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

To use shells when revoling about a vertical line, we need to take our representative slices vertically, so the thickness is #dx#. And we need the boundaries written as functions of #x#.

#y = sqrtx#, #y=-sqrtx#, #y=x-2#

Here is a picture of the region to be revolved about the #y# axis. I have included two representative slices of the region.

The curves intersect at #(1,-1)# and at #(4,2)#

The radii of our cylindrical shells will be #x# from #x=0# to #x=4#.
The thickness is #dx# and
the height of each shell is the greater #y# at #x# - lesser #y# at #x#.
(The #y# above - #y# below#)

The greater value of #y# can be found using #y=sqrtx# for all #x# from #0# to #4#.
The lesser value of #y# changes at #x=3#. So we'll need to evaluate two integrals.

From #x=0# to #x=1#, the lesser #y# value is #-sqrtx#, so the height is #sqrtx-(-sqrtx) = 2sqrtx#

To get that part of the volume, we need to evaluate:

#int_0^1 2pirhdx = 2pi int_0^1 x(2sqrtx)dx = 4pi int_0^1 x^(3/2) dx = 8/5pi#

From #x=1# to #x=4#, the lesser #y# value is #x-2#, so the height is #sqrtx-(x-2) = sqrtx-x+2#

To get that part of the volume, we need to evaluate:

#int_1^4 2pirhdx = 2pi int_1^4 x(sqrtx-x+2)dx = 2pi int_1^4 (x^(3/2) - x^2+2x) dx = 64/5pi#

The total volume is found by adding these two results, to get:

#72/5pi#