MATH 265 Section A
Spring 2005
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EXAM 3
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1. (15 points) Find the volume under the surface z = x2 + xy and above
the rectangle R defined by 1 ≤ x ≤ 4, 0 ≤ y < 1.
Solution:
Z 4Z
1
1
(x2 + xy) dy dx =
0
99
4
2. (20 points) For the iterated integral
Z 2 Z 4−x2
x dydx
0
4−2x
(a) Sketch the region of integration.
(b) Give an equivalent integral with the order of integration reversed.
(c) Evaluate either integral.
Solution:
Z 4Z
√
4−y
x dx dy =
0
2− y2
4
3
3. (20 points) Let S denote the region of the xy plane bounded by y = 0
and y = cos x for − π2 ≤ x ≤ π2 . If the density is δ(x, y) = 1, find the
center of mass.
Solution:
π
(x̄, ȳ) = (0, )
8
MATH 265 Section A
Spring 2005
EXAM 3
4. (15 points) Set up, but do not evaluate, the triple integral for the volume
above the paraboloid z = x2 + 4y 2 and below the plane z − 2x = 15.
Solution: Two of the possible 6 orders of integration are
q
Z
5
Z
16−(x−1)2
4
and
5
Z
2x+15 Z
x2
−3
2x+15
dz dy dx
q
2
− 16−(x−1)
4
−3
Z
Z
x2 +4y 2
q
z−x2
4
q
2
− z−x
4
dy dz dx
5. (15 points) Find the surface area of the portion of the surface
z = 9 − x2 − y 2
which is in the first octant.
Solution:
π
2
Z
Z 3p
1 + 4r2 r dr dθ =
0
0
π
(373/2 − 1)
24
RRR 2
6. (15 points) Evaluate the integral
S z dV where S is the region be2
2
2
tween the spheres x + y + z = 1 and x2 + y 2 + z 2 = 4.
Solution:
Z
0
2π
Z
0
π
Z
2
ρ4 cos2 φ sin φ dρ dφ dθ =
1
Page 2
124π
15