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.
Best Answer
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.