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Homework 4

# Homework 4 - tgo72 Homework 04 Gompf(58370 This print-out...

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tgo72 – Homework 04 – Gompf – (58370) 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 Find the volume, V , of the solid obtained by rotating the region bounded by y = 4 x , x = 2 , x = 6 , y = 0 about the x -axis. 1. V = 16 3 π correct 2. V = 16 3 3. V = 4 3 π 4. V = 4 3 5. V = 8 3 6. V = 8 3 π Explanation: The volume of the solid of revolution ob- tained by rotating the graph of y = f ( x ) on [ a, b ] about the x -axis is given by volume = π integraldisplay b a f ( x ) 2 dx . When f ( x ) = 4 x , a = 2 , b = 6 , therefore, V = π integraldisplay 6 2 16 x 2 dx . Consequently, V = π bracketleftbigg - 16 x bracketrightbigg 6 2 = 16 3 π . 002 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 2 , x = y 4 about the x -axis. 1. V = 7 15 π cu. units correct 2. V = 1 2 cu. units 3. V = 5 12 cu. units 4. V = 5 12 π cu. units 5. V = 1 2 π cu. units 6. V = 7 15 cu. units Explanation: Since the graphs of y = x 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

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tgo72 – Homework 04 – Gompf – (58370) 2 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 / 4 ) 2 - ( x 2 ) 2 bracerightBig dx = π integraldisplay 1 0 braceleftBig x 1 2 - x 4 bracerightBig dx = π bracketleftbigg 2 3 x 3 2 - 1 5 x 5 bracketrightbigg 1 0 . Consequently, V = π parenleftBig 2 3 - 1 5 parenrightBig = 7 15 π cu. units . 003 10.0 points Find the volume, V , of the solid formed by rotating the region bounded by the graphs of y = x + 5 , y = 5 , x = 0 , x = 1 about the line y = 4. 1. V = 5 3 π cu. units 2. V = 4 3 π cu. units 3. V = 7 6 π cu. units 4. V = 3 2 π cu. units 5. V = 11 6 π cu. units correct Explanation: The region bounded by the given curves is shown as the shaded area in 5 4 1 y x (not drawn to scale) and the line about which it is rotated is shown as the dotted line. Thus the volume generated when rotating this shaded region about the dotted line y = 4 is given by V = π integraldisplay 1 0 braceleftBig ( y - 4) 2 - (5 - 4) 2 bracerightBig dx = π integraldisplay 1 0 braceleftBig ( x + 1) 2 - 1 bracerightBig dx = π integraldisplay 1 0 braceleftBig x + 2 x bracerightBig dx . Consequently, V = π bracketleftbigg 1 2 x 2 + 4 3 x 3 / 2 bracketrightbigg 1 0 = 11 6 π . 004 10.0 points A cap of a sphere is generated by rotating the shaded region in y 1 3 about the y -axis. Determine the volume of this cap when the radius of the sphere is 3 inches and the height of the cap is 1 inch.
tgo72 – Homework 04 – Gompf – (58370) 3 1. volume = 10 3 π cu. ins 2. volume = 3 π cu. ins 3. volume = 7 3 π cu. ins 4. volume = 8 3 π cu. ins correct 5. volume = 2 π cu. ins Explanation: Since the sphere has radius r = 3 inches, we can think of this sphere as being generated by rotating the circle x 2 + y 2 = 3 2 about the y - axis. The cap of the sphere is then generated by rotating the the graph of x = f ( y ) = radicalbig 3 2 - y 2 on the interval [2 , 3] about the y -axis. Thus the volume of the cap is V = π integraldisplay 3 2 f

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Homework 4 - tgo72 Homework 04 Gompf(58370 This print-out...

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