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Let $S$ be given by

$$S= \left[(x,y,z) \in \Bbb{R}\;|\; x^2+y^2+z^2+xy+xz+yz=\frac12 \right]$$

and $$\omega = xdy \wedge dz\, -\, \frac {2z}{y^3} \, dx\wedge dy \,+\, \frac1{y^2}dz\wedge dx $$

Explain how to give an orientation for $S$ and compute $\int_S \omega$ with respect to that orientation.

I know it involves Stokes's theorem and that $d\omega = dx \wedge dy \wedge dz$ but I'm kind lost about the orientation and actually computing it.

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If in fact $\mathrm{d}\omega = dx \wedge dy \wedge dz$ then by Stokes' Theorem, you are computing the volume of the region enclosed by $S$. This region happens to be an ellipsoid, tilted with respect to the given axes. You can figure out how to obtain the dimensions and orientation of the axes here. It turns out that this ellipsoid has semiaxes of length $1$, $1$, and $1/2$, so the volume of the ellipsoid is $2\pi/3$, and the value of the integral is $(2\pi/3) dx \wedge dy \wedge dz$.

That said, I am not so sure that $\mathrm{d}\omega = dx \wedge dy \wedge dz$ as you state, because the 2-forms are in cyclic order and will not change sign upon differentiation. In that case, the terms with $y$ will not cancel. The resulting integral, however, would be a mess and if this is homework, I'd be surprised if that was the intention.

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I think it's fine, unless I messed up somewhere while computing $d\omega$, thank you for your answer in any case. – Mike Feb 13 '13 at 23:51
I was just checking something... isn't the volume $\frac23 \pi$? – Mike Feb 14 '13 at 2:37
Unless my numbers were off, I got $\pi/2$ based on the volume being $\pi$ times the product of the semi-axes. The semi-axes are the square root of the eigenvalues of the inverse of the coefficient matrix (see the link I provided), and my calculations provided semi-axes of $1$, $1$, and $1/2$. – Ron Gordon Feb 14 '13 at 2:42
Yeah, I agree about the semiaxes, but isn't the volume what you said times $\frac43$? – Mike Feb 14 '13 at 2:51
Oh man, you are right - how humbling. Of course there's a 4/3 factor! I will fix in the solution. – Ron Gordon Feb 14 '13 at 2:53

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