# How do you find the volume of the solid generated by revolving the region bounded by the curves y=x^3 and y=x^4 rotated about the y-axis?

Aug 26, 2015

This is something that can be done two ways---the earlier way and the Shell Method. The Shell Method isn't too difficult to apply.

In this method, since you are rotating about the y-axis, your thickness $f \left(x\right)$ is the function farther right minus the function farther left. If you get a negative volume at the end, you know you did it backwards.

graph{(x^3 - y)(x^4 - y)sqrt(0.5^2 - (x-0.5)^2)/sqrt(0.5^2 - (x-0.5)^2) <= 0.00 [-1, 2, -0.095, 2]}

from $x = 0$ to $x = 1$.

The formula uses the idea of circumference (the $2 \pi x$) with the varying thickness $f \left(x\right)$ of the solid along the vertical direction to build the solid by stacking shells vertically, while the radius $x$ indicates the distance from the axis of rotation; since the axis of rotation is the y-axis, the radius is simply $x$, spanning $0$ to $\pm 1$.

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

$= 2 \pi \int x \left({x}^{3} - {x}^{4}\right) \mathrm{dx}$

$= 2 \pi \int {x}^{4} - {x}^{5} \mathrm{dx}$

$= 2 \pi \left[\frac{1}{5} {x}^{5} - \frac{1}{6} {x}^{6}\right] {|}_{0}^{1}$

$= 2 \pi \left[\left(\frac{1}{5} - \frac{1}{6}\right) - 0\right]$

$= \textcolor{b l u e}{\frac{\pi}{15} {\text{u}}^{3}}$