University:
Amherst College
Course:
MATH 121 | Intermediate Calculus
Academic year:

2010
Views:

284

Pages:

5
Author:

Rotporti8p

Math 13 Fall 2010: Exam 3 Tuesday April 27, 2010 Name: Instructions: There are 4 questions on this exam each of which is scored out of 8 points for a total of 32 points. You may not use any outside materials(eg. notes or calculators). You have 50 minutes to complete this exam. Remember to fully justify your answers. Score: 1 Problem 1. Evaluate Z 0 1 Z √1−y2 0 1 dxdy 1 + x2 + y 2 Proof. Changing to polar we have Z 1 Z √1−y2 Z π/2 Z 1 Z π/2 ¯ r 1 ¯¯ 1 dxdy = drdθ = ln 1 + r2 ¯|10 dθ 2 2 2 1+x +y 2 0 0 1+r 0 0 0 Z π/2 ln 2 π ln 2 = dθ = . 2 4 0 2 Problem 2. Consider the region R that lies outside the cone z 2 = 3(x2 + y 2 ) and inside the sphere x2 + y 2 + z 2 = 4 in the first octant. Set up the integral for the volume of R in rectangular, cylindrical, and spherical coordinates. Proof. The curve of intersection is given by 4(x2 + y 2 ) = 4 and √ hence x2 + y 2 = 1. We also need to know the cone equation is spherical coordinates, which is tan φ = 1/ 3 and so φ = π6 . So we have Z 1 √ 1−x2 Z Z √3(x2 +y2 ) V = Z 2 dzdydx + 0 Z 0 π/2 Z 1 Z 0 √ r 3 V = Z π/2 Z 0 2 Z rdzdrdθ + 0 Z π/2 Z 0 0 π/2 Z 2 V = 0 π/6 Z √ 4−x2 √ 1−x2 √ 4−r 2 Z √4−x2 −y2 dzdydx 0 rdzdrdθ 0 1 0 ρ2 sin φdρdφdθ. 0 or we can subtract the area inside the cone from the sphere. We have Z 1 Z √1−x2 Z √4−x2 −y2 4 V = π− √ 2 2 dzdydx 3 0 0 3(x +y ) Z π/2 Z 1 Z √4−r2 4 rdzdrdθ V = π− √ 3 0 0 r 3 Z π/2 Z π/6 Z 2 4 V = π− ρ2 sin φdρdφdθ. 3 0 0 0 3 Problem 3. Consider the region R above the upper sheet of the hyperboloid x2 + y 2 = z 2 − 1 and below the plane z = 3 with density δ(x, y, z) = 2z. 1. Find the mass of R. 2. Set-up but do not evaluate the integral for the moment of inertia about the z-axis for R. Proof. We have that the vertex of the upper sheet is given by z = 1 at (0, 0, 1) and at z = 3 we have x2 + y 2 = 8. 1. For the mass (using cylindrical coordinates) we have Z 2π Z √ 8 M= Z 0 0 2π Z = Z 0 Z √ √ 8 Z 3 2π 0 Z √ 8 2rzdzdrdθ = 1+r 2 0 9r − r(1 + r2 )drdθ 0 8r − r3 drdθ 0 = 2π Z √ 4r2 − r4 /4]0 8 dθ = 2π 32 − 16dθ 0 = 32π. 2. For the moment of inertia we have Z 2π √ Z 8 Iz = 0 Z 3 √ 1+r 2 0 4 2r3 zdzdrdθ. Problem 4. Let R be the parallelogram bounded by the lines x+y = 1, x+y = 2, 2x−3y = 2, 2x−3y = 4. Evaluate ZZ 5xdA. R Proof. Set u = x + y and v = 2x − 3y and so we have x = 15 (3u + v) and y = 51 (2u − v). The jacobian of the transformation is given by ¯ µ3 ¶¯ 1 ¯ ¯ 1 5 5 ¯ ¯ J = ¯det 2 1 ¯= . − 5 5 5 So we have ZZ Z 2 Z 5xdA = R 1 5 Z = 1 2 2 4 1 (3u + v)dvdu 5 6u + 6du 1 1 (9 + 6) = 3. 5 5

Math 13 Fall 2010 Exam 3

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