Math 52 - Winter 2006 - Midterm Exam I
Problem 1. (8 pts.) Let R be a region of R2 symmetrical about the y-axis, and R+ be
the “half” of the region R contained in the half plane x ≥ 0. Mark as TRUE/FALSE the
following statements:
ZZ
ZZ
dA = 2 ·
a)
dA
TRUE
R+
R
ZZ
ZZ
ZZ
x dA = 2 ·
b)
x dA
R
R+
R
ZZ
x dA = 0.)
FALSE (in fact
ZZ
sin(y) dA = 2 ·
c)
R
sin(y) dA
TRUE
R+
ZZ
ZZ
2
x dA = 2 ·
d)
R
x2 dA
TRUE
R+
ZZZ
Problem 2. (12 pts.)
x2 dV , where T is the tetrahedron
Evaluate the integral
T
bounded by coordinate planes and the plane 2x + 3y + z = 6.
Solution:
ZZZ
Z 3 Z 2− 2 x Z 6−2x−3y
Z 3 Z 2− 2 x
3
3
2
2
x dV =
x dz dy dx =
x2 (6 − 2x − 3y) dy dx =
0
T
Z
=
0
3
0
0
0
0
2
Z 3 2
3 2
2
2 2
27
2
x (6 − 2x) 2 − x − x 2 − x
dx =
x 6 − 4x + x
dx =
3
2
3
3
5
0
2
Problem 3. (12 pts.) Setup, but do not evaluate the triple integral representing the volume
of the bounded solid W bounded by the graphs of the cylinders x2 + z 2 = 1, y 2 + z 2 = 1 that
is contained the first octant of the coordinate system.
Solution: First octant implies x ≥ 0, y ≥ 0 and z ≥ 0. We consider W as z-simple. Then
the projection is the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, and the bounds for z are:
√
p
2
2
0 ≤ z ≤ min
1−x , 1−y
1
√
thus the upper bound is
1 − x2 if
p
√
1 − x2 ≤ 1 − y 2
if 1 − x2 ≤ 1 − y 2
if y 2 ≤ x2
which for x ≥ 0, y ≥ 0 means y ≤ x. Thus:
1
Z
x
Z
√
Z
1−x2
V =
1
Z
y
Z
Z √1−y2
dz dy dx +
0
0
dz dx dy
0
0
0
0
Note: because of symmetries you could take twice of any one of the above integrals.
Problem 4. (12 pts.) Evaluate
Z
e
Z
1
Solution:
Z e Z 1
1
ln y
1 x2
· e dx dy =
y
Z
0
1
Z
1
ex
1
ln y
1 x2
· e dx dy
y
1 x2
· e dy dx =
y
1
Z
0
Z
e x
e ln y dx =
x2
1
1
0
1
2
ex x dx = (e − 1)
2
Problem 5. (12 pts.) Change the order of integration in
Z
2
|x|
Z
sin(x + 3y) dy dx
−1
0
DO NOT EVALUATE THE INTEGRAL
Solution:
Z Z
−y
1
Z
2
Z
2
sin(x + 3y) dx dy
sin(x + 3y) dx dy +
0
−1
0
y
Problem 6.
1. (4 pts.) Sketch the region in the xy−plane described by the inequality
|x| + |y| ≤ 1
2. (6 pts.) Show that
Z1
ZZ
f (x + y) dx dy =
f (t) dt,
−1
|x|+|y|≤1
for any continuous real-valued function f .
Hint: Perform some change of variables...
2
Solution: Perform the change of variables u = x + y, v = x − y. Then
∂(u, v)
=2
∂(x, y)
thus:
ZZ
Z
1
f (x + y) dx dy =
|x|+|y|≤1
−1
Z
1
1
f (u) · dv du =
2
−1
Z
1
f (u) du
−1
Problem 7. (10 pts.) Consider the change of variables u = xy, v = yz, w = xz.
∂(x, y, z)
a) Find the Jacobian
.
∂(u, v, w)
Solution:
y x 0 ∂(u, v, w)
= det 0 z y = 2xyz
∂(x, y, z)
z 0 x ∂(x,y,z)
1
so ∂(u,v,w)
= 2xyz
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.
Hint: Use the fact that uvw = x2 y 2 z 2 .
Solution: With the change of variables described in a):
Z 4Z 9Z 4
√ 4 √ 9 √ 4
1
√
V =
dw dv du = 2 u1 · 2 v 4 · 2 w1 = 72
uvw
1
4
1
Problem 8. a) (8 pts.) Set up an integral in cylindrical coordinates that represents
the volume of the region in R3 bounded by the two surfaces z = x2 and z = 9 − y 2 .
Solution: The inequalities representing the region are x2 ≤ z ≤ 9 − y 2 . Thus the projection
onto x − y plane is the region x2 + y 2 ≤ 9, i.e. the disk of radius 9 centered at the origin.
Hence the volume is:
Z 2π Z 3 Z 9−y2
Z 3
81π
r dz dr dθ = 2π
(9 − r2 )r dr =
2
0
0
x2
0
b) (4 pts.) evaluate the integral in part a).
Problemp9. (12 pts.) Let S be the part of the ball x2 + y 2 + z 2 ≤ 4 that lies above the
cone z = x2 + y 2 . Find the volume of S.
Hint: use spherical coordinates.
Solution:
√ !
Z 2π Z π/4 Z 2
8
16π
2
π/4
V =
ρ2 sin φ dρ dφ dθ = 2π · · (− cos φ)|0 =
1−
3
3
2
0
0
0
3