How to find $\int_{S^2}f \cdot n \ \text{d}S$ if $f(x,y,z):=(x^3,y^3,z^3)^T$ With $\mathbb{S}^2$ being the unit sphere, how to find 
$$\int\limits_{\mathbb{S}^2} \vec{f} \cdot \vec{n} \ \text{d}S$$ 
if $\vec{f}(x,y,z):=(x^3,y^3,z^3)^T$?
Apparently, we need to use Gauss. With $B$ being the unit ball we get:
$$\int\limits_{\mathbb{S}^2} \vec{f} \cdot \vec{n} \ \text{d}S = \int\limits_B \operatorname{div} \vec{f} \ \text{d}V$$ 
with $\operatorname{div}f=3x^2+3y^2+3z^2$.
Yet I'm not quite sure how to move further. I've tried the transformation theorem:
$$\int_{\phi(K)} f = \int_{K} f(\phi(x)) |\det \phi'(x)| \ \text{d}x$$
with $\phi(r,\alpha,\beta):=(r \cos\alpha, r \sin\alpha \cos\beta, r \sin\alpha \sin\beta)$ and got stuck right away.
 A: We have $\operatorname{div} \vec{f} = 3x^2+3y^2+3z^2$, which in spherical coordinates
$$
\begin{align}
x & = r \cos \theta \sin \varphi, \\
y & = r \sin \theta \sin \varphi, \\
z & = r \cos \varphi,
\end{align}
$$
for $0 \leq \theta \leq 2 \pi$ and $0 \leq \varphi \leq \pi$, becomes $\operatorname{div} \vec{f} = 3 r^2.$
Therefore, using Gauss's Theorem we obtain
$$
\begin{align}
\int\limits_{\mathbb{S}^2} \vec{f} \cdot \vec{n} \, \text{d}S & = \int\limits_{B} \operatorname{div} \vec{f} \, \text{d}V \\
 & = \int_0^{2\pi} \hspace{-5pt} \int_0^{\pi} \hspace{-5pt} \int_0^1 (3r^2) \cdot (r^2 \sin \varphi) \, \text{d} r \, \text{d} \varphi \, \text{d} \theta \\
 & = 3 \int_0^{2\pi} \hspace{-5pt} \text{d} \theta \int_0^{\pi} \hspace{-5pt} \sin \varphi \, \text{d} \varphi \int_0^1 \hspace{-3pt} r^4 \, \text{d} r \\
 & = \frac{12 \pi}{5}.
\end{align}
$$
Edit: Elaborating as the OP asked, we have 
$$\operatorname{div} \vec{f} = 3x^2+3y^2+3z^2 =3(x^2+y^2+z^2)$$ 
in cartesian coordinates. When we apply the spherical coordinate changes we have 
$$x^2+y^2+z^2=r^2,$$ 
therefore 
$$\operatorname{div} \vec{f} = 3(x^2+y^2+z^2) = 3r^2.$$ 
