MATH 2263 Spring 2012
Quiz 7(Mar. 29, 2012)
Name:
Student ID #:
Section #: 12 / 14
? Please show your work.
1. (5 points) Please evaluate the integral
x − 2y
dA
R 3x − y
by making an appropriate change of variables. Here R is the parallelogram encolosed by the lines
x − 2y = 0, x − 2y = 4, 3x − y = 1, and 3x − y = 9.
Answer: Set
ZZ
u = x − 2y, v = 3x − y.
(1)
Then we have
2
3
1
1
x = − u + v, y = − u + v,
5
5
5
5
and hence the Jacobian
∂(x, y) =
∂(u, v) ∂x
∂u
∂y
∂u
∂x
∂v
∂y
∂v
1
−
= 53
−
5
2
5
1
5
1
= .
5
We also know that the region S on uv-plane that corresponds to R is
S = (u, v) ∈ R2 : 0 ≤ u ≤ 4, 1 ≤ v ≤ 9 .
Hence we have
ZZ
R
u ∂(x, y) dA
v ∂(u, v) Z ZS
u
=
dA
S 5v
Z 9Z 4
u
=
du dy
1
0 5v
Z 9
8
=
dy
5v
1
8
= (ln 9 − ln 1)
5
16 ln 3
.
=
5
x − 2y
dA =
3x − y
ZZ
1
2
2. (5 points) Please Evaluate
R
C
F · dr, where
F(x, y, z) = (x + y) i + (y + z) j + (x + z) k,
C : r(t) = t i + 2t j + t2 k, 0 ≤ t ≤ 1.
Answer: Along the curve C, we have
F = ht + 2t, 2t + t2 , t + t2 i = h3t, t2 + 2t, t2 + ti
r0 = h1, 2, 2ti.
Hence
Z
Z
1
h3t, t2 + 2t, t2 + ti · h1, 2, 2ti dt
F · dr =
C
0
Z
1
3t + 2t2 + 4t + 2t3 + 2t2 dt
=
0
Z
=
1
2t3 + 4t2 + 7t dt
0
1 4 4 3 7 2
t + t + t
2
3
2
16
= .
3
1
=
0