Two form surface integral over sphere I'm trying to compute $\int_M \omega$ with
\begin{align*}
\omega &= x^4 dy \wedge dz + y^4 dz \wedge dx + z^4 dx \wedge dy, \\
M &: x^2 + y^2 + z^2 = R^2. 
\end{align*}
I have done this in two ways: Stokes' theorem and direct computation of the wedge product via a spherical coordinate transformation. Using $\phi$ as azimuthal and $\theta$ as polar angle and applying Stokes I obtain
\begin{align*}
4 \int_0^{2 \pi}\int_0^{\pi} (x^3 + y^3 + z^3) R^2 \sin \theta d\theta d\phi = 0,
\end{align*}
where I used the standard
\begin{align*}
x &= R \sin \theta \cos \phi \\
y &= R \sin \theta \sin \phi \\
z &= R \cos \theta.
\end{align*}
By directly computing the wedge products and plugging them in I obtain
\begin{align*}
dy \wedge dz &= R^2 \sin^2 \theta \cos \phi d\theta \wedge d\phi \\
dz \wedge dx &= R^2 \sin^2 \theta \sin \phi d\theta \wedge d\phi \\
dx \wedge dy &= R^2 \cos \theta \sin \theta d\theta \wedge d\phi \\
\int_M x^4 dy &\wedge dz + y^4 dz \wedge dx + z^4 dx \wedge dy = 0.
\end{align*}
The solution given by my professor was $\frac{12}{5} \pi^2 R^5$, which neither method agrees with. Then I repeat this procedure but instead of taking a sphere, I take a hemisphere. Changing the limits in the $\theta$ integrals to $[0, \frac{\pi}{2}]$, the Stokes' method yields $2 \pi R^5$ and the direct wedge computation yields $\frac{\pi R^6}{3}$. The solution given then is just $\frac{6}{5} \pi^2 R^5$, which also does not agree with my solutions. Any help with my mistakes would be greatly appreciated.

In case of sphere, the surface integral is indeed zero.
Please note in below working of mine, I use $\theta$ as azimuthal angle and $\phi$ as polar. I think you have the other way round. So want to call out to avoid confusion.
In case of hemisphere above $z = 0$,
Parametrization of sphere is $r(\phi, \theta) = (R \cos \theta \sin \phi, R \sin\theta \sin\phi, R \cos\phi)$
For outward normal vector, $\displaystyle r'_{\phi} \times r'_\theta = R^2 \sin\phi(\cos\theta \sin\phi, \sin\theta \sin\phi, \cos\phi)$
The integral becomes,
$\displaystyle \small \int_0^{2\pi}\int_0^{\pi/2} R^2 \sin\phi \ (R^4 \cos^4\theta\sin^4\phi, R^4 \sin^4\theta\sin^4\phi, R^4 \cos^4\phi) \cdot (\cos\theta \sin\phi, \sin\theta \sin\phi, \cos\phi) \ d\phi \ d\theta$
$= \displaystyle \frac{\pi R^6}{3} \ $, which is same as your working.
In case of applying divergence theorem, we close the surface with a disk at $z = 0$.
$\small \nabla \cdot (x^4, y^4, z^4) = 4(x^3+y^3+z^3) = 4r^3(\cos^3\theta \sin^3\phi+ \sin^3\theta \sin^3\phi + \cos^3\phi)$
where $0 \leq r \leq R$.
So volume integral is,
$\displaystyle \small \int_0^{2\pi} \int_0^{\pi/2} \int_0^R 4r^5 \sin\phi (\cos^3\theta \sin^3\phi+ \sin^3\theta \sin^3\phi + \cos^3\phi) \ dr \ d\phi \ d\theta = \frac{\pi R^6}{3}$.
Now note the flux through the bottom disk is zero so flux through spherical surface is $\frac{\pi R^6}{3}$, the same answer we received through direct surface integral.
