# How do you find the volume of the solid generated by revolving the region bounded by the graphs x=y^2, x=4, about the line x=6?

Mar 31, 2017

#### Explanation:

The graph of the region is shown below. A representative slice of thickness $\mathrm{dy}$ has been taken at height $y$. The radii of rotation are shown as dashed and dotted red lines.

When revolved about the line $x = 6$ the resulting slice has volume

$\left(\pi {R}^{2} - \pi {r}^{2}\right) \cdot \text{thickness}$ where $R$ is the greater radius and $r$ the lesser.

In this problem

$R = 6 - {y}^{2}$ and
$r = 6 - 4 = 2$ and
$\text{thickness} = \mathrm{dy}$.

$y$ varies from $- 2$ to $2$. (Or go from $0$ to $2$ and double the result.)

So the volume of the solid of revolution is

$V = \pi {\int}_{-} {2}^{2} \left({\left(6 - {y}^{2}\right)}^{2} - {\left(2\right)}^{2}\right) \mathrm{dy} = \pi {\int}_{-} {2}^{2} \left(32 - 12 {y}^{2} + {y}^{4}\right) \mathrm{dy}$

$= \frac{384 \pi}{5}$

or about $241.3$.