I'm trying to learn about calculus on manifolds and read that $ \omega = x dy \wedge dz – y dx \wedge dz + z dx \wedge dy$ gives the area on the sphere. I tried to find the integral $\int_{S^2}\omega$ by using the parametrization $\phi : [0, 2 \pi] \times [0, \pi] \rightarrow S^2$ given by $\phi(u, v) = (\sin(v)\cos(u),\,\sin(v)\sin(u),\,\cos(v))$.
I calculated:
$dx \wedge dy = -\sin(v)\cos(v)du \wedge dv $,
$dx \wedge dz = \sin^2(v)\sin(u) du \wedge dv$ and
$dy \wedge dz = -\sin^2(v)\cos(u) du \wedge dv$.
When I integrated I got:
\begin{align} \int_S \omega &= \int_{S^2}x dy \wedge dz – y dx \wedge dz + z dx \wedge dy \\\\ &= \int_0^{2 \pi} \int_0^{\pi} \sin(v)\cos(u)[-\sin^2(v)\cos(u)] – \sin(v)\sin(u)[\sin^2(v)\sin(u)] + \cos(v)[-\sin(v)\cos(v)] dvdu \\\\ &= – 2 \pi \int_0^{\pi}\sin(v)dv \\\\ &= -4 \pi \end{align}
Since the area should be $4 \pi$, I was wondering what my mistake was.
Best Answer
You need to write the pullback as a function multiplying $dv\wedge du$, not $du\wedge dv$, and then do the integration. Your order of variables is not the usual one to get the outward-normal pointing actually outward. When you switch the order of your $2$-forms as I suggest, that will take care of your negatives. The crucial point is that your must have an orientation-preserving parametrization, and the given $2$-form $\omega$ has it built in that $(x,y,z)$ is the unit outward-pointing normal. (You might find some of my YouTube lectures on differential forms and surface integrals helpful. See the link in my profile.)