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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... –  Guess who it is. 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... –  Guess who it is. May 2 '12 at 17:51
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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

2 Answers 2

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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The reason that the method is not working is if you are looking to integrate around a horizontal line, the function must be shifted upward (or downward depending on the value of $c$).

In order to shift a function upward or downward, a function transformation of the general form $f(x) + a $ is required.

First, consider the case of integration around the y-axis, i.e., $y = 0$, if we assume that the area of integration is bounded above by $f(x)$ and bounded below by $g(x)$, then we can use the disk method which takes the general form:

$$\pi \int{A(x)}\ \text{d}x\ \text{where}\ A = f(x)^2 - g(x)^2$$

Second, assuming that the axis of revolution is now set at $y = c$, the upper and lower bounding functions must be changed to $f(x) + c$ and $g(x) + c$ respectively. NOT $f(x+c)$ which would result in a horizontal shift.

An additional wrinkle must be dealt with in this case. Namely, neither the upper nor lower bounding functions can be less than the axis of revolution. This would be satisfied by the condition $\forall x \in [a,b], f(x) + c \ge c\ \text{and}\ g(x) + c \ge c $.

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