# How do you find the volume of the solid generated by revolving the region bounded by the curves x=y-y^2 rotated about the y-axis?

Oct 26, 2015

See the explanation section below.

#### Explanation:

Here is a picture of the region with a representative slice taken perpendicular to the axis of rotation. This is a set-up to use disks to find the volume. The thickness is $\mathrm{dy}$

As $y$ varies from $0$ to $1$, the disk at $y$ has radius $y - {y}^{2}$ and thickness $\mathrm{dy}$.

The volume of the representative disk is
$\pi \times \text{radius"^2xx d"thickness} = \pi {\left(y - {y}^{2}\right)}^{2} \mathrm{dy}$.

So, to find the volume of the solid, we need to evaluate

${\int}_{0}^{1} \pi {\left(y - {y}^{2}\right)}^{2} \mathrm{dy} = \pi {\int}_{0}^{1} \left({y}^{2} - 2 {y}^{3} + {y}^{4}\right) \mathrm{dy}$.

This integral evaluates to $\frac{\pi}{30}$.

Bonus

Here is a link to the same volume problem using shells instead of disks.

http://socratic.org/questions/how-do-you-use-the-shell-method-to-set-up-and-evaluate-the-integral-that-gives-t-38179893