am2 - PRINT NAME: SOLUTIONS Calculus IV [2443002] Midterm...

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Unformatted text preview: PRINT NAME: SOLUTIONS Calculus IV [2443002] Midterm II Q1]...[10 points] Consider the double integral Z 1 Z 2- 2 y- 1- y f ( x,y ) dxdy Sketch the region of integration. Soln. The limits x = 2- 2 y and x =- 1- y tell us that the region is bounded on the right by the line x + 2 y = 2 and on the left by the parabola (left half) y = 1- x 2 . The limits y = 0 and y = 1 tell us the upper and lower bounds for this region. We see that the parabola and line already intersect at y = 1, so the region is drawn as shown. . . . . . . . .... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . x + 2 y = 2 x y y = 1- x 2 Reverse the order of integration. Soln. Note that reversing the order of integration means building up the region using vertical strips. There are two different tops on this region; the parabola top on the left side of the y-axis, and the straight line top on the right side. Thus, we have to divide the region into two pieces along the y-axis. So our answer will be a sum of two iterated integrals as shown. Z- 1 Z 1- x 2 f ( x,y ) dy dx + Z 2 Z (2- x ) / 2 f ( x,y ) dy dx Q2]...[10 points] Consider the following polar coordinates double integral Z / 4- / 4 Z sec r 3 dr d Sketch the region of integration....
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am2 - PRINT NAME: SOLUTIONS Calculus IV [2443002] Midterm...

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