Question

What is the volume of the solid produced by revolving `f(x)=x^3, x in [0,3]`around `y=-1`?

Answer 1

See explanation below

Explanation:

We have a revolution solid with a hole inside because the spin axis is y=-1.
Consider a volume element with a width of `dx` and two radios:
`R= 1 + x^3` and `r=1`.
So, the volume of this volume element is `dV=pi(R^2-r^2)dx`
The sbstitution in that formula give us:

`dV=pi((1+x^3)^2-1^2)dx`

Integrating between 0 and 3 wil give us the volume requested

`int_0^3 pi(1+2x^3+x^6-1)dx=int_0^3pi(2x^3+x^6)dx`

solving this integral we have:

`V=2pix^4/4+x^7/7}_0^3=81pi/2+2187pi/7=4941/14pi`