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?


  • $\begingroup$ Do you know Pappus' theorem on volumes of rotation? $\endgroup$ – Andrew D. Hwang Nov 28 '15 at 17:53
  • $\begingroup$ 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! $\endgroup$ – HLP Nov 28 '15 at 18:09
  • $\begingroup$ 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 $(2\pi h)(\pi ab)$. $\endgroup$ – Andrew D. Hwang Nov 28 '15 at 18:18
  • $\begingroup$ 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? $\endgroup$ – HLP Nov 28 '15 at 19:00
  • $\begingroup$ 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? $\endgroup$ – Andrew D. Hwang Nov 28 '15 at 19:16

(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.

Overlap in revolving a tilted ellipse

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.

  • $\begingroup$ So we used the spherical coordinates of the ellipse? Why couldn't we have used polar or cartesian coordinates? Isn't there another way to utilise them to take care of the are swept twice? If we couldn't have done so, how do we add a vertical translation across the x axis for spherical coordinates? for a general equation it'd be (x+h)^2, how would what translate to spherical coordinates? Thanks so much for the help! $\endgroup$ – HLP Dec 1 '15 at 17:47
  • $\begingroup$ Converting to spherical coordinates is just one approach, which happens to avoid the overlap issue. Accounting for the area of the overlap and integrating in Cartesian coordinates is another; I couldn't see a nice way of doing so, but didn't look especially hard. If you translate the ellipse away from the axis (but the ellipse still crosses the axis), spherical coordinates are probably not a good strategy. $\endgroup$ – Andrew D. Hwang Dec 1 '15 at 18:34
  • $\begingroup$ So what would the overlap area look like? I'm sorry, but I'm having a bit of trouble visualising it. If I know it's shape I can probably integrate to find it. Thanks! $\endgroup$ – HLP Dec 1 '15 at 22:31
  • $\begingroup$ For the ellipse with its center on the axis, the overlap is swept by the curved diamond shape in the middle. $\endgroup$ – Andrew D. Hwang Dec 1 '15 at 23:27
  • $\begingroup$ I see. since the intersection comes after half s turn would the shape swept out be three dimensional? How would I start by finding the volume of the area? Do I try to find a function whose solid of revolution produces that diamond like shape? $\endgroup$ – HLP Dec 2 '15 at 9:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.