# How do you find the volume of the solid generated by revolving the region bounded by the curves y = x² and y =1 rotated about the y=-2?

Sep 29, 2015

See the explanation below.

#### Explanation:

Here is a picture of the region and a vertical slice. The picture is set up to use washers (disks).
Thickness is $\mathrm{dx}$
$x$ values go from $- 1$ to $1$
the radius of the larger washer is the greater $y$ minus $- 2$ (the line we are revolving about is $y = - 2$)
$R = 1 - \left(- 2\right) = 3$
the radius of the smaller washer is the lesser $y$ minus $- 2$
$r = {x}^{2} - \left(- 2\right) = {x}^{2} + 2$

The representative slice has volume $\pi \left({R}^{2} - {r}^{2}\right) \mathrm{dx}$.

We need to evaluate the integral

int_-1^1 pi((3^2-(x^2+2)^2)dx=pi int_-1^1(5-4x^2-x^4)dx

$= \frac{104}{15} \pi$

Shells Method
If we had taken a slice horizontally: This is set up to use cylindrical shells of thickness $\mathrm{dy}$

The volume of each shell is $2 \pi \left(\text{radius")("height")("thickness}\right)$ with thickness $\mathrm{dy}$, that is $2 \pi \cdot r h \cdot \mathrm{dy}$
In the region, the $y$ values go from $0$ to $1$.
The radius of the representative shell is $y + 2$ (dotted line in picture)
The height goes from the greater $x$ (the one on the right) to the lesser $x$ (the one on the left).
We need to rewrite the boundary as functions of $y$ instead of $x$. $y = {x}^{2}$ becomes the two functions $x = - \sqrt{y}$ (on the left) and $x = \sqrt{y}$ (on the right). The height of the shell is $\sqrt{y} - \left(- \sqrt{y}\right) = 2 \sqrt{y}$

The representative shell has volume $2 \pi \left(y + 2\right) \left(2 \sqrt{y}\right) \mathrm{dy}$.

The solid has volume

$V = {\int}_{0}^{1} 2 \pi \left(y + 2\right) \left(2 \sqrt{y}\right) \mathrm{dy} = 4 \pi {\int}_{0}^{1} \left({y}^{\frac{3}{2}} + 2 {y}^{\frac{1}{2}}\right) \mathrm{dy}$

$= 4 \pi \left(\frac{26}{15}\right) = \frac{104}{15} \pi$