Let $S = \{(x, y, z) \in \Bbb R^3:\frac{x^2}{4}+\frac{y^2}{9}+\frac{z^2}{16}=1\}$ (oriented by outward normal). I want to compute the surface integral $\displaystyle\int_S \dfrac{(x-1)~dy\wedge dz+y~dz\wedge dx+z~dx\wedge dy}{((x-1)^2+y^2+z^2)^{3/2}}$.
I first tried to use parametrization $(0,\pi)\times (0,2\pi)\to S$, $(u,v)\mapsto (2\sin u\cos v, 3\sin u \sin v , 4\cos u)$, but this calculation does not work well. Can I get a hint?
Hint: Use the Divergence Theorem to change the integral from the ellipsoid to a sphere centered at $(1,0,0)$.
The 2-form is closed (equivalently, the associated field is incompressible). Hence if $S'$ is a small enough sphere centered at $(1,0,0)$, then the integral over $S'$ (oriented outward) is the same as the integral over $S$ - that is because the triple integral of the divergence over the region bounded by the two surfaces is zero.