Let $M$ be the $3-$ball in $\Bbb{R}^3$ $\{(x,y,z):x^2+y^2+z^2 \leq 1\}$. Let $\omega$ be the $2-$form given by
$$\omega = x \, dy \wedge dz + y \, \wedge dz \wedge dx + z \,dx \wedge dy.$$
Verify stokes theorem by verifying both RHS and LHS of the theorems statement.
So for the LHS, i computed $d \omega$ to be
$$d \omega = 3 \, dx \wedge dy \wedge dx.$$
So can I just say since
$$\int_M dx \wedge dy \wedge dz$$
is the volume of the $3-$ball that the final value is $3$ times the volume of a $3-$ball in $3$ space which is $4\pi$. Then for the RHS, I know the boundary of the $3-$ball is the unit $2$ ball $S^2$, which is all $(x,y,z) \in \Bbb{R}^3$ such that $x^2+y^2+z^2=1$. Then I need to compute
$$\int_{S^2} x \, dy \wedge dz + y \, dz \wedge dx + z\, dx \wedge dy$$
Which I still dont know how to compute integrals with wedge products, so is this a sum of $3$ triple integrals? Or could i swap to spherical coodrinates by setting
$$x = r \sin \theta \cos \phi, y = r \sin \theta \sin \phi, z = r \cos \theta$$
Where my $r=1$ Then my integral becomes
$$\int_0^{2 \pi} \int_0^{2 \pi} \int_0^1 f(r,\theta,\phi) r^2 \sin \theta d r d \theta d \phi$$
Where $f(r,\theta,\phi)$ is $\omega$ with spherical coordinates. But that would get rather messy, no? So what’s the fastest method for computing the LHS and is my RHS ok as far as justification goes? With the polar swap I get
\begin{align}
\omega&= (\sin \theta \cos \phi)d(\sin \theta \sin \phi) \wedge d(\cos \theta) + (\sin \theta \sin \phi ) d(\cos \theta) \wedge d(\sin \theta \cos \phi)+(\cos \theta) d(\sin \theta \cos \phi) \wedge d( \sin \theta \sin \phi)\\
&=\sin^3 \theta \cos \phi +\sin^4 \theta \sin \phi \cos^2 \phi d \theta \wedge d \phi \wedge \cos \theta \cos \phi + \cos^2 \theta \cos \phi d \theta – (2 \sin \theta \sin \phi) d \phi
\end{align}
Best Answer
By symmetry the answer will be $6$ times the integral of $z \,dx \wedge dy$ over the upper half of the sphere. Since $z = \sqrt{1 - x^2 - y^2}$, you can just do this in coordinates as the integral $$\int_{x^2 + y^2 < 1} \sqrt{1 - x^2 - y^2} \,dA$$ The above integral is easily determined to be ${\displaystyle {2\pi \over 3}}$ after a routine change to polar coordinates. Multiplying this by $6$ gives the desired answer of $4\pi$.
What Stokes theorem is basically doing here is turning the integral of the constant function $3$ over the interior of the sphere into the sum of three integrals of the constant function $1$, and then integrating each of these first in one variable and then in the remaining two variables. The above calculation is one of the two integrals you have left when you integrate in $z$ first.