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I was interested by this question in finding the volume of a solid of revolution using a "more accurate" cone frustum instead of the usual cylindrical disk.

So, starting with the volume of a cone frustum $ V = \frac{\pi}{3} ( R^2 + r^2 + Rr ) $, compute:

conical frusta $ \frac{\pi}{3}( (r+ \Delta y)^2 + r^2 + (r+ \Delta y)(r) ) \Delta x $

$ = \frac{\pi}{3}( 3r^2 + 3 r \Delta y + \Delta y^2 ) \Delta x $

Which is interesting, because the regular disk formula comes out if you ignore the $ \Delta y $ terms, which means you're assuming the cone is actually a cylinder.

I'm not sure how to finish this though! How can I work with $ \Delta y $ to create an integral formula?

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    @Nils your comment can be the answer2011-07-19

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Just assuming that $r$ varies continuously with $x$, you get that $\Delta y \to 0$ as $\Delta x \to 0$.

So in the limit you get $ \frac{dV}{dx} = \lim_{\Delta x \to 0} \frac{\pi ( 3r^2 + 3 r \Delta y + \Delta y^2) \Delta x}{3 \Delta x} = \pi r^2 $