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