Using the shell method, find the volume of the solid of revolution obtained by rotating the region bounded byy=xandy=x2 about thex-axis

<a> pi/2

<b>2pi/5

<c>pi/15

<d>2pi/5

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- Mar 18th 2007, 12:04 PMharryvolume of rotation
Using the shell method, find the volume of the solid of revolution obtained by rotating the region bounded by

*y*=*x*and*y*=*x*2 about the*x*-axis

<a> pi/2

<b>2pi/5

<c>pi/15

<d>2pi/5 - Mar 18th 2007, 01:03 PMJhevon
since we are rotating about the x-axis, we have to do it with respect to y.

y = x

=> x = y

y = x^2

=> x = sqrt(y)

what are our limits of integration?

they are where

sqrt(y) = y

=> y = y^2

=> y^2 - y = 0

y(y - 1) = 0

=> y = 0, y = 1 ......these are our limits.

now see the graph below, we have to look at it sideways now.

note that the height of the piece we want to rotate is the height of the higher graph minus the height of the lower graph. so height = sqrt(y) - y

by the shell method

V = int{(circumference)(height)}dy where circumference is 2pi*r, our radius here is just y

so V = int{2pi*y*h}dy

=> V = int{2pi*y*(sqrt(y) - y)}dy

=> V = 2pi*int{y(y^(1/2) - y}dy

=> V = 2pi*int{y^(3/2) - y^2}dy

=> V = 2pi*[(2/5)y^(5/2) - (1/3)y^3] evaluated between 0 and 1, the limits we found above

=> V = 2pi*[2/5 - 1/3]

=> V = 2pi/15 - Mar 18th 2007, 01:06 PMJhevon
for practice, try doing this by the disk (washer) method to verify the answer.

washer method says:

V = int{pi*(outer radius)^2 - pi*(inner radius)^2}dx

P.S. you've called several threads "multiple choice" now. why not shake things up and call it something more exciting, like "AAAAHHHHH! HEELLPP!!! Shell Method! God Save Us All!" - Mar 18th 2007, 01:16 PMJhevon
as always, if there's anything you don't get, say so, and myself or someone else will help you.