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!
(I've changed notation to conform to the mathematics convention of spherical coordinates.)
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.