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 xaxis.

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))^2dxpi*Int(0 to 1)(x^2)^2dx=
pi(1/21/5)

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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