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=x^2# rotated about the x axis?

Graph the region, including the points of intersection, #(-1,1)# and #(2,4)#.

In order to use shells, we must take our representative slices parallel to the axis of rotation. So the thickness of each shell will be #dy#.

The thickness of each shell will be #dy# and we will be integrating with respect to #y#. So, we need to express the boundaries as function of #y#:

#x=y-2# and #x= +- sqrty#

And the limits of integration are #0# to #4#

The volume of the representative shell is #2pi*"radius"*"height"*dy#

The radius of each shell will be #y#

The "height" of each cylindrical shell will be the distance between the #x# values. This will be the #x# on the right (#x_"right"#) minus the #x# on the left (#x_"left"#).

Notice in the picture, that as #y# goes from #0# to #4#, the values of #x_"right"# and #x_"left"# change.

From #y=0# to #y=1# , we have:
#"height" = x_"right" - x_"left" = sqrty - (-sqrty) = 2sqrty = (8pi)/5#

So we need

#int_0^1 2 pi y (2sqrty) dy = 4pi int_0^1 y^(3/2) dy#

From #y=1# to #y=4# , we have:
#"height" = x_"right" - x_"left" = sqrty - (y-2)#

So we need

#int_1^4 2 pi y (sqrty-y+2) dy = 2pi int_1^4 (y^(3/2)-y^2+2y) dy = (64pi)/5#

The total volume is #(8pi)/5 + (84pi)/5 = 72/5 pi#