HW14-solutions-1

# HW14-solutions-1 - zakaria(mmz255 HW14 gilbert(55485 1 This...

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Unformatted text preview: zakaria (mmz255) HW14 gilbert (55485) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Find the volume of the solid in the first octant bounded by the cylinders x 2 + y 2 = 9 , y 2 + z 2 = 9 . Hint: in the first octant the cylinders are shown in x y z 3 3 1. volume = 16 cu. units 2. volume = 20 cu. units 3. volume = 17 cu. units 4. volume = 19 cu. units 5. volume = 18 cu. units correct Explanation: As the figure shows, the solid in the first octant bounded by the cylinders x 2 + y 2 = 9 , y 2 + z 2 = 9 is the solid below the graph of z = radicalbig 9- y 2 above that part of the circle x 2 + y 2 = 9 lying in the first quadrant of the xy-plane. Thus the volume of the solid is given by the double integral V = integraldisplay integraldisplay A radicalbig 9- y 2 dxdy where A is the region in the first quadrant of the x- y plane bounded by the quarter-circle braceleftBig ( x, y ) : 0 x radicalbig 9- y 2 , y 3 bracerightBig , and so V can be represented as the iterated integral V = integraldisplay 3 braceleftBig integraldisplay 9- y 2 radicalbig 9- y 2 dx bracerightBig dy . In this case, V = integraldisplay 3 bracketleftBig x radicalbig 9- y 2 bracketrightBig 9- y 2 dy = integraldisplay 3 (9- y 2 ) dy . Consequently, V = bracketleftBig 9 y- 1 3 y 3 bracketrightBig 3 = 18 cu. units . 002 10.0 points Evaluate the double integral I = integraldisplay integraldisplay D 4 (1 + x + y ) 3 / 2 dA when D is the region in the first quadrant bounded by y = 3 x + 3 , x = 3 as well as the x and y-axes. zakaria (mmz255) HW14 gilbert (55485) 2 1. I = 8 correct 2. I = 10 3. I = 9 4. I = 11 5. I = 12 Explanation: The region of integration D is the region shaded in 3 1 y = 3 x + 3 the vertical line interior to the region showing that we should integrate first with respect to y . Then I becomes the repeated integral I = integraldisplay 3 parenleftBig integraldisplay 3 x +3 4 (1 + x + y ) 3 / 2 dy parenrightBig dx. Now the inner integral is bracketleftBig- 8 (1 + x + y ) 1 / 2 bracketrightBig 3 x +3 = 4 (1 + x ) 1 / 2 , and so I = integraldisplay 3 4 (1 + x ) 1 / 2 dx = bracketleftBig 8(1 + x ) 1 / 2 bracketrightBig 3 . Consequently, I = 8 . 003 10.0 points Reverse the order of integration in the inte- gral I = integraldisplay 4 1 parenleftBig integraldisplay ln y f ( x, y ) dx parenrightBig dy , but make no attempt to evaluate either inte- gral. 1. I = integraldisplay 4 parenleftBig integraldisplay 4 e x f ( x, y ) dy parenrightBig dx 2. I = integraldisplay ln 4 parenleftBig integraldisplay 4 e x f ( x, y ) dy parenrightBig dx correct 3. I = integraldisplay ln 4 parenleftBig integraldisplay e x 1 f ( x, y ) dy parenrightBig dx 4. I = integraldisplay ln 4 parenleftBig integraldisplay 4 e y f ( x, y ) dy parenrightBig dx 5. I = integraldisplay ln 4 parenleftBig integraldisplay e y 1 f ( x, y ) dy parenrightBig dx 6.6....
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## This note was uploaded on 06/05/2011 for the course MATH 408 D taught by Professor Gilbert during the Spring '11 term at University of Texas.

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HW14-solutions-1 - zakaria(mmz255 HW14 gilbert(55485 1 This...

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