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 = 1 + x^2#, #y = 0#, #x = 0#, #x = 2# rotated about the x-axis?

Here is the region to be revolved about the #x# axis:

If we must use shells, we will need to take our representative slices horizontally and use thickness #dy#.

We need to rewrite the boundaries as functions of #y# so the curve becomes #x=sqrt(y-1)#.
Note that, in the region #y# varies from #0# to #5#. The radius of the shells will involve #y# values.
For the 'height' of the shell (which is lying on its side), we will use the #x# on the right minus the #x# on the left.

For #y = 0# to #1#, the 'height' is #2-0=2#, The resulting solid is a cylinder with radius #1# and height #2#. Its volume is #2pi#.
(We could do an integral for this, but we learned the volume of a cylinder in geometry class.)

From #y=1# to #y=5#, the 'height' is #2-sqrt(y-1)#.
The radius of the representative shell is #y# and the thickness is #dy#

We integrate:

#int 2 pi rh dy = 2 pi int_1^5 y(2-sqrt(y-1))dy = (176pi)/15#

Adding the two volumes, we get:

#V = 2pi+176/15 pi = 206/15 pi#

Note
If we had been allowed to use disks, we would integrate:

#pi int_0^2 (1+x^2)^2 dx = pi int_0^2 (1+2x^2+x^4) dx#

# = pi [x+(2x^3)/3 + x^5/5]_0^2#

# = pi [2+16/3+32/5] = pi [(30+80+96)/15] = 206/15 pi#

Of course, we get the same answer, but we only need to do one calculation and the integral is (I think) simpler.