# 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?

Sep 26, 2015

See the explanation section below.

#### Explanation:

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

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

The thickness of the shell is $\mathrm{dy}$

The curve $y = \sqrt{x}$ 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).

$h = \left(4 - {y}^{2}\right)$

The volume of a representative shell is $2 \pi r h \mathrm{dy}$.

So we need to integrate:

${\int}_{0}^{2} 2 \pi \left(6 - y\right) \left(4 - {y}^{2}\right) \mathrm{dy}$

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