Download Math 53 Final Exam Solutions and more Exams Vector Analysis in PDF only on Docsity! Math 53. Multivariable Calculus. Final Exam. 12.15.15. Solutions 1. Change the order of integration: ∫ 1 0 ∫ √ z 0 ∫ √ y 0 f(x, y, z) dxdydz = ∫ ? ? ∫ ? ? ∫ ? ? f(x, y, z) dydzdx . Solution. The region is 0 ≤ x4 ≤ y2 ≤ z ≤ 1. Respectively the integral is equal to ∫ 1 0 ∫ 1 x4 ∫ √ z x2 f(x, y, z) dydzdx. 2. In what proportion is the surface of the sphere of radius r divided by a plane passing at the distance d < r from the center? Justify your answer. Solution. There are several ways to compute the area of the part of the sphere x2 + y2 + z2 = r2 above the plane z = d. A. The area of the graph of the function z(x, y) = √ r2 − x2 − y2 over the domain x2 + y2 ≤ r2 0 := r2 − d2 is found by integrating √ 1 + z2x + z2y = √ 1 + x2 r2 − x2 − y2 + y2 r2 − x2 − y2 = r √ r2 − x2 − y2 . In polar coordinates δ = √ x2 + y2, θ = arctan y x , we find: r ∫ 2π 0 ∫ r0 0 δ dδ dθ√ r2 − δ2 = −2πr 2 ∫ r0 0 d(r2 − δ2)√ r2 − δ2 = −2πr √ r2 − δ2 ∣ ∣ δ=r0 δ=0 = 2πr(r − d), since r2 0 = r2 − d2, and hence √ r2 − r2 0 = d. B. Alternatively, in spherical coordinates the surface are of this piece of the sphere is given by the formula r2 ∫ 2π 0 ∫ arccos d/r 0 sinφ dφdθ = 2πr2 ( 1− d r ) . C. Perhaps, the most illuminating way is based on representing the sphere as the surface of revolution about the z-axis with the cylindrical equation δ(z) = √ r2 − z2, whose surface area is given by the integral 2π ∫ r d √ 1 + δ2z δ(z) dz = 2πr ∫ r d dz = 2πr(r − d). The rest of the sphere has the area 2πr(r + d). Thus, the required ratio is (r − d)/(r + d). 2 3. To a table’s surface with coordinates (x, y), a square piece of rubber 0 ≤ u, v ≤ 1 is glued according to the rule x(u, v) = 3u+ v2, y(u, v) = v + (3u+ v2)2. Find the area of the region covered by this piece of rubber. Solution. The Jacobian ∣ ∣ ∣ ∣ xu xv yu yv ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ 3 2v 2(3u+ v2)3 1 + 2(3u+ v2)2v ∣ ∣ ∣ ∣ = 3, and hence the area of the image of the square is 3 times the area of the square, i.e. 3. 4. Find the maximum and minimum values of the function 2x2 + 4xy − y2 on the unit circle x2 + y2 = 1. Solution. By the Lagrange multiplier method, we get the system of two linear equations 4x+ 4y = 2λx, 4x− 2y = 2λy, which has solutions under the constraint x2 + y2 = 1 only when ∣ ∣ ∣ ∣ 2− λ 2 2 −1− λ ∣ ∣ ∣ ∣ = λ2 − λ− 6 = 0, i.e. λ = 3 or λ = −2. The solutions have x/y = 2 and y/x = −2 re- spectively. The respective critical points are: (x, y) = ±(2/ √ 5, 1/ √ 5) and (x, y) = ±(1/ √ 5,−2/ √ 5). The corresponding critical values are 3 and −2. Therefore the maximum value is 3, and the minimum −2. 5. Does there exist a function f(x, y) satisfying fxx(0, 0) = 8, fxy(0, 0) = 4, fyy(0, 0) = 2 which at (x, y) = (0, 0) has: (a) a local minimum? (b) a local maximum? (c) a critical point which is neither local minimum nor local maximum? In each case (a),(b),(c), if the an- swer is “no” explain “why”, if “yes”, give an example of such a function and explain why the required property holds. Solution. (a) g(x, y) := 4x2+4xy+y2 = (2x+y)2 has the required values of gxx, gxy, gyy, and a local minimum at the origin. (b) Since fxx(0, 0) = 4 > 0, the function f(x, 0) has an isolated local minimum at x = 0, and hence f(x, y) cannot have a local maximum at the origin. (c) f(x, y) := (2x + y)2 − (x4 + y4) on the line 2x + y = 0 is negative unless (x, y) = (0, 0), and hence f has neither minimum nor maximum at the origin.
