2022 Fall MAS102 Calculus 2
Quiz 7, 10AM
1. Use the transformation x = 2uv, y = u2 + v2 to evaluate the integral
8 points
Z
6
Z
1.5
x2
36
+9
1
p
5x
3
y 2 − x2
dydx.
(Hint: Draw the region bounded by v = 3u, v = 3 and uv = 3/4 and show its relevance
with the problem.)
Solution. Let T be the region bounded by v = 3u, v = 3 and uv = 3/4 on the uv-plane.
v
v = 3u
3
T
1.5
2uv = x
uv =
0
0.25 0.5
3
4
1
u
Note that each curve 2uv = x intersects T if and only if 1.5 ≤ x ≤ 6. (+2 points) Moreover,
along each curve 2uv = x restricted to the region T , the value of y = u2 + v 2 is strictly increasing
in v. Its minimum is achieved at the intersection of 2uv = x and v = 3u, which has the solution
r
r
x
3x
5x
u=
, v=
, u2 + v 2 =
.
6
2
3
Meanwhile, its maximum is achieved at v = 3 and u = x/6, where the value of y is x2 /36 + 9.
In summary, the transformation is a 1-1 mapping between T and the desired xy-region of the
integration. (+2 points)
Also observe that
y 2 − x2 = (u4 + 2u2 v 2 + v 4 ) − 4u2 v 2 = u4 − 2u2 v 2 + v 4 = (v 2 − u2 )2 .
Given this, we observe
Z 6 Z x2 /36+9
0
5x/3
1
p
dydx =
y 2 − x2
ZZ
T
1
|J| dudv, (+2 points)
|v 2 − u2 |
where J is the Jacobian of the transformation. We calculate
|J| =
∂x
∂u
∂y
∂u
Hence, the desired integral is
ZZ
Z 3 Z v/3
Z
4 dudv =
4 dudv = 4·
T
1.5
3/4v
∂x
∂v
∂y
∂v
3
1.5
=
3
v
−
3 4v
2v
2u
2u
2v
dv = 4
1
= 4|v 2 − u2 |.
v2
3
− ln v
6
4
3
=
1.5
9
−3 ln 2.(+2 points)
2
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2022 Fall MAS102 Calculus 2
Quiz 7, 10AM
2. Consider the curve C parametrized by r(t) = (r cos t, r sin t, t) for t ∈ [0, 2π]. Calculate
4+3 the center of mass and the moment Iz of inertia about the z-axis for the coil lying
points over C with density δ(t) = 2t.
Solution. Since v(t) = r0 (t) = (−r sin t, r cos t, 1), ds = |v|dt =
mass is
Z
2π
Z
M=
2t ·
δds =
√
r2 + 1dt. (+1 point) So the
p
p
r2 + 1dt = 4π 2 r2 + 1.
(+1 point)
0
C
The coordinates of the center of mass are given by
Z
Z 2π
p
1
1
x̄ =
2rt cos t · r2 + 1dt = 0,
xδds =
M C
M 0
ȳ =
1
M
z̄ =
Hence the center of mass is
Z
yδds =
C
1
M
1
M
Z
zδds =
C
(0, − πr , 4π
3 ).
Z
2π
2rt sin t ·
p
0
1
M
2π
Z
2t2 ·
p
r
r2 + 1dt = − ,
π
r2 + 1dt =
0
(+2 points)
The moment of inertia about the z-axis is given by
Z
Iz =
(x2 + y 2 )δds
C
Z 2π
=
2r2 t ·
p
4π
.
3
r2 + 1dt = 4π 2 r2
p
r2 + 1.
(+2 points)
(+1 point)
0
• For the calculations of x̄, ȳ and z̄, (+1 point) is assigned for the correct definitions and
another (+1 point) is assigned for the correct calculations.
2
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