# Find the volume of the solid by rotating the region

Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line.

$$\left\{\begin{array}{ll}x&=y-y^2\\x&=0\end{array}\right..$$ About the $y$-axis

I'm not sure where to start, so I don't have a specific question. The reversal of the varibles is throwing me off.

• It will be the same as the volume of the solid obtained by rotating the region bounded by the curves $y=x-x^2$ and $y=0$ about the $x$-axis. – Henry Nov 28 '15 at 21:23

Well, the curve you have described looks like this

And the idea in these kind of problems, is that you take that plot and rotate it around the indicated axis. The 'tail' left by the curve in the rotation, is the surface that encloses the solid. If that is not intuitive, here is a picture

So, how to write the region of integration?. Using cylindrical coordinates, $(x,y,z)=(r cos(\theta), y, r sin(\theta))$, $V = \{(r,\theta,y): 0 \leq \theta \leq 2 \pi , 0\leq y \leq 1, 0\leq r \leq y-y^2\}$.

Volume = $\int_{V} 1 dV$ = $\int_0^{2 \pi} \int_0^1 \int_0^{y-y^2} 1 r$ $dr dy d\theta$ = $2 \pi$ $\int_0^1 \frac{r^2}{2} \bigg|_{r=0}^{y-y^2}dz$ = $\frac{\pi}{30}$

These problems require a bit of visualization. So image the point $(x,y)$. Now imagine rotating it about the $y$-axis.

What shape will it make?

A circle

$x$

Now imagine we take a rectangular strip with width $x$ and height $\Delta y$ and rotate it about the $y$-axis. It will make a cylinder, and It's volume will be:

$$\pi x^2 \Delta y$$

Now we can approximate the volume of rotating $x=f(y)$ about the $y$ axis by rotating some (appropriate) strips and summing up the volumes made. If we make $\Delta y$ Infinitely small and sum all the volumes of these cross sections. This is nothing but:

$$\int \pi x^2 dy$$

Now for your question you need to know over what interval of $y$ are you going to be integrating. Look at your region and see that when $y=0$ and when $y=1$, $x=0$.

Thus the integral for your volume is:

$$v=\int_{0}^{1} \pi x^2 dy$$

$$v=\int_{0}^{1} \pi (y-y^2)^2 dy$$

The curve $x = y - y^2$ crosses the $x$-axis twice, at $y = 0$ and $y = 1$, so these will be the endpoints of the object we have by rotating our curve. The volume is $$\pi \int^{1}_{0} (y - y^2)^2 dy = \pi \left (\frac {y^3} {3} - \frac {y^4} {2} + \frac {y^5} {5} \right) |^1_0 = \frac {\pi} {30}.$$