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?

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.