Solutions11 - Math 601 Solutions to Homework 11 1 Consider...

Math 601 Solutions to Homework 11
1. Consider the curve given by the following parametric equations:
x = cos( t ) and y = sin(2 t )
Let D be the region inside this curve and to the right of the y -axis:
(a) Use Green’s Theorem to compute
ZZ
D
x
2
dA .
(b) Use Green’s Theorem to find the area of the region D .
Answer:
(a) We need to find a vector field F such that rot( F ) = x
2
. The
vector field F = - yx 2 i will work (as would
1
3
x 3 j or lots of other
possibilities). Then, by Green’s Theorem:
ZZ
D
x
2
dA =
I
∂D
- yx
2
dx
From the parametrization of the curve, we have:
x = cos t
y = sin(2 t )
dx = - sin t dt
dy = 2 cos(2 t ) dt
The bounds for t are -
π
2
≤ t ≤
π
2
(we can see this by noting
that x and y are both 0 when t is a multiple of π/ 2, and then
by plotting a few points of the parametric curve to make sure we
choose the right-side of the lemniscate).
We can now compute the integral (recall that sin(2 t ) = 2 sin t cos t ):
ZZ
D
x
2
dA =
I
∂D
- yx
2
dx
=
Z
π/ 2
- π/ 2
- sin(2 t ) cos
2
t ( - sin t ) dt
=
Z
π/ 2
- π/ 2
2 sin
2
t cos
3
t dt
=
Z
π/ 2
- π/ 2
2 sin
2
t (1 - sin
2
t ) cos t dt

We use the substitution u = sin t , du = cos t dt :
ZZ
D
x
2
dA =
Z
1
- 1
2 u
2
(1 - u
2
) du
=
•
2
3
u
3
-
2
5
u
5
‚
1
- 1
=
8
15
Note: If we had used a different vector field, we would have
obtained the same answer at the end.
(b) To compute the area we need to evaluate the integral
Z
D
dA , so
we need to find a vector field F such that rot( F ) = 1. The vector
field F = - y i will work (as would x j or lots of other possibilities).
Then, by Green’s Theorem:
ZZ
D
dA =
I
∂D
- y dx
As in part (a), we have:
x = cos t
y = sin(2 t )
dx = - sin t dt
dy = 2 cos(2 t ) dt
with -
π
2
≤ t ≤
π
2
.
We can now compute the integral (recall that sin(2 t ) = 2 sin t cos t ):
ZZ
D
dA =
I
∂D
- y dx
=
Z
π/ 2
- π/ 2
- sin(2 t )( - sin t ) dt
=
Z
π/ 2
- π/ 2
2 sin
2
t cos t dt
We use the substitution u = sin t , du = cos t dt :
ZZ
D
dA =
Z
1
- 1
2 u
2
du
=
•
2
3
u
3
‚
1
- 1
=
4
3

2. Consider the surface integral
ZZ
S
z dA
where S is the upper half ( z ≥ 0) of the sphere x 2 + y 2 + z 2 = 1.
(a) Set up this integral using a parametrization where t = x and u = y .
(b) Set up this integral using a parametrization where t = r and u = θ .
(c) Set up this integral using a parametrization where t = θ and u = φ .
Answer:
(a) With t = x and u = y , we have the parametric equations:
x = t
y = u
z =
√
1 - t 2 - u 2
-
√
1 - u
2
≤ t ≤
√
1 - u
2
- 1 ≤ u ≤ 1
Then, the tangent vectors to the surface are
T
t
= 1 , 0 ,
- t
√
1 - t
2
- u
2
¶
T
u
= 0 , 1 ,
- u
√
1 - t 2 - u 2
¶
And, the normal vector to the surface is
T
t
× T
u
=
fl
fl
fl
fl
fl
fl
fl
fl
i j k
1 0
- t
√
1 - t
2
- u
2
0 1
- u
√
1 - t
2
- u
2
fl
fl
fl
fl
fl
fl
fl
fl
=
t
√
1 - t
2
- u
2
¶
i +
u
√
1 - t
2
- u
2
¶
j + k
Thus:
dA = k T
t
× T
u
k dt du =
s
t
2
1 - t
2
- u
2
+
u
2
1 - t
2
- u
2
+ 1 dt du
=
r
1
1 - t 2 - u 2
dt du
Thus:
ZZ
S
z dA =
Z
1
- 1
Z √
1 - u 2
-
√
1 - u 2
√
1 - t
2
- u
2
k T
t
× T
u
k dt du
=
Z
1
- 1
Z √
1 - u 2
-
√
1 - u 2
√
1 - t
2
- u
2
r
1
1 - t
2
- u
2
dt du
=
Z
1
- 1
Z
√
1 - u 2
-
√
1 - u 2
dt du

(b) With t = r and u = θ , we have the parametric equations:
x = t cos u
y = t sin u
z =
√
1 - t 2
0 ≤ t ≤ 1
0 ≤ u ≤ 2 π
Then, the tangent vectors to the surface are
T
t
= cos u, sin u,
- t
√
1 - t 2
¶
T
u
= ( - t sin u, t cos u, 0)
And, the normal vector to the surface is
T
t
× T
u
=
fl
fl
fl
fl
fl
fl
fl
fl
i j k
cos u sin u - t
√
1 - t 2
- t sin u t cos u 0
fl
fl
fl
fl
fl
fl
fl
fl
=
t 2 cos u
√
1 - t
2
¶
i +
t 2 sin u
√
1 - t
2
¶
j + t k
Thus:
dA = k T
t
× T
u
k dt du =
s
t 4 cos 2 u
1 - t 2
+
t 4 sin
2
u
1 - t 2
+ t 2 dt du
=
s
t 2
1 - t
2
dt du
Thus:
ZZ
S
z dA =
Z
2 π
0
Z
1
0
√
1 - t
2
k T
t
× T
u
k dt du
=
Z
2 π
0
Z
1
0
√
1 - t 2
s
t 2
1 - t 2
dt du
=
Z
2 π
0
Z
1
0
t dt du
(c) Recall that x = ρ sin φ cos θ , y = ρ sin φ sin θ and z = ρ cos φ . On
the sphere, we have ρ = 1. Thus, if we use the parameters t = θ
and u = φ , we have the parametric equations:
x = sin u cos t
y = sin u sin t
z = cos u
0 ≤ t ≤ 2 π
0 ≤ u ≤
π
2

Then, the tangent vectors to the surface are
T
t
= ( - sin u sin t, sin u cos t, 0)
T
u
= (cos u cos t, cos u sin t, - sin u )
And, the normal vector to the surface is
T
t
× T
u
=
fl
fl
fl
fl
fl
fl
fl
fl
i j k
- sin u sin t sin u cos t 0
cos u cos t cos u sin t - sin u
fl
fl
fl
fl
fl
fl
fl
fl
= - sin
2
u cos t i - sin
2
u sin t j - sin u cos u k
Thus:
dA = k T
t
× T
u
k dt du =
p
sin
4
u cos 2 t + sin
4
u sin
2
t + sin
2
u cos 2 u dt du
=
p
sin 4 u + sin 2 u cos
u
dt du
=
p
sin
2
u dt du = sin u dt du
Thus:
ZZ
S
z dA =
Z
π/ 2
0
Z
2 π
0
cos u k T
t
× T
u
k dt du
=
Z
π/ 2
0
Z
2 π
0
cos u sin u dt du
=
Z
π/ 2
0
Z
2 π
0
cos u sin u dt du