M408L hwk4 solutions

# M408L hwk4 solutions - Pham Quoc Homework 4 Due 3:00 am...

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Pham, Quoc – Homework 4 – Due: Feb 13 2007, 3:00 am – Inst: Eric Katerman 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Find the volume of the paraboloid gener- ated by rotating the parabola y = 6 x about the x -axis between x = 0 and x = 2. 1. V = 70 π cu.units 2. V = 73 π cu.units 3. V = 71 π cu.units 4. V = 69 π cu.units 5. V = 72 π cu.units correct Explanation: The volume, V , of the solid of revolution generated by rotating the graph of y = f ( x ) about the x -axis between x = a and x = b is given by V = π Z b a f ( x ) 2 dx. When f ( x ) = 6 x and a = 0 , b = 2, there- fore, V = π Z 2 0 36 x dx = π 2 h 36 x 2 i 2 0 . Conequently, V = 72 π cu.units . keywords: volume, integral, solid of revolu- tion 002 (part 1 of 1) 10 points Find the volume, V , of the solid obtained by rotating the region bounded by y = x 2 , x = 0 , y = 16 about the y -axis. (Hint: as always graph the region first ). 1. V = 128 π cu. units correct 2. V = 128 cu. units 3. V = 256 3 π cu. units 4. V = 64 cu. units 5. V = 256 3 cu. units 6. V = 64 π cu. units Explanation: The region rotated about the y -axis is sim- ilar to the shaded region in 16 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 Z 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 Z 16 0 y dy = π 1 2 y 2 16 0 .

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Pham, Quoc – Homework 4 – Due: Feb 13 2007, 3:00 am – Inst: Eric Katerman 2 Consequently, V = 128 π . keywords: volume, integral, solid of revolu- tion 003 (part 1 of 1) 10 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 3 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 correct 4. V = 7 20 cu. units 5. V = 2 5 cu. units 6. V = 13 30 π cu. units Explanation: Since the graphs of y = x 3 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 Z 1 0 n ( x 1 / 3 ) 2 - ( x 3 2 ) 2 o dx = π Z 1 0 n x 2 3 - x 3 o dx = π 3 5 x 5 3 - 1 4 x 4 1 0 . Consequently, V = π 3 5 - 1 4 · = 7 20 π cu. units . keywords: volume, volume of revolution, ra- tional powers, disc method, washer method, 004 (part 1 of 1) 10 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 + 1 = 0 . 1. volume = 29 15 π 2. volume = 44 15 π 3. volume = - 1 15 π 4. volume = 14 15 π correct 5. volume = - 16 15 π Explanation: The solid is obtained by rotating the shaded region about the line x + 1 = 0 as shown in
Pham, Quoc – Homework 4 – Due: Feb 13 2007, 3:00 am – Inst: Eric Katerman 3 1 x + 1 = 0 (not drawn to scale). To compute the volume of this solid we use the washer method.

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