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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?

Thanks.

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    $\begingroup$ 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
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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

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

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