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

Answer 1

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.

Answer 2

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)

Answer 3

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)

Answer 4

Thank you so much that makes alot of since.