ π, 0 ≤ θ ≤ 2π
Then
rφ = (2 cos φ cos θ)i + (2 cos φ sin θ)j − (2 sin φ)k
rθ = (−2 sin φ sin θ)i + (2 sin φ cos θ)j
and
i
j
k
rφ × rθ = 2 cos φ cos θ 2 cos φ sin θ −2 sin φ
−2 sin φ sin θ 2 sin φ cos θ
0
= (4 sin2 φ cos θ)i + (4 sin2 φ sin θ)j + (2 sin 2φ)k
q
⇒ |rφ × rθ | = 16 sin4 φ cos2 θ + 16 sin4 φ sin2 θ + 4 sin2 2φ = 4 sin φ
Therefore,
¨
¨
ˆ 2π
ˆ π
2
G(x, y, z)dσ =
y dσ =
4 sin2 φ sin2 θ|rφ × rθ |dφdθ
S
S
0 0
ˆ 2π
ˆ π
ˆ π
ˆ 2π
2
2
2
= 16
sin φ sin θ sin φdφdθ = 16
(1 − cos φ) sin φdφ
sin2 θdθ
0 0
0
0
ˆ −1
ˆ 2π
h
i
h
i2π 64π
3
−1
1 − cos 2θ
z
θ 1
2
= 16
(z − 1)dz
dθ = 16
−z
− sin 2θ
=
2
3
2 4
3
1
0
1
0
4. Use a parametrization to find the flux of F = xyi − zk outward
p (normal away form the z-axis) through the cone z =
2 x2 + y 2 , 0 ≤ z ≤ 2.
[like §16.6 #41]
p
Sol. Let x = r cos θ, y = r sin θ, then z = 2 x2 + y 2 = 2r.
Since 0 ≤ z ≤ 2 ⇒ 0 ≤ r ≤ 1. Thus the cone can be
parametrized as
S : r(r, θ) = (r cos θ)i + (r sin θ)j + 2rk, 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π
Note that
rr = (cos θ)i + (sin θ)j + 2k, rθ = (−r sin θ)i + (r cos θ)j
So the outward normal vector on S is
i
j
k
rθ × rr = −r sin θ r cos θ 0 = (2r cos θ)i + (2r sin θ)j − rk
cos θ
sin θ 2 Therefore, the flux of F outward through S is
ˆ 2π
ˆ 1
¨
rθ × rr
F·
F · ndσ =
|rθ × rr |drdθ
|rθ × rr |
0 0
S
ˆ 2π
ˆ 1
=
[(xyi − zk) · (2r cos θ)i + (2r sin θ)j − rk)]drdθ
0 0
ˆ 2π
ˆ 1
ˆ 2π
ˆ 1
=
(2xyr cos θ + zr)drdθ =
(2r3 cos2 θ sin θ + 2r2 )drdθ
0 0
0 0
ˆ 2π h
ˆ 2π i
1
1
1 4
2 3
2
2
2
=
r cos θ sin θ + r dθ =
cos θ sin θ +
dθ
2
3 0
2
3
0
0
h 1
2 i2π 4π
= − cos3 θ + θ
=
6
3 0
3
5. Use the surface integral in Stokes’ Theorem to calculate the
circulation of the field F = (y 2 + z 2 )i + (x2 + z 2 )j + (x2 + y 2 )k
around the curve C : The boundary of the triangle cut from
the plane x + y + z = 2 by the first octant, counterclockwise
when viewed from above.
[like §16.7 #4]
Sol. Note that the surface S enclosed by C has parametric
equations as
s(x, y) = xi + yj + (2 − x − y)k, 0 ≤ x ≤ 2, 0 ≤ y ≤ 2
And observe that
i
j
k
∂
∂
∂
curl F = ∇×F = ∂x
∂y
∂z
y 2 + z 2 x2 + z 2 x2 + y 2
= (2y−2z)i+(2z−2x)j+(2x−2y)k
√ . Hence the circulation of F
and clearly we have n = i+j+k
3
around C counterclockwise is
ffi
¨
F · dr =
curl F · ndσ
C
S
¨
¨
1
=√
[(2y − 2z) + (2z − 2x) + (2x − 2y)]dσ =
0dσ = 0
3 S
S
¨
∇ × (yi) · ndσ, where S is the hemisphere x2 + y 2 +
6. Evaluate
S
z 2 = 1, z ≥ 0.
