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How do I calculate the volume of a solid revolution when the axis of revolution is NOT the x or y axis? I thought you do \begin{equation} π∫_a^b f^2(x-c) - g^2(x-c) dx \end{equation} where y=c (a horizontal line) is the axis of revolution, but it doesn't always work. It seems like sometimes I'm supposed to do (c-x) instead, but I can't figure out why. Can anyone explain this to me?

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You can always rotate your coordinate system so that the axis of revolution is a coordinate axis... – J. M. May 2 '12 at 17:41
@J.M.: I think that SRK's problem is with an axis of the form $x=c$ when $c\ne 0$, not with an oblique axis of revolution. – Brian M. Scott May 2 '12 at 17:45
SRK, if it is as @Brian describes, then things are easier; you can just shift your axis... – J. M. May 2 '12 at 17:51
You seem to be looking for a general formula. It can be given, but understanding logic of a few concrete cases is much more useful. – André Nicolas May 2 '12 at 17:53
I am guessing that you are trying to revolve around $y=c$? If this is the case, you need to shift $f,g$ instead, with appropriate reality checks? – copper.hat May 2 '12 at 19:07

I am guessing you want $$\pi \int_a^b (f(x)-c)^2 - (g(x)-c)^2 \; dx$$ instead? (You need, presumably, to check that $f(x)\geq c$, and $g(x)\geq c$ for $x \in [a,b]$.)

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