calculus
posted by Anonymous .
Find the volume of the solid obtained by rotating the region bounded by y= x^2 and x=y^2 about the xaxis.

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. 
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) 
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) 
Thank you so much that makes alot of since.