[like §16.7 #10]
Sol. Let f (x, y, z) = x2 + y 2 + z 2 , then the hemisphere is the
level surface f (x, y, z) = 1, z ≥ 0. Then
p
∇f = 2xi + 2yj + 2zk ⇒ |∇f | = 2 x2 + y 2 + z 2 = 2
∇f
⇒n=
= xi + yj + zk
|∇f |
Also we have that
i
j
k
∂ ∂ ∂ ∇ × (yi) = ∂x ∂y ∂z = −k
y 0 0 p
Since z ≥ 0, so z = 1 − x2 − y 2 . Thus the hemisphere is
defined over the region R = {(x, y)x2 + y 2 ≤ 1}. By taking
p = k, then
¨
¨
|∇f |
∇ × (yi) · ndσ =
(−k) · (xi + yj + zk)
dA
|∇f · p|
S
R
¨
ˆ 2π h 2 i1
ˆ 2πˆ 1
1
r
z dA = −
=−
rdrdθ = −
dθ
2 0
R z
0
0 0
ˆ
1 2π
dθ = −π
=−
2 0
7. Use the surface integral in Stokes’ Theorem to calculate the flux
of the curl of the field F = x2 yi + 2y 3 zj + 3zk across the surface
S : r(r, θ) = (r cos θ)i + (r sin θ)j + rk, 0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π
in the direction of the outward unit normal n.
[like §16.7 #15]
Sol. Observe that
i
j
k ∂
∂
∂ 3
2
curl F = ∇ × F = ∂x
∂y
∂z = −2y i − x k
x2 y 2y 3 z 3z and rr = (cos θ)i + (sin θ)j + k, rθ = (−r sin θ)i + (r cos θ)j, so
the outward normal vector of S is
i
j
k
rθ × rr = −r sin θ r cos θ 0 = (r cos θ)i + (r sin θ)j − rk
cos θ
sin θ 1 Hence the flux of curl F across S is
¨
ˆ 2π
ˆ 2
rθ × rr
curl F · ndσ =
curl F ·
|rθ × rr |drdθ
|rθ × rr |
S
0 0
ˆ 2π
ˆ 2
=
[(−2y 3 i − x2 k) · ((r cos θ)i + (r sin θ)j − rk)]drdθ
0 0
ˆ 2π
ˆ 2
ˆ 2π
ˆ 2
3
2
=
(−2ry cos θ + rx )drdθ =
(r3 cos2 θ − 2r4 cos θ sin3 θ)drdθ
0 0
0 0
ˆ 2π h
ˆ 2π i2
1 3
2
64
2
5
3
=
r cos θ − r cos θ sin θ dθ =
4 cos2 θ −
cos θ sin3 θ dθ
4
5
5
0
0
0
ˆ 2π
ˆ 0
h
i
2π
1 + cos 2θ
64
1
1
=4
dθ −
z 3 dz = 4 θ + sin 2θ
= 4π
2
5 0
2
4
0
0
8. Use the Divergence Theorem to find the outward flux of F =
2xzi − xyj − z 2 k across the boundary of the wedge cut from the
first octant by the plane y + z = 5 and the elliptical cylinder
x2 + 2y 2 = 8.
[like §16.8 #11]
Sol.
Denote the boundary of the wedge D by S. Then the flux of F
across S is
˚ ˚
¨
∂
∂
∂
∇ · FdV =
F · ndσ =
(2xz) +
(−xy) + (−z 2 ) dV
∂y
∂z
D ∂x
D
S
ˆ 2√ˆ2 √4− 1 xˆ2 5−y
˚
2
(2z − x − 2z)dV = −
=
xdzdydx
D
0
0
0
ˆ 2√ˆ2 √4− 1 x2
ˆ 2√ˆ2 √4− 1 x2
5−y
2
2
=−
xz 0 dydx = −
(5x − xy)dydx
0
0
0
0
√
ˆ 2√2 h
1 2 i 4− 12 x2
dx
=−
5xy − xy
2
0
0
!
r
ˆ 2√2
1 3
x2
=−
−2x + x + 5x 4 −
dx
4
2
0
h
i2√2
1 4
5
68
2
2 32
= − − x + x − √ (8 − x )
=−
16
3
0
3 2