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 = sqrt(x)#; #y = 0#; and #x = 4# rotated about #y=6#?

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Answer 1 · 2015-09-26T18:19:23

See the explanation section below.

Here is a picture of the region and the line #y=6#:

enter image source here

I've tried to show a representative slice to use cylindrical shells.

The thickness of the shell is #dy#

The curve #y=sqrtx# needs to be re-described as #x=y^2#

In the region, #y# varies from #0# to #2#.

At each value of #y#, the shell has radius #6-y#

and the 'height' of the shell (it is lying on its side) will be the greater #x# (the one on the right) minus the lesser #x# (on the left).

#h = (4-y^2)#

The volume of a representative shell is #2 pi r h dy#.

So we need to integrate:

#int_0^2 2 pi (6-y)(4-y^2)dy#

Expand the polynomial and integrate term by term to get #V=56pi#