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HW05-solutions

# HW05-solutions - griffin(ang684 HW05 Gilbert(56650 This...

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griffin (ang684) – HW05 – Gilbert – (56650) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Determine the volume of the right circular cone generated by rotating the line x = 2 y about the y -axis between y = 0 and y = 4. 1. V = 244 3 π cu.units 2. V = 253 3 π cu.units 3. V = 256 3 π cu.units correct 4. V = 247 3 π cu.units 5. V = 250 3 π cu.units Explanation: The volume, V , of the solid of revolution generated by rotating the graph of x = f ( y ) about the y -axis between y = a and y = b is given by V = π integraldisplay b a f ( y ) 2 dy. When f ( y ) = 2 y and a = 0 , b = 4, therefore, V = π integraldisplay b a 4 y 2 dx = π bracketleftBig 4 3 y 3 bracketrightBig 4 0 . Consequently, V = 256 3 π cu.units . 002 10.0 points Find the volume, V , of the solid obtained by rotating the region bounded by y = x 2 , x = 0 , y = 3 about the y -axis. 1. V = 9 2 π cu. units correct 2. V = 3 π cu. units 3. V = 3 cu. units 4. V = 9 4 cu. units 5. V = 9 4 π cu. units 6. V = 9 2 cu. units Explanation: The region rotated about the y -axis is sim- ilar to the shaded region in 3 y x (not drawn to scale). Now the volume of the solid of revolution generated by revolving the graph of x = f ( y ) for a y b about the y -axis is given by volume = π integraldisplay b a f ( y ) 2 dy . To apply this we have first to express x as a function of y since initially y is defined in terms of x by y = x 2 . Thus after taking square roots we see that V = π integraldisplay 3 0 y dy = π bracketleftbigg 1 2 y 2 bracketrightbigg 3 0 . Consequently, V = 9 2 π .

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griffin (ang684) – HW05 – Gilbert – (56650) 2 003 10.0 points Find the volume, V , of the solid obtained by rotating the bounded region in the first quadrant enclosed by the graphs of y = x 5 2 , x = y 3 about the x -axis. 1. V = 13 30 cu. units 2. V = 2 5 π cu. units 3. V = 7 20 π cu. units 4. V = 13 30 π cu. units correct 5. V = 2 5 cu. units 6. V = 7 20 cu. units Explanation: Since the graphs of y = x 5 2 , x = y 3 intersect at (0 , 0) and at (1 , 1) the bounded region in the first quadrant enclosed by their graphs is the shaded area shown in 1 1 Thus the volume of the solid of revolution generated by rotating this region about the x -axis is given by V = π integraldisplay 1 0 braceleftBig ( x 1 / 3 ) 2 - ( x 5 2 ) 2 bracerightBig dx = π integraldisplay 1 0 braceleftBig x 2 3 - x 5 bracerightBig dx = π bracketleftbigg 3 5 x 5 3 - 1 6 x 6 bracketrightbigg 1 0 . Consequently, V = π parenleftBig 3 5 - 1 6 parenrightBig = 13 30 π cu. units . 004 10.0 points Find the volume, V , of the solid generated by rotating about the x -axis the region en- closed by the graphs of y = sec x, x = 0 , y = 0 , x = π 3 . 1. V = π cu. units 2. V = π 3 cu. units correct 3. V = 8 π 3 cu. units 4. V = π 3 cu. units 5. V = π ln parenleftbigg 1 2 + 3 parenrightbigg cu. units Explanation: The solid is generated by rotating the re- gion π /3
griffin (ang684) – HW05 – Gilbert – (56650) 3 about the x -axis. The cross-section perpen- dicular to the x -axis has area A ( x ) = πy 2 = π sec 2 x .

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