MATH 52 MIDTERM 1. April 23, 2004

Transcription

1 MATH 5 MIDTERM April 3, 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 bo around your final answer to each question. There are questions. Point values are given in parentheses. You have 5 minutes to answer all the questions. Question Score Maimum 3 Total

2 . a) 5 points) Sketch the region of integration in the double integral: + y 3 dy d. 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. + y 3 dy d y + y 3 d dy + y 3 [ y + y 3 dy 9 + y3 ) 3 9 ] y dy

3 . points) Let R be the first-quadrant region bounded by the curve + y y and the line y. Use polar coordinates to evaluate: R + + y ) da. Solution: + y y + y ) y y + y ) We can now write this in polar coordinates as r tan θ. In the first quadrant, this reduces to r tan θ. The line y can be rewritten as y, which is just tan θ in polar coordinates. In the first quadrant, this reduces to θ π. Therefore, in polar coordinates, the region R is the region lying between the curve r tan θ and the ray θ π. Picture: We can now compute R + + y ) da tan θ r dr + r ) [ ] tan θ + r ) + tan θ) + ) cos θ + ) [ + cosθ)] + ) θ 8 sinθ) π π 6 8

4 3. points) Set up, but DO NOT EVALUATE a triple integral in rectangular coordinates that represents the moment of inertia about the z-ais of the region T bounded by the cone z + y and the elliptical cylinder + z 3). 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 we can integrate the and z coordinates inside the ellipse + z 3), and then integrate y from one bottom side of the cone to the top side of the cone. Picture: To integrate over the ellipse, we integrate from to and then integrate z from 3 to 3 +. This comes from solving for z in the ellipse equation.) Net, we integrate y from z to z. This comes from solving for y in the cone equation.) Of course, one must check that z everywhere in the ellipse. Finally, to get the moment of inertia about the z-ais, the integrand must be + y. Putting it all together, we get: 3+ z 3 z + y ) dy dz d

5 . points) Let R be a region in the first quadrant which satisfies the following conditions: R is symmetric with respect to the line + y 5. The volume obtained by revolving R about the -ais is twice the volume obtained by revolving R around the y-ais. Find the coordinates, y) of the centroid of R. Eplain your reasoning. Solution: The first statement implies that + y 5. Convince yourself that this is true.) The second statement relates two volumes that we will call V and V y. It says that V V y. Now let A be the area of the region R. The First Theorem of Pappus says that V πya and V y πa. So we see that y. Intersecting this line with the line + y 5, we see that, y), ).