Question

How do you use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region `y=x^2` and `y^2=x` rotated about the x-axis?

Answer 1

To use shells, we take our representative slices parallel to the line we are revolving around. So the thickness will be `dy`, the radius will be `y`.
The height will be the greater `x` (the one on the right) minus the lesser `x` (the one on the left). We need both curves with `x` as a function of `y`.
The limits of integration will be `y` values. That won't matter in this case, because the curves intersect at `(0,0)` and `(1,1)`.

The two curves are: `y=x^2` and `x=y^2`

Solving the first equation for `x`, we get `x=+-sqrty` which does not give `x` as a function of `y` (functions don't say "or"). Don't panic. A quick sketch of the curves shows us that the bounded region has only positive `x` values, so we will use the function: `x=sqrty`.
The second equation already gives `x` as a function of `y`.
Looking again at our sketch of the region, we see that the curve `x=sqrty` is on the right and `x=y^2` is on the left.
The height is `sqrty-y^2`

The volume of the representative shell is:
`2 pi r h * "thickness" = 2 pi y (sqrty-y^2) dy`
And the limits of the region are `y=0` to `y=1`.

`V = int_0^1 2 pi y (sqrty-y^2) dy = 2 pi int_0^1 (y^(3/2)-y^3) dy`

I am sure that you can finish from here.