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#?

To find the volume of the solid obtained by rotating the region bounded by the curves $y=\sqrt{x}$ and $y=\frac{x}{3}$ around the line $x=-1$, you can use the method of cylindrical shells.

The limits of integration will be determined by the points where the two curves intersect. Setting $\sqrt{x} = \frac{x}{3}$, you can solve for $x$ to find the intersection point.

$\sqrt{x} = \frac{x}{3}$
$x = \frac{x^2}{9}$
$9x = x^2$
$x^2 - 9x = 0$
$x(x - 9) = 0$

$x = 0$ or $x = 9$

So, the limits of integration will be $x = 0$ to $x = 9$.

The radius $r$ of the cylindrical shell at $x$ is $x + 1$, because the axis of rotation is $x = -1$. The height $h$ of the cylindrical shell at $x$ is the difference between the $y$-values of the upper and lower curves, which is $\sqrt{x} - \frac{x}{3}$.

The volume $V$ of each cylindrical shell is given by $V = 2\pi rh$.

Therefore, the total volume is given by the integral:

$V = \int_{0}^{9} 2\pi(x + 1)\left(\sqrt{x} - \frac{x}{3}\right) dx$