Question

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?

Answer 1

`(8pi)/3`

Explanation:

use disc method to integrate the volume:
volume = `piint_a^br(x)dx` where `r(x)` 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(x)` is the difference between `x^2-2` and `-2`, so `r(x)=x^2-2-(-2)=x^2`

plugging in: volume = `piint_0^2x^2dx`
`pi(F(2)-F(0))`, where `F(x)=1/3x^3`, or the integral of `x^2`
`=pi(1/3(2)^3-1/3(0)^3)`
`=pi(8/3)=(8pi)/3`