# Use the washer method to find the volume of an area rotated about y=4

Here's my problem:

Find the volume of the solid generated by revolving the "triangular" region bounded by the curve $y=\frac{4}{x^3}$ and the lines $x=1$ and $y=1/2$ about $y=4$.

I have the graph drawn up and I know I would like to use the washer method (to stay in terms of $x$) but I don't know what my inner and outer radii would be. I think the inner radius would be $4-\frac{4}{x^3}$, and the integral would be from 1 to 2 but I honestly can't figure out what needs to happen for the outer radius. Please help!

You must draw the region to understand how to set up the integral. You should also draw the region rotated $180^\circ$ about the line of revolution as I have done in the following desmos.com graph.
Here we can clearly see that the inner radius is $r=4-\dfrac{4}{x^3}$ and that the outer radius is $R=4-\frac{1}{2}=\frac{7}{2}$.
You are correct about the interval of integration being $[1,2]$.
$$V=\int_1^2\pi\left(R^2-r^2\right)\,dx$$
• Wouldn't the inner radius be $r$, and the outer radius $R$ so that the integral would be: $V=\int_1^2\pi\left((\frac{7}{2})^2-(4-\frac{4}{x^3})^2\right)\,dx$. I thought you subtracted the inner from the outer? – calmcalculus Feb 7 '17 at 23:32
• Yes, I accidentally reversed the $R$ and $r$ in the line below the graph. Will make that correction. – John Wayland Bales Feb 8 '17 at 1:53