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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

{[ snackBarMessage ]}

Page1 / 13

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online