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Can we derive a formula for the volume of revolution in polar coordinates?

My attempt was so consider a very small piece of a circle that is bounded by the counterclockwise angles $\theta$ and $\theta+d \theta$ in the $xy$ plane. Then the lines forming those angles are $y=(\tan \theta) x$ and $y=\tan (\theta+d \theta) x$. So I suppose to get the (approximate) volume obtained by rotating this little about the $x$ axis. We should consider,

$$\int\limits_{0}^{r \cos \theta} (\tan (\theta+d \theta) x)^2-(\tan (\theta) x)^2 dx$$

This is a bit sloppy and even so, I'm not sure where to go from here.

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Can't say I have seen it come up befor, but:

$V = \pi \int_{x_0}^{x_1} y^2 dx$

$y = r\sin\theta\\ x = r\cos\theta \\ dx = -r\sin\theta\ + \cos\theta \frac {dr}{d\theta}d\theta $

$v = \pi \int_{0}^{\pi} r^3\sin^3\theta\ - r^2\sin^2\theta\cos\theta \frac {dr}{d\theta} d\theta$

The limits of integration might not necessarily be $0$ to $\pi.$ But frequently enough it would be.

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  • $\begingroup$ Thanks! ${}{}{}{}{}{}{}{}$ $\endgroup$ – Ahmed S. Attaalla Mar 23 '17 at 1:57

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