Math 52 - Autumn 2006 - Midterm Exam I
Problem 1.
Z2 Z2
a) (10 points) Sketch the region of integration of
0
(0,2)
cos(x2 ) dx dy
y
y=x
(2,0)
Z2 Z2
cos(x2 ) dx dy
b) (10 points) Evaluate
0
y
Solution: After the change of order of integration we get:
2
Z 2
Z2 Zx
1
sin 4
2
2
2 x · cos(x ) dx = sin(x ) =
cos(x ) dy dx =
2
2
0
0
0
0
Problem 2. (20 points) Let W be the solid defined by the inequalities:

2

 z ≤1−y

 z ≥ y2 − 1

x+z ≤1



x≥0
ZZZ
For the above solid W write
f (x, y, z)dV as a triple iterated integral over an x−simple
W
region. (there are still two possible choices of the order of the integrals, so just choose one
of them...)
Solution:
Z1 1−y
Z 2 Z1−z
f (x, y, z) dx dz dy
−1
y 2 −1
0
Problem 3. (20 points) Find the moment of inertia with respectto the line x = 1 of a thin
sheet of constant density ρ = 1 bounded by the curve y = sin2 x /(x − 1)2 , the x-axis and
the strip π ≤ x ≤ 2π.
1
You may use the formula:
Z2π
sin2 x dx =
π
2
π
Solution:
2
sin2 x/(x−1)
Z
Z2π
2
Z2π
(x − 1) dx =
Ix=1 =
π
0
sin2 x dx =
π
2
π
Problem 4. (20 points) Consider the change of variables u = xy, v = yz, w = xz.
∂(x, y, z)
a) Find the Jacobian
.
∂(u, v, w)
Hint: Use the fact that uvw = x2 y 2 z 2 . Solution:
y x 0 ∂(u, v, w)
= det 0 z y = 2xyz
∂(x, y, z)
z 0 x ∂(x,y,z)
1
1
so ∂(u,v,w)
= 2xyz
= 2√uvw
b) Find the volume of the region in the first octant enclosed by the hyperbolic cylinders
xy = 1, xy = 4, xz = 1, xz = 4, yz = 4, yz = 9. Solution:
With the change of variables described in a):
Z 4Z 9Z 4
1 √ 4 √ 9 √ 4
1
√
dw dv du = 2 u1 · 2 v 4 · 2 w1 = 4
V =
2
1
4
1 2 uvw
Problem 5. (20 points) Evaluate the integral:
Za
−a
√
2
2
Za −y
ex
2 +y 2
dx dy
0
Hint: Use polar coordinates. Solution: In polar coordinates:
Zπ/2 Za
−π/2
0
a
π 2
1 r2 e r dr dθ = π · e = · ea − 1
2
2
0
r2
2