am2 - PRINT NAME SOLUTIONS Calculus IV[2443002 Midterm II...

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PRINT NAME: SOLUTIONS Calculus IV [2443–002] Midterm II Q1]...[10 points] Consider the double integral Z 1 0 Z 2 - 2 y - 1 - y f ( x, y ) dx dy 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 0 - 1 Z 1 - x 2 0 f ( x, y ) dy dx + Z 2 0 Z (2 - x ) / 2 0 f ( x, y ) dy dx Q2]...[10 points] Consider the following polar coordinates double integral Z π/ 4 - π/ 4 Z sec θ 0 r 3 dr dθ Sketch the region of integration. Soln. Note that the lines θ = π/ 4 and θ = - π/ 4 correspond to the cartesian lines y = x and y = - x respectively. Also the curve r = sec θ is just r = 1 cos θ which rewrites as r cos θ = 1 or x = 1. [Gotta hate those trig functions!] Thus we get the following region.
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