Integrate the 2-form $\omega=\mathrm{d}x\wedge\mathrm{d}y$ over an ellipsoid $\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$.

Here is my solution. First, parametrize the ellipsoid. We get a map $$F:D=[0,\pi]\times[0,2\pi]\rightarrow\mathbb{R^3}$$ $$(\varphi,\theta)\mapsto(a\sin\varphi\cos\theta,b\sin\varphi\sin\theta,c\cos\varphi)$$ Now compute the pullback:

$$F\mbox{*}\omega=F\mbox{*}(\mathrm{d}x\wedge\mathrm{d}y)=F\mbox{*}\mathrm{d}x\wedge F\mbox{*}\mathrm{d}y=\mathrm{d}F\mbox{*}x\wedge\mathrm{d}F\mbox{*}y=\\ =(a\cos\varphi\cos\theta\mathrm{d}\varphi-a\sin\varphi\sin\theta\mathrm{d}\theta)\wedge(b\cos\varphi\sin\theta\mathrm{d}\varphi+b\sin\varphi\cos\theta\mathrm{d}\theta)=\\ =(ab\cos\varphi\sin\varphi\cos^2\theta+ab\cos\varphi\sin\varphi\sin^2\theta)\mathrm{d}\varphi\wedge\mathrm{d}\theta=\\ =ab\cos\varphi\sin\varphi\mathrm{d}\varphi\wedge\mathrm{d}\theta=\\ =\frac{ab}{2}\sin 2\varphi\mathrm{d}\varphi\wedge\mathrm{d}\theta$$

Finally, $$\int_E\omega=\int_DF\mbox{*}\omega=\int_0^{2\pi}\int_0^{\pi}\frac{ab}{2}\sin 2\varphi\mathrm{d}\varphi\mathrm{d}\theta=\\=ab\int_0^{2\pi}-\cos 2\varphi\big|_0^{\pi}\mathrm{d}\theta=0$$

Is this correct? If so, is there an easier way to see that it is 0?


  • 1
    $\begingroup$ If $(x,y,z)$ is a point on the ellipsoid, $(\pm x,\pm y,\pm z)$ are also points on the same ellipsoid. Such symmetry grants complete cancellation. $\endgroup$ – Jack D'Aurizio Feb 22 '17 at 2:00

I haven't checked your calculations but the answer is definitely zero by Stokes' theorem. We have $d\omega = 0$ and if we denote the full ellipsoid $x^2 + y^2 + z^2 \leq 1$ by $M$ then by Stokes' theorem

$$ 0 = \int_M d\omega = \int_{\partial M} \omega = \int_E \omega. $$


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.