# How do you use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by y=8-x^2, y=x^2 revolved about the x=2?

Nov 16, 2016

#### Explanation:

In the following picture, the region is shown in blue. The points of intersection of the two curves are $\left(\pm 2 , 4\right)$. The axis of rotation (the line $x = 2$) is in red.

To set up for the method of shells, a slice has been taken parallel to the axis of rotation. The thin side (the thickness of the shell) is $\mathrm{dx}$. The slice has black borders and the radius of rotation is shown as a dashed black line. The Volume of a representative shell is

$2 \pi r h \text{thickness}$

Since the thickness is
$\mathrm{dx}$

(the variable we will work with is $x$) we note that $x$ varies from $- 2$ to $2$.

The radius of the shell is
$r = 2 - x$.

The height is the upper curve minus the lower curve

$h = \left(8 - {x}^{2}\right) - \left({x}^{2}\right) = 8 - 2 {x}^{2}$

So the volume of the solid of revolution is

$V = {\int}_{-} {2}^{2} 2 \pi {\underbrace{\left(2 - x\right)}}_{r} {\underbrace{\left(8 - 2 {x}^{2}\right)}}_{h} {\overbrace{\mathrm{dx}}}^{\text{thickness}}$

$= 2 \pi \left[\frac{128}{3}\right] = \frac{256 \pi}{3}$