Using the Divergence Theorem on the surface of a unit sphere 
Using the Divergence Theorem, evaluate $\int_S F\cdot dS$ , where  $F=(3xy^2 , 3yx^2 , z^3)$, where $S$ is the surface of the unit sphere.

My Attempt
$$ \text{div} F = \left(\frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z} \right) = 3y^{2} + 3x^{2} + 3z^{3} = 3(y^{2} + x^{2} +z^{2}).$$
                Let us use spherical coordinates 
                $$ x = \rho \sin \varphi \cos \theta \quad, y = \rho\sin\varphi \sin\theta \quad, z = \rho\cos\varphi. $$
                We get ,
                \begin{align*}
      \int\limits_{S} \vec{F} \cdot ds  &= 3 \int_{0}^{\pi}\int_{0}^{2\pi} \int_{0}^{1} (\rho^{2} \sin^{2}\varphi \sin^{2} \theta + \rho^{2} \sin^{2}\varphi \rho  \cos^{2}\theta + \rho^{2} \cos^{2} \varphi ) \rho^{2} \sin\varphi \ d\rho d\theta d\varphi \\
      &= 3 \int_{0}^{\pi}\int_{0}^{2\pi} \int_{0}^{1} \rho^{4} (\sin^{3}\varphi (\sin^{2} \theta + \cos^{2}\theta) + \cos^{2}\varphi \sin\varphi) \ d\rho d\theta d\varphi \\
      &= \frac35 \int_{0}^{\pi} \int_{0}^{2\pi} \sin^{3}\varphi  + \sin\varphi \cos^{2}\varphi \ d\theta d\varphi \\
      &= \frac35 \int_{0}^{\pi} \int_{0}^{2\pi} \sin\varphi \ d\theta d\varphi \\
      &= \frac{6\pi}{5} \int_{0}^{\pi} \sin\varphi \ d\varphi = \frac{12\pi}{5}.
     \end{align*}
Could someone confirm if my reasoning is correct ? I feel like I have missed a step somewhere in the beggining.

It is correct, in fact, it is even more simple because you could have directly substituted $\rho^{2}=x^{2}+y^{2}+z^{2}$ 
