Exam 3 Math 211 Multivariable Calculus

Exam 3 Math 211 Multivariable Calculus, 2019 Name: Question: 1 2 3 4 5 Total Points: 17 17 17 17 32 100 Score: INSTRUCCIONS: • In this exam you are will be asked to a) set up only but not evaluate integrals or b) set up and evaluate integrals. suggestion: If you have to evaluate integrals, I suggest you set up the integral in the proper coordinate system and move on to the next question. Then do the evaluation of the integral after you finish setting up all the other integrals that do not require evaluations.) • This exam has a total of (5) questions and (8) pages, including cover. • Please write your name clearly. • This is is a closed book, closed notebook exam. No calculators nor electronics are allowed. • Solve all the (5) questions on this test. • Justify your work. Providing just an answer without justification will earn you little or no credit. • You are not allowed to communicate the content of this exam to anyone that has not taken the test yet, until the test is discussed in class. • Trust in what you have learned! Z 2 Z x2 1. Given the integral f (x, y) dy dx 0 −x2 (a) (7 points) Sketch the region of integration. Solution: The answer is the region below and above the positive x-axis in the following sketch 4 y x2 −x2 2 x −2 −1 1 2 −2 −4 (b) (10 points) Rewrite the integral with the order of integration reversed. Solution: Z 0 Z 2 Z 4 Z 2 f (x, y) dx dy + −4 −x2 f (x, y) dx dy 0 x2 Page 2 ZZ 4xy dA, where D is the region in the plane inside the circle 2. (17 points) Compute D x2 + y 2 = 4x and above the line y = x. Solution: 4 y f g 2 x −2 −1 1 2 3 4 −2 Note that using polar coordinates the line y = x becomes r sin θ = r cos θ → θ = π/4. Also, the equation x2 + y 2 = 4x becomes r = 4 cos θ. Z Z Z π/2 Z 4 cos θ 16 So we have: 4xy dA = 4r cos θ · r sin θ · r dr dθ = 3 π/4 0 D Page 3 3. Let f (x, y, z) = x, be the mass density function of the solid E. And let E be the solid that is p x2 + y 2 and • outside the cone z = • inside the sphere x2 + y 2 + z 2 = 4 and • above the first and second quadrant of the xy plane Suppose you are asked to compute the mass of E. (a) (13 points) Choose the best suitable coordinate system (the easiest one) to evaluate this integral and set up but do not evaluate the integral to compute the mass of E. Solution: The projection of the solid of interest on the yz plane, is the upper portion of the 2 z f g 1 y −2 −1 1 2 −1 −2 following graph: Note that the cone is given by the equation φ = π/4. In this case we would have the integral, in spherical coordinate: Z π/2 π/4 Z 0 π Z 2 ρ sin φ cos θ · ρ sin2 φ dρ dθ dφ 0 (b) (4 points) Write the name of the coordinate system you chose in part (a), and In one single sentence, explain why the system you chose in part (a) is the easiest one to solve this integral. Solution: Spherical coordinate. This coordinate system simplifies calculations b/c, among other things, it gives easier limits of integration. Page 4 4. Let E be the solid in the first octant enclosed by the paraboloids z = 4x2 + 4y 2 and z = 5 − x2 − y 2 , and with the mass density function given by ρ(x, y, z) = x2 + y 2 . 5π (ii.e. you don’t need to calculate the mass, just use Assume that the mass of E is 24 this letter value). Suppose you are asked to compute the z-coordinate of the center of mass of E. (a) (13 points) Choose the best suitable coordinate system (the easiest one) to evaluate this integral and set up but do not evaluate the integral to compute the z coordinate of the center of mass of E. Solution: In Cylindrical Coordinates we find that these two paraboloids intersect when 4r2 = 5 − r2 , that is r = 1. Their intersection is given the curve in the first quadrant 1 z f g 0.8 0.6 0.4 0.2 y 0.2 0.4 0.6 0.8 1 In Cylindrical we would have 24 5π Z 0 π/2 Z 0 1 Z 5−r2 zr3 dz dr dθ 4r2 (b) (4 points) Write the name of the coordinate system you chose in part (a), and In one single sentence, explain why the system you chose in part (a) is the easiest one to solve this integral. Solution: Cylindrical coordinate. This coordinate system simplifies calculations b/c, among other things, it gives easier limits of integration. Page 5 5. Let E be the solid bounded by the surfaces y = z 2 , y − x = −1,and x + 2y = 4. Suppose you are asked to compute the volume of the solid E. (a) (11 points) In rectangular coordinate, and taking D to be the projection of E onto the yz plane, set up but do not evaluate the integral to compute the volume of E, starting from this region D. Solution: Note that the projection of this solid in the yz plane, is 1 z f f 0.5 y 0.2 0.4 0.6 0.8 1 −0.5 −1 In this case the volume is given by: 1 Z Z 0 √ y √ − y Z 4−2y dx dz dy y+1 (b) (11 points) Still in rectangular coordinate, rewrite the integral in part (a) changing the order of integration. Any order-change is acceptable. Solution: Using the same graph as before we get another reordering Z 1 Z 1 Z 4−2y dx dy dz −1 z2 y+1 Page 6 (c) (10 points) Still in rectangular coordinate, re-write the integral in an order different than the one you used in parts (a) and (b). Solution: Note that in the xy plane the projection is the triangle whose base is the segment [1, 4]. 3 z x−1 4−x 2 2 1 y −1 1 2 3 4 −1 −2 We thus have Z 0 1 Z 4−2y √ Z y √ − y y+1 dz dx dy Bonus Question (3 points) Rewrite the integral in question (5) in three different orders of integration not already used in your answer to that question. • for 1 more order of integration get zero points •for 2 more orders of integration get 2 points •3 more orders of integration get the full 3 points) Solution: The remaining orders of integration are Z 2 Z x−1 Z √y Z dz dy dx + √ 1 − y 0 4 Z 2 (4−x)/2 √ Z y √ − y 0 dz dy dx The other orders are based on the graph below, that is, the portion enclosing the interval [1, 4] in the x-axis. Z 1 Z 2 Z x−1 Z 1 Z 4−2z2 Z (4−x)/2 dy dx dz + dy dx dz z 2 +1 −1 z2 −1 z2 2 The final order of integration is Z 1 2 √ Z x−1 √ − x−1 Z x−1 Z 4 Z √(4−x)/2 Z dy dz dx + z2 2 Page 7 − √ (4−x)/2 (4−x)/2 dy dz dx z2 √ x−1 √ − x−1 p (4 − x)/2 p − (4 − x)/2 z 1 y 1 2 3 4 −1 Page 8