# How do you use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by y^2=8x and x=2 revolved about the x=4?

May 21, 2018

#### Explanation:

Here is a sketch of the region. To use shells, we'll take a representative slice parallel to the axis of rotation. (Parallel to the line $x = 4$.) The slice is taken at some value of $x$.

The volume of the representative shell is $2 \pi r h \times \text{thickness}$

In this case,

$\text{thickness} = \mathrm{dx}$

the radius $r$ is shown as a dotted black line segment from the slice at $x$ to the line at $4$. So, $r = 4 - x$

The height of the slice is the upper $y$ value minus the lower $y$ value.
Solving ${y}^{2} = 8 x$, we see that ${y}_{\text{upper}} = \sqrt{8 x}$ and ${y}_{\text{lower}} = - \sqrt{8 x}$.
So, $h = \sqrt{8 x} - \left(- \sqrt{8 x}\right) = 2 \sqrt{8 x}$

The volume of the representative shell is $2 \pi \left(4 - x\right) \left(2 \sqrt{8 x}\right) \mathrm{dx}$

$x$ varies from $0$ to $2$, so the volume of the solid is

$V = {\int}_{0}^{2} 2 \pi \left(4 - x\right) \left(4 \sqrt{2 x}\right) \mathrm{dx}$

$= 8 \pi \sqrt{2} {\int}_{0}^{2} \left(4 - x\right) \sqrt{x} \setminus \mathrm{dx}$

$= \frac{896}{15} \pi$