# How do you find the volume of the solid obtained by rotating the region bounded by the curves y=sqrtx and y=x/3 rotated around the x=-1?

Oct 9, 2015

See the explanation section, below.

#### Explanation:

Here is a picture of the region (in blue) and the line $x = - 1$. It shows a representative slice of thickness $\mathrm{dx}$. The slice is taken parallel to the axis of rotation, so when we rotate we'll get cylindrical shell.
The volume of the shell is $2 \pi r h \mathrm{dx}$. The radius, $r$, is shown by the dotted red horizontal line. Its length is $x + 1$.
The height of the shell is the greater $y$ value minus the lesser $y$ value. That is: $h = \sqrt{x} - \frac{x}{3}$

We also need the point of intersection where $\frac{x}{3} = \sqrt{3}$. Which happens at $0$ and at $9$

The integral we need to evaluate is:

${\int}_{0}^{9} 2 \pi \left(x + 1\right) \left(\sqrt{x} - \frac{x}{3}\right) \mathrm{dx}$

Expand the integrand to get 4 terms in powers of $x$ which is straightforward to evaluate.

$2 \pi {\int}_{0}^{9} \left(x + 1\right) \left(\sqrt{x} - \frac{x}{3}\right) \mathrm{dx} = \frac{207}{5} \pi$