# Washers and Integrals

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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The inner radius is $0$, there is no "hole." –  André Nicolas Dec 15 '12 at 18:32

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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It just sounded funny how you said that the function is "clearly" defined on $[-1,1]$. :P But good answer. +1 –  Patrick Da Silva Dec 15 '12 at 18:37
@PatrickDaSilva: Yeah, now that I read it again I agree that sounds a bit strange. I guess the fraction just made me automatically make the mental note that it actually is defined on this interval :) –  Thomas Dec 15 '12 at 18:39

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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The left graph is misleading in that it doesn't actually include the $x$ axis (which is to be rotated about). The right one appears not to have anything to do with the problem. –  Henning Makholm Dec 15 '12 at 19:31
@HenningMakholm This might be the plot you are looking for. –  Fabian Dec 15 '12 at 19:37
Sorry! I misread the axis of revolution as "y axis". I will fix the pic. –  JohnD Dec 15 '12 at 19:43
Can you tell me the program you use to get the visualization please? –  extremez Dec 2 '13 at 1:03
@extremez: I did these in Mathematica. –  JohnD Dec 2 '13 at 2:40