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This is a problem from a past qualifying exam that seems a bit too calculation intensive so I was hoping that someone might see a better way to approach this.

We are given the 2-form $$\omega = \frac{xdy\wedge dz + y dz\wedge dx + z dx\wedge dy}{(x^2 + y^2 + z^2)^{3/2}},$$ and asked to verify that it is a closed 2-form on $\mathbb{R}-\{0\}.$ This is straight forward. Then we are asked to evaluate the the integral $\int_T \omega$ where $T\subset \mathbb{R}^3-\{0\}$ is the torus $$\left(\sqrt{(x-2)^2 + y^2}-2\right)^2 + z^2 = 1$$obtained by rotating the unit circle in the $xz$-plane about the line $x=2,$ $y=0$.

I approached this by finding the parametric equations for the torus, namely $$F(u,v) = (2 + (2+\cos v)\cos u, (2+\cos v)\sin u, \sin v),$$ and calculating $\int_0^{2\pi}\int_0^{2\pi} F^*\omega.$ This is pretty time-intensive and there is a lot of room for computational error, so I figured there may be a different way to approach this.

Bohring
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    Shouldn't the integral just be zero because of Stokes' theorem and the fact that the torus is the boundary of the solid torus, which is a subset of $\mathbb{R}^3\backslash{0}$? – Abel Sep 19 '13 at 07:49
  • @Abel Thanks! I am a bit rusty I guess... Good thing the test isn't for a few months. – Bohring Sep 19 '13 at 08:29
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    @Abel Actually there is a slight issue here... the point ${0}$ is actually inside the torus. So I think we can apply Stoke's to the solid torus minus a small ball around ${0}$ and then find the integral of the sphere around ${0}$. – Bohring Sep 25 '13 at 20:50
  • You're right. I seem to have mistakenly thought that $0$ was outside of your torus. You're right about the sphere around $0$ idea. It should work and it should simplify your calculations considerably since on such a sphere the denominator of $\omega$ is constant. – Abel Sep 25 '13 at 20:57

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For completeness, here is the computation of integral over the unit sphere $S^2$ (which gives the same result as integration over torus, due to the form being closed in $\mathbb R^3\setminus \{0\}$).

Let $\zeta=x\,dy\wedge dz+y\,dz\wedge dx+z\,dx\wedge dy$. Then $$ \int_{S^2} \omega = \int_{S^2} \zeta = \int_{B^3} d\zeta = \int_{B^3} 3\,dx\wedge dy\wedge dz = 3\mathrm{vol}\,(B^3)=4\pi $$ where the second step is Stokes and $B^3$ is the unit ball.

user98130
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