Let $$\omega = \frac{1}{z}\mathrm{d}x\wedge\mathrm{d}y - \frac{1}{y}\mathrm{d}x\wedge\mathrm{d}z + \frac{1}{x}\mathrm{d}y\wedge\mathrm{d}z$$ defined on the open subset of $\mathbb{R}^3$ $\{(x,y,z)\mid xyz \neq 0\}$. Find the integral of $\omega$ on the ellipsoid $$\frac{x^2}{a^2}+\frac{y^2}{c^2}+\frac{z^2}{c^2}=1$$

Now what I did was taking the following parametrization of the ellipsoid: $$x = a\cos\theta\sin\phi,\; y= b\sin\theta\sin\phi,\; z=c\cos\phi$$ where $(\theta, \phi) \in [0,2\pi]\times [0,\pi]$. Now something that is annoying me is the fact that the parametrization is not injective in $[0,2\pi]\times [0,\pi]$, so I thought about removing a parallel and a meridian of the ellipsoid, that is, taking $(0,2\pi)\times(0,\pi)$ instead, since what I'm removing has measure zero and therefore doesn't contribute at all to the value of the integral (I'm not really sure if this is how I should proceed or if it doesn't make sense at all, so some help clarifying that would be helpful too). Now I compute $\mathrm{d}x, \mathrm{d}y$ and $\mathrm{d}z$ in terms of $\theta, \phi$, and I get $$\begin{align*} \mathrm{d}x &= -a\sin\theta\sin\phi\mathrm{d}\theta + a\cos\theta\cos\phi\mathrm{d}\phi\\ \mathrm{d}y &= b\cos\theta\sin\phi\mathrm{d}\theta + b\sin\theta\cos\phi\mathrm{d}\phi\\ \mathrm{d}z &=-c\sin\theta\mathrm{d}\phi \end{align*}$$ and computing the wedge products, I finally get $$\varphi^{*}\omega = -\left(\frac{ab}{c} + \frac{ac}{b} + \frac{bc}{a} \right)\sin\phi \mathrm{d}\theta\wedge\mathrm{d}\phi$$ so if we let $M$ denote the ellipsoid with the meridian and parallel remove as explained above, then $$\begin{align*} \int_M \omega & = \int_U \varphi^*\omega = -\left(\frac{ab}{c} + \frac{ac}{b} + \frac{bc}{a} \right)\int_0^{2\pi}\int_0^{\pi}\sin\phi\mathop{d}\theta\mathop{d}\phi\\ & = -4\pi \left(\frac{ab}{c} + \frac{ac}{b} + \frac{bc}{a} \right) \end{align*}$$

So is this correct? Is there any way to simplify the calculations and arrive to the same result more easily? Thank you!

  • $\begingroup$ I would remark that there are a few careless errors in your typing. In the final integral, you should have $d\phi\,d\theta$. $dz=-c\sin\phi\,d\phi$, and to get the usual outward orientation on the ellipsoid you want to have the $2$-form $d\phi\wedge d\theta$, so there's no negative in the answer. (That's good, because the $2$-form is a sum of positive $2$-forms.) And you're really only removing one half-meridian (including the north and south pole). $\endgroup$ – Ted Shifrin Feb 8 '17 at 22:20

Here's a more conceptual (but no quicker) approach to the computation. Start by observing that on the unit sphere, the area $2$-form $$dA = \frac{dx\wedge dy}z = \frac{dz\wedge dx}y = \frac{dy\wedge dz}x.\tag{$\star$}$$ You can check this in coordinates in spherical coordinates (computing as you did) or as follows: Since the unit normal of the (unit) sphere is $(x,y,z)$, we have $dA = x\,dy\wedge dz + y\,dz\wedge dx+z\,dx\wedge dy$. On the other hand, on the sphere, we have $x\,dx+y\,dy+z\,dz=0$ (why?), so we can substitute and deduce ($\star$).

Next, the linear map $T(x,y,z)=(ax,by,cz)$ maps the unit sphere diffeomorphically to our ellipsoid (and preserves the outward orientations, assuming $a,b,c>0$). Note that $T^*(\frac{dx\wedge dy}z) = \frac{ab}c\cdot\frac{dx\wedge dy}z$, $T^*(\frac{dz\wedge dx}y) = \frac{ac}b\cdot\frac{dz\wedge dx}y$, and $T^*(\frac{dy\wedge dz}x) = \frac{bc}a\cdot\frac{dy\wedge dz}x$. Thus, $T^*\omega = \left(\frac{ab}c+\frac{ac}b+\frac{bc}a\right)dA$. The result is now immediate, since we know that the surface area of the unit sphere is $4\pi$.

  • $\begingroup$ Thank you for both your answer and your comment! Just one small question: how do I know that I want $d\phi \wedge d\theta$ instead of $d\theta \wedge d\phi$? Do I need to calculate the integral first and then just switch depending on whether the result is positive or negative? $\endgroup$ – user313212 Feb 9 '17 at 15:03
  • $\begingroup$ No, this is a standard thing you always have to check. If you evaluate the $2$-form on a right-handed basis for a tangent space, you must get a positive answer. The vector tangent to a $\phi$-curve followed by a vector tangent to a $\theta$-curve gives a right-handed basis. $\endgroup$ – Ted Shifrin Feb 9 '17 at 15:59

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