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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