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

##### 1 Answer
Feb 7, 2018

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 $\mathrm{dx}$ and two radios:
$R = 1 + {x}^{3}$ and $r = 1$.
So, the volume of this volume element is $\mathrm{dV} = \pi \left({R}^{2} - {r}^{2}\right) \mathrm{dx}$
The sbstitution in that formula give us:

$\mathrm{dV} = \pi \left({\left(1 + {x}^{3}\right)}^{2} - {1}^{2}\right) \mathrm{dx}$

Integrating between 0 and 3 wil give us the volume requested

${\int}_{0}^{3} \pi \left(1 + 2 {x}^{3} + {x}^{6} - 1\right) \mathrm{dx} = {\int}_{0}^{3} \pi \left(2 {x}^{3} + {x}^{6}\right) \mathrm{dx}$

solving this integral we have:

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