I'm struggling a lot with an exercise from my class.
If $M=\{x \in \mathbb{R}^n: |x| = 1, x_n \ge 0\}$, then how can we calculate $\int_M \omega$, where $\omega = x_n \ dx_2\wedge…\wedge dx_n$?
I suspect this will be $0$, because $x_n = 0$ on the boundary of $M$. I tried using Stokes theorem but I couldn't crack this. I could find $\theta$ such that $d\theta = \omega$, but then using Stokes' theorem I got stuck at $\int_{M} \omega = \int_{M} d\theta = \int_{\partial M} \theta$. Is this the correct approach ? How can I solve this exercise?
Best Answer
There are a few $\theta$'s that satisfy $d\theta = \omega$. One choice is $$\theta = (-1)^{n}\tfrac 1 2 x_n^2 \ dx_2 \wedge \dots \wedge dx_{n-1}.$$
It's good that you spotted that $x_n = 0$ on $\partial M$. Notice that this tells you that $(-1)^{n}\tfrac 1 2 x_n^2 = 0$ on $\partial M$, and therefore, $\theta = 0$ on $\partial M$.
So $$ \int_M \omega = \int_{\partial M} \theta = 0.$$
Here's an alternative approach. Let $ M' = \{ x \in \mathbb R^n : |x| \leq 1, x_n = 0 \}$ be the unit disk in the $x_n = 0$ plane, and let $N = \{ x \in \mathbb R^n : |x| \leq 1, x_n \geq 0 \} $ be the upper half of the unit ball. Then the boundary of $N$ is the union of $M$ and $M'$. By Stoke's theorem, we have $$ \int_N d\omega = \int_M \omega - \int_{M'} \omega.$$ (The minus sign in front of $\int_{M'} \omega$ is due to the fact that the orientation of the boundary of $N$ is opposite to the natural orientation on $M'$.)
But $d\omega = 0$, so $\int_N d\omega = 0$.
Also, $x_n = 0$ on $M'$, so $\omega = 0$ on $M'$. Therefore, $\int_{M'} \omega = 0$.
Hence $\int_M \omega = 0$.