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MATH 52 MIDTERM 1 April 23, 2004 Name: Student ID: Signature: Instructions: Print your name and student ID number and write your signature to indicate that you accept the honor code. During the test, you may not use notes, books, or calculators. Read each question carefully, and show all your work. Put a box around your ﬁnal answer to each question. There are 4 questions. Point values are given in parentheses. You have 50 minutes to answer all the questions. Question Score Maximum 1 10 2 10 3 10 4 10 Total 40 Name: 1. (a) (5 points) Sketch the region of integration in the double integral: 4 2 x 1 + y 3 dy dx. x 0 2 Solution: (b) (5 points) Evaluate the integral in part (a) by changing the order of integration. Solution: Using the solution to part (a), we see how to change the order of integration for the triangular region. 4 2 2 2y x 1 + y 3 dy dx = x 1 + y 3 dx dy x 0 2 0 0 2 2y 1 2 = 1 + y3 x dy 0 2 0 2 = 2y 2 1 + y 3 dy 0 2 4 3 = (1 + y 3 ) 2 9 0 104 = 9 Name: 2. (10 points) Let R be the ﬁrst-quadrant region bounded by the curve x4 + x2 y 2 = y 2 and the line y = x. Use polar coordinates to evaluate: 1 dA. R (1 + x2 + y 2 )2 Solution: x4 + x2 y 2 = y 2 x2 (x2 + y 2 ) = y 2 y 2 x2 + y 2 = x We can now write this in polar coordinates as r2 = tan2 θ. In the ﬁrst quadrant, this y reduces to r = tan θ. The line y = x can be rewritten as x = 1, which is just tan θ = 1 in polar coordinates. In the ﬁrst quadrant, this reduces to θ = π . Therefore, in polar 4 coordinates, the region R is the region lying between the curve r = tan θ and the ray θ = π . Picture: 4 We can now compute π tan θ 1 4 1 dA = r dr dθ R (1 + x2 + y 2 )2 0 0 (1 + r2 )2 π tan θ 4 −1 = dθ 0 2(1 + r2 ) 0 π 4 −1 1 = 2 + dθ 0 2(1 + tan θ) 2 π 4 1 1 = − cos2 θ + dθ 0 2 2 π 4 1 1 = − [1 + cos(2θ)] + dθ 0 4 2 π 1 1 4 = θ − sin(2θ) 4 8 0 π 1 = 16 − 8 Name: 3. (10 points) Set up, but DO NOT EVALUATE a triple integral in rectangular coordinates that represents the moment of inertia about the z-axis of the region T bounded by the cone 2 z 2 = x2 + y 2 and the elliptical cylinder x2 + (z−3) = 1. 4 Solution: The important thing is to treat the y-coordinate as the “third” coordinate. That is, we turn the problem on its side so that the cylinder is “standing” up. Then 2 we can integrate the x and z coordinates inside the ellipse x2 + (z−3) = 1, and then 4 integrate y from one “bottom” side of the cone to the “top” side of the cone. Picture: To integrate over the √ √ ellipse, we integrate x from −1 to 1 and then integrate z from 2. 3 − 2 1 − x2 to 3 + 2 1 − x√ (This comes from solving for z in the ellipse equation.) √ Next, we integrate y from − z 2 − x2 to z 2 − x2 . (This comes from solving for y in the cone equation.) Of course, one must check that z 2 ≥ x2 everywhere in the ellipse. Finally, to get the moment of inertia about the z-axis, the integrand must be x2 + y 2 . Putting it all together, we get: √ √ 1 3+2 1−x2 z 2 −x2 √ √ (x2 + y 2 ) dy dz dx −1 3−2 1−x2 − z 2 −x2 Name: 4. (10 points) Let R be a region in the ﬁrst quadrant which satisﬁes the following conditions: • R is symmetric with respect to the line x + 2y = 5. • The volume obtained by revolving R about the x-axis is twice the volume obtained by revolving R around the y-axis. Find the coordinates (x, y) of the centroid of R. Explain your reasoning. Solution: The ﬁrst statement implies that x + 2y = 5. (Convince yourself that this is true.) The second statement relates two volumes that we will call Vx and Vy . It says that Vx = 2Vy . Now let A be the area of the region R. The First Theorem of Pappus says that Vx = 2πyA and Vy = 2πxA. So we see that y = 2x. Intersecting this line with the line x + 2y = 5, we see that (x, y) = (1, 2).

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posted: | 3/25/2011 |

language: | English |

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