# How do you find the volume of the solid if the region in the first quadrant bounded by the curves x=y-y^2 and the y axis is revolved about the y axis?

##### 1 Answer
May 7, 2015

I've edited your question to what I think you meant to ask. If I'm mistaken, please accept my apologies.

The curve $x = y - {y}^{2}$ is a parabola that opens to the left. It has $y$ intercepts $0$ and $1$.

Using discs, we see that the radius is $x$ or $y - {y}^{2}$, the thickness is $\mathrm{dy}$, so the volume or a representative disc is

$\pi {r}^{2} d 9 y = \pi {\left(y - {y}^{2}\right)}^{2} \mathrm{dy}$

Integrate from $y = 0$ to $y = 1$

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

Which I believe is $\frac{17 \pi}{60}$