Posted by **Anonymous** on Saturday, April 23, 2011 at 4:41pm.

Find the volume of the solid obtained by rotating the region bounded by y= x^2 and x=y^2 about the x-axis.

- calculus -
**drwls**, Saturday, April 23, 2011 at 4:53pm
The bounded region is x =0 to x = 1, so those are your limits of integration.

The integral is of

pi*(x^2 - sqrt x)^2 dx.

Visualize it as a stack of washers with holes in the middle.

- calculus -
**Mgraph**, Saturday, April 23, 2011 at 5:17pm
I think that

the volume=pi*Int(from 0 to 1) (sqrt(x))^2dx-pi*Int(0 to 1)(x^2)^2dx=

pi(1/2-1/5)

- calculus -
**drwls**, Saturday, April 23, 2011 at 5:23pm
Mgraph is correct. I should have written

Integral of pi[(sqrtx)^2 - (x^2)^2] dx

0 to 1

which is pi*(1/2 - 1/5)

- calculus -
**Anonymous**, Saturday, April 23, 2011 at 5:26pm
Thank you so much that makes alot of since.

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