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Unformatted text preview: Rehman (aar638) – HW05 – sachse – (56620) 1 This printout should have 20 questions. Multiplechoice 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 yaxis between y = 0 and y = 3. 1. V = 35 π cu.units 2. V = 39 π cu.units 3. V = 37 π cu.units 4. V = 36 π cu.units correct 5. V = 38 π cu.units Explanation: The volume, V , of the solid of revolution generated by rotating the graph of x = f ( y ) about the yaxis 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 = 3, therefore, V = π integraldisplay b a 4 y 2 dx = π bracketleftBig 4 3 y 3 bracketrightBig 3 . Consequently, V = 36 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 = 4 about the yaxis. 1. V = 8 cu. units 2. V = 8 π cu. units correct 3. V = 4 π cu. units 4. V = 16 3 π cu. units 5. V = 4 cu. units 6. V = 16 3 cu. units Explanation: The region rotated about the yaxis is sim ilar to the shaded region in 4 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 yaxis 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 4 y dy = π bracketleftbigg 1 2 y 2 bracketrightbigg 4 . Consequently, V = 8 π . 003 10.0 points Rehman (aar638) – HW05 – sachse – (56620) 2 Find the volume, V , of the solid obtained by rotating the bounded region in the first quadrant enclosed by the graphs of y = x 3 2 , x = y 4 about the xaxis. 1. V = 7 15 π cu. units 2. V = 7 15 cu. units 3. V = 5 12 π cu. units correct 4. V = 5 12 cu. units 5. V = 1 2 cu. units 6. V = 1 2 π cu. units Explanation: Since the graphs of y = x 3 2 , x = y 4 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 xaxis is given by V = π integraldisplay 1 braceleftBig ( x 1 / 4 ) 2 ( x 3 2 ) 2 bracerightBig dx = π integraldisplay 1 braceleftBig x 1 2 x 3 bracerightBig dx = π bracketleftbigg 2 3 x 3 2 1 4 x 4 bracketrightbigg 1 . Consequently, V = π parenleftBig 2 3 1 4 parenrightBig = 5 12 π cu. units . 004 10.0 points Find the volume, V , of the solid generated by rotating about the xaxis the region en closed by the graphs of y = sec x, x = 0 , y = 0 , x = π 3 ....
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This note was uploaded on 02/01/2011 for the course MATH 408L taught by Professor Gogolev during the Fall '09 term at University of Texas at Austin.
 Fall '09
 GOGOLEV
 Calculus, Cone

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