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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!

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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.

solid of revolution

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 $[1,2]$.

Then of course you use

$$ V=\int_1^2\pi\left(R^2-r^2\right)\,dx $$

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  • $\begingroup$ 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? $\endgroup$ – calmcalculus Feb 7 '17 at 23:32
  • $\begingroup$ Yes, I accidentally reversed the $R$ and $r$ in the line below the graph. Will make that correction. $\endgroup$ – John Wayland Bales Feb 8 '17 at 1:53

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