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HW 05 solution

# HW 05 solution - gutierrez(ig3472 HW05 Neitzke(56585 This...

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gutierrez (ig3472) – HW05 – Neitzke – (56585) 1 This print-out should have 22 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 = 3 4 y about the y -axis between y = 0 and y = 4. 1. V = 10 π cu.units 2. V = 13 π cu.units 3. V = 12 π cu.units correct 4. V = 9 π cu.units 5. V = 11 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 ) = 3 4 y and a = 0 , b = 4, therefore, V = π integraldisplay b a 9 16 y 2 dx = π bracketleftBig 3 16 y 3 bracketrightBig 4 0 . Consequently, V = 12 π cu.units . 002 10.0 points Find the volume of the paraboloid gener- ated by rotating the graph of y = 6 x be- tween x = 0 and x = 2 about the x -axis. 1. volume = 70 π cu.units 2. volume = 71 π cu.units 3. volume = 68 π cu.units 4. volume = 72 π cu.units correct 5. volume = 69 π cu.units Explanation: The solid of revolution generated by rotat- ing the graph of y = f ( x ) about the x -axis between x = a and x = b has volume = π integraldisplay b a f ( x ) 2 dx . When f ( x ) = 6 x, a = 0 , b = 2 , therefore, V = π integraldisplay 2 0 36 x dx = π 2 bracketleftBig 36 x 2 bracketrightBig 2 0 . Consequently, V = 72 π cu.units . keywords: volume, integral, solid of revolu- tion 003 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 y -axis. (Hint: as always graph the region first ). 1. V = 4 cu. units 2. V = 16 3 cu. units 3. V = 16 3 π cu. units 4. V = 8 cu. units

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gutierrez (ig3472) – HW05 – Neitzke – (56585) 2 5. V = 4 π cu. units 6. V = 8 π cu. units correct Explanation: The region rotated about the y -axis 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 ) on the interval [ a, b ] on the y -axis 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 . But after taking square roots we see that x = y 1 / 2 . Thus V = π integraldisplay 4 0 y dy = π bracketleftbigg 1 2 y 2 bracketrightbigg 4 0 . Consequently, V = 8 π . 004 10.0 points Let A be the bounded region enclosed by the graphs of f ( x ) = x , g ( x ) = x 4 . Find the volume of the solid obtained by ro- tating the region A about the line x + 4 = 0 . 1. volume = 11 15 π 2. volume = 71 15 π 3. volume = 41 15 π correct 4. volume = 26 15 π 5. volume = 56 15 π Explanation: The solid is obtained by rotating the shaded region about the line x + 4 = 0 as shown in 1 x + 4 = 0 (not drawn to scale). To compute the volume of this solid we use the washer method. For this we have to express f and g as functions of y : y = f ( x ) = x = x = y , while y = g ( x ) = x 4 = x = y 1 / 4 .
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