How do you find the volume of a solid that is enclosed by y=x^2-2, y=-2, and x=2 revolved about y=-2?

Dec 24, 2017

$\frac{8 \pi}{3}$

Explanation:

use disc method to integrate the volume:
volume = $\pi {\int}_{a}^{b} r \left(x\right) \mathrm{dx}$ where $r \left(x\right)$ is the distance from a certain point on $y = {x}^{2} - 2$ to the axis of rotation, $y = - 2$

to find a and b, find where $y = {x}^{2} - 2$ intersects with $y = - 2$:
${x}^{2} - 2 = - 2 , {x}^{2} = 0 , x = 0$ this means the vertex of the parabola $y = {x}^{2} - 2$ lies on $y = - 2$.
one of the bounds is $x = 2$ as given from the problem, so the values for a and b are: $a = 0$ and $b = 2$

$r \left(x\right)$ is the difference between ${x}^{2} - 2$ and $- 2$, so $r \left(x\right) = {x}^{2} - 2 - \left(- 2\right) = {x}^{2}$

plugging in: volume = $\pi {\int}_{0}^{2} {x}^{2} \mathrm{dx}$
$\pi \left(F \left(2\right) - F \left(0\right)\right)$, where $F \left(x\right) = \frac{1}{3} {x}^{3}$, or the integral of ${x}^{2}$
$= \pi \left(\frac{1}{3} {\left(2\right)}^{3} - \frac{1}{3} {\left(0\right)}^{3}\right)$
$= \pi \left(\frac{8}{3}\right) = \frac{8 \pi}{3}$