MATH 201 Final, Fall 2007

American University of Beirut · Calculus and Analytical Geometry III · 11 pages

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FINAL EXAMINATION MATH 201 January 24, 2008, 11:30-1:30pm Name: Student number: Section number (Encircle): 21 22 23 Instructions: • No calculators are allowed. • There are two types of questions: Part I consists of ten multiple choice questions out of 5 points each with exactly one correct answer. Part II consists of five work-out problems out of 10 points each. Give a detailed solution for each of these problems. Part I: 1 - The series �∞ n=2 (−1)n nlnn : a. Converges absolutely b. Converges conditionally c. Diverges 2. The interval of convergence of the power series �∞ n=1 (x+π)n √n is: a. ] − 1 − π, 1 − π[ b.] − 1 − π, 1 − π] c. [−1 − π, 1 − π[ d. [−1 − π, 1 − π] e. None of the above 3. The Maclaurin series of the function � x 1 ln(1−t) t dt is: a. − �∞ n=0 xn+1 (n+1)2 d. − �∞ n=1 xn+1 n(n+1) b. − �∞ n=1 xn+1 (n+1)2 e. None of the above. c. − �∞ n=0 xn n(n+1) 4. The function f(x, y, z) at a point P decreases most rapidly in the direction of v = i − 2j + 3k. In this direction the value of the derivative is −5 √ 14. The ∇f is: a. −5 √ 14i + 10 √ 14j − 15 √ 14k b. 5 √ 14i − 10 √ 14j + 15 √ 14k c. −5i + 10j − 15k d. 5i − 10j + 15k e. None of the above 5. The critical point(s) of f(x, y) = e(x2+y2)(2x2 − y2) is (are): a. (0, 0) b. (0, 0), (+ − 1 √ 2, 0) c. (0, 0), (0,+ − √ 2) d. (0, 0), (+ − 1 √ 2, 0), (0,+ − √ 2) e. None of the above 6. Let C be a simple closed curve, then � C(x+y)dx+(x2−y2)dy is the expression of: a. The counterclockwise circulation of the field F = (x2 − y2)i + (x + y)j around C. b. The counterclockwise circulation of the field F = (x2 − y2)i − (x + y)j around C. c. The outward flux of the field F = (x + y)i + (x2 − y2)j across C d. The outward flux of the field F = (x2 − y2)i − (x + y)j across C. e. None of the above 7. Use the chain rule to compute ∂w ∂s when w = f(x, y, z), x = 2s + t2, y = tu, and z = u2 − s2: a. ∂w ∂s = (2s + t2)∂w ∂x + tu∂w ∂y + (u2 − s2)∂w ∂z b. ∂w ∂s = 2∂w ∂x − 2s∂w ∂z c. ∂w ∂s = t∂w ∂y + 2u∂w ∂z d. ∂w ∂s = tu∂w ∂y + (u2 − s2)∂w ∂z e. None of the above 8. The value of the double integral � 2 0 � 1 y/2 3yex3dxdy is: a. 2(e − 1) b. 9 2(e − 1) c. 25 2 (e − 1) d. 8(e − 1) e. None of the above 9. Let D be the region bounded by the paraboloids z = 8−x2−y2 and z = x2 + y2. The volume of D in rectangular coordinates is expressed by the following iterated integral: a. � 4 0 � √z −√z � √ 8−z−x2 − √ 8−z−x2 dydxdz + � 8 4 � √8−z −√8−z � √ z−x2 − √ z−x2 dydxdz b. � 4 0 � √8−z −√8−z � √ z−x2 − √ z−x2 dydxdz + � 8 4 � √z −√z � √ 8−z−x2 − √ 8−z−x2 dydxdz c. � 4 0 � √8−z −√8−z � √ 8−z−x2 − √ 8−z−x2 dydxdz + � 8 4 � √z −√z � √ z−x2 − √ z−x2 dydxdz d. � 4 0 � √z −√z � √ z−x2 − √ z−x2 dydxdz + � 8 4 � √8−z −√8−z � √ 8−z−x2 − √ 8−z−x2 dydxdz e. None of the above 10. The value of the integral I = � ∞ 0 e−x2dx is: (Hint: Compute I2 = ( � ∞ 0 e−x2dx)( � ∞ 0 e−x2dx)). a. π 4 b. √π 2 c. �π 2 d. π 2 e. None of the above Part II I- (10 points) Use the transformation u = x − y and v = x + y to evaluate the double integral: � � R(x − y)2cos2(x + y)dA over the region R bounded by the lines: x − y = 1, x − y = −1, x + y = 1 and x + y = 3. II- (10 points) Use the method of Lagrange Multipliers to find the dimensions of the rectangle of greatest area that can be inscribed in the ellipse x2 16 + y2 9 = 1. III- (10 points) Let D be the region bounded by below by the xy-plane, laterally by the cylinder x2 + y2 = 9, and above by the sphere x2 + y2 + z2 = 25. Set up, but do not evaluate, the triple integral in cylindrical and spherical coordinates that represents the volume of D. IV- (10 points) Consider the vector field F = (y + z)i + (x − 2y)j + xk a- Verify that the field is conservative. b- Find a potential function f(x, y, z) for F. c- Use part b- to evaluate the work over C where C is a smooth curve whose inital point is (3,-2,1) and the terminal point is (-1,1,2) d- Evaluate the work by finding parametrization for the segment that make up C. V- (10 points) Let C be the closed curve in the plane starting from (0, 0) which first goes to (1, 1) along the parabola y = x2 and then returns to (0, 0) along the line y = x. Compute � C ydx + eydy twice: a) directly b) using Green’s theorem