Math 251
Sample Exam 2 Answers
1. D
2. B
3. A
4. D
5. E
6. C
7. C
8. You need to solve fx = y − x2 = 0 and fy = x − 2y = 0, and then you find that (0, 0)
and ( 21 , 14 ) are the critical points of f . Use the second derivative test to find that (0, 0)
is a saddle point and ( 21 , 14 ) is a local maximum.
9. (a) You intersect the two surfaces and find that they are sitting
above the disk x2 +y 2 ≤
Z 2 Z √4−x2
(8 − 2x2 − 2y 2 ) dy dx.
4 in the xy-plane. Setting up the integral yields V =
√
− 4−x2
−2
Z
2π
Z
(b) V =
0
2
(8 − 2r2 )r dr dθ
0
(c) Evaluate the integral in (b). The value is 16π.
10. (a) The region is the triangle in xy-plane whose vertices are (0, 0), (0, 4), and (2, 4).
(b) To evaluate the integral you either need to do a trigonometric substitution (not
recommended!) or simply change the order of integration. Changing the order of
integration we find that
Z 4 Z 1y
Z 2Z 4
2
1
1
p
p
dy dx =
dx dy.
1 + y2
1 + y2
0
0
2x
0
√
This integral can be evaluated easily, and the answer is
17
2
− 12 .
11. After looking at these two surfaces for a moment, you realize that the cone is above
the paraboloid. Also, they sit above the unit disk in the xy-plane.
Z 1 Z √1−x2 Z √x2 +y2
(a) V =
dz dy dx
√
−1 − 1−x2
2π Z 1 Z r
Z
(b) V =
0
0
r2
x2 +y 2
3r3 z cos(θ) sin(θ) dz dr dθ