# How do you find the volume of the solid generated by revolving the region bounded by the curves y = x-2, the x-axis, x=2, and x=4 rotated about the x=-1?

Aug 19, 2015

Let's see what this looks like:

graph{(y+2-x)(sqrt(2-(x-2)))/(sqrt(2-(x-2))) <= 0 [-5, 6, -0.3, 6]}

Revolving around $x = - 1$ means doing the shell method makes this easier (you're revolving to make a shell, and you don't have to convert to $f \left(y\right)$ functions).

$V = 2 \pi {\int}_{a}^{b} x f \left(x\right) \mathrm{dx}$

where:
$x$ is the radius of the shell
$f \left(x\right)$ is the thickness of the shell
$2 \pi$ indicates that the shell is fully revolved ($2 \pi$ radians = 1 full revolution)

Now it's just a matter of figuring out what your equations are.

$f \left(x\right) = x - 2$
(the thickness is from the inside surface of the shell to the outside surface, which is what $f \left(x\right)$ from 2 to 4 revolves to become)

Interval: $\left[2 , 4\right]$
(implied by $x = 2$ and $x = 4$ in the question)

The radius of the shell is determined by the distance from the axis of $x = - 1$ to the current value of $x$ in the interval $\left[2 , 4\right]$. Therefore, the radius is $x - \left(- 1\right) = x + 1$.

Overall, we get:

$V = 2 \pi {\int}_{2}^{4} \left(x + 1\right) \left(x - 2\right) \mathrm{dx}$

$= 2 \pi {\int}_{2}^{4} {x}^{2} - x - 2 \mathrm{dx}$

$= 2 \pi \left[{x}^{3} / 3 - {x}^{2} / 2 - 2 x\right] {|}_{2}^{4}$

$= 2 \pi \left[\left({4}^{3} / 3 - {4}^{2} / 2 - 2 \left(4\right)\right) - \left({2}^{3} / 3 - {2}^{2} / 2 - 2 \left(2\right)\right)\right]$

$= 2 \pi \left[\left(\frac{64}{3} - \frac{16}{2} - 8\right) - \left(\frac{8}{3} - \frac{4}{2} - 4\right)\right]$

$= 2 \pi \left[\left(\frac{64}{3} - \frac{48}{3}\right) - \left(\frac{8}{3} - \frac{18}{3}\right)\right]$

$= 2 \pi \left[\frac{64}{3} - \frac{48}{3} - \frac{8}{3} + \frac{18}{3}\right]$

$= 2 \pi \left[\frac{26}{3}\right]$

$\textcolor{b l u e}{= \frac{52 \pi}{3} {\text{u}}^{3}}$