So I'm working on washers and I was given the equation of $$1/\sqrt{1+x^2}$$ and I am supposed to rotate the solid around the $x$-axis on the interval of $[-1,1]$. I know that I am supposed to use washers, but I can't figure out how to find the equation for the outer radius and the inner radius.
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You have the functions $$f(x) = \frac{1}{\sqrt{1 + x^2}}.$$ The function is clearly defined on the interval $[-1,1]$. When you rotate the graph around the $x$-axis, then get a solid. If you try to draw a picture it might help to see that for a fixed $x$ the "inner radius" in this case is simply $0$. The outer radius is $f(x)$. So the volume is $$ \int_{-1}^{1} \pi f(x)^2 \; dx. $$ Unless you ask, I will let you find this integral. |
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Be sure you have a clear visualization of what you are working with (and trying to accomplish):
Rotate the region on the left about the $x$ axis to generate the solid on the right. (It is drawn to look hollow, but imagine it as a solid region.) The arrows on the left indicate some representative radii of the disks you get when you rotate about the $x$ axis to generate the solid. As mentioned above, this is really using disks rather than washers since there is no hole in the solid of revolution. |
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