# Volume of Revolution of a Transformed Ellipse?

I'm looking at an ellipse (a bunch of them actually) transformed by $h$ on the $x$-axis away from the center and rotated by an angle of $Q$ from the $xy$ axis.

I got the following equation: the $x$ is transformed as $(x + h)$ and the rotation is done by $x\cos Q + y\sin Q$ and $x\sin Q - y\cos Q$.

Now I want to try and find the volume of solid of revolution. I'm not quite sure where I should begin. Is it even possible to do this? Or should I do the integration for the new $x$ and $y$ axis?

Thanks!

• Do you know Pappus' theorem on volumes of rotation? – Andrew D. Hwang Nov 28 '15 at 17:53
• Not quite familiar with it, so pardon if I say something a tad bit insane. But I'm having a bit of trouble visualising how it would apply to an ellipse that's been rotated and moved away from the centre. Would I use the same equation of A.d for the normal ellipse but add the transformations alongside it? Thanks! – HLP Nov 28 '15 at 18:09
• As I understand it, you have an ellipse with semi-axes $a$ and $b$ (at arbitrary angular orientation), the center of the ellipse lies at distance $h$ from an axis of rotation $\ell$ that does not touch the ellipse, and you want the volume swept out under rotation about $\ell$. Assuming that's right, by Pappus' theorem, the volume is $(2\pi h)(\pi ab)$. – Andrew D. Hwang Nov 28 '15 at 18:18
• Hmm..well what about rotating the transformed ellipse across the x-axis? The shape should be different from an oblate spheroid, right? So i'd have to integrate it by hand? – HLP Nov 28 '15 at 19:00
• It's not entirely clear to me what your picture is: Does the ellipse cross the axis of rotation (here, the $x$-axis), so that revolving "covers" part of the swept region more than once? – Andrew D. Hwang Nov 28 '15 at 19:16

For definiteness, the issue is that when the ellipse $$\frac{x^{2}}{a^{2}} + \frac{z^{2}}{b^{2}} = 1$$ is rotated through an angle $\phi_{0}$ and revolved about the $z$-axis, the "profile" intersects itself after half a turn.
As indicated by the radial segments in the diagram, however, the volume swept out can be expressed conveniently in spherical coordinates. The unrotated ellipse satisfies $$\frac{\rho^{2} \sin^{2} \phi}{a^{2}} + \frac{\rho^{2} \cos^{2} \phi}{b^{2}} = 1,$$ or after rotation by $\phi_{0}$ and rearrangement, $$\rho = R(\phi) = \frac{ab}{\sqrt{b^{2} \sin^{2}(\phi - \phi_{0}) + a^{2} \cos^{2}(\phi - \phi_{0})}}.$$ The solid swept out by revolving about the $z$-axis is described by the inequalities $$0 \leq \theta \leq 2\pi,\quad 0 \leq \phi \leq \pi,\quad 0 \leq \rho \leq R(\phi).$$ The volume swept out is $$2\pi \int_{0}^{\pi} \int_{0}^{R(\phi)} \rho^{2} \sin\phi\, d\rho\, d\phi = \frac{4\pi (ab)^{3}}{3} \int_{0}^{\pi/2} \frac{\sin\phi\, d\phi}{\bigl[b^{2} \sin^{2}(\phi - \phi_{0}) + a^{2} \cos^{2}(\phi - \phi_{0})\bigr]^{3/2}}.$$ Offhand this looks elementary (i.e., "possible to evaluate in closed form"), but I don't see a good way of integrating.