# Method of cylindrical shells

Use the method of cylindrical shells to find the volume V generated by rotating the region bounded by the given curves about the specified axis.

$y=32-x^2, \ y=x^2$ about the line $x=4$

My confusion is that what will be the radius 'x' of cylinder shells which we have to put in the integral.

Let $-4\le x\le 4$. Draw a thin vertical strip of width "$dx$" at $x$. For the picture, let $x$ for example be $1.5$.
Rotate this thin strip about the line $x=4$. We get a cylindrical shell. This shell has height $(32-x^2)-x^2$. Its distance from the line $x=4$ is $4-x$. That is the radius of the cylindrical shell. It follows that the volume is equal to $$\int_{-4}^4 2\pi(4-x)(32-2x^2)\,dx.$$
• You are welcome. You drew almost the right picture. You needed in addition to draw or visualize the thin strip that would be rotated. Then you can more or less read off the radius. It is the distance from the strip to the line you are rotating around. To protect against minor errors, check whether the formula you come up with, in our case $4-x$, works at some typical values of $x$ such as $x=0$, or $x=-4$. The height of the shell can also be more or less read off from the picture. Commented Dec 20, 2015 at 6:38
Cylindrical shell: consider the volume element $$dV = 2 \pi h\,dr = 2\pi\,2y\,dx = 4\pi\,x^2dx$$ Then integrate $$V = \int_{x=-4}^4 dV = 4\pi \int_{-4}^4 x^2dx$$
Cylindrical disk: Cut your solid in half so that you only have to consider the bottom part. Then integrate the volume element $$dV = \pi x^2 dy = \pi (2\sqrt{y})^2 dy$$ $$V = \int_{y=0}^{16} dV = 4\pi \int_0^{16} y\,dy$$