Evaluate surface integral over unit sphere 
Evaluate $\iint (ax^2+by^2+cz^2)dS$, where $S$ is surface of unit sphere $x^2+y^2+z^2 = 1$.

My attempt. Using Gauss divergence theorem, $F.n = (ax^2+by^2+cz^2) \implies F=\langle ax,by,cz\rangle$
$\text{div} F = (a+b+c)$. by gauss divergence theorem, $$\iint_S F.n ds = \iiint_V \text{div} F dV = \iiint (a+b+c)dV = \frac{4}{3}\pi(a+b+c)$$
But if I calculate by surface integral I am getting wrong answer. 
My try:
Changing to spherical coordinates $x=\sin \phi \cos \theta, y = \sin \phi \sin \theta, z= \cos \phi$ (I am taking radius $r = 1$ over sphere)
$$\iint_{S: \phi: 0 - \pi,\; \theta:0-2\pi} (a\sin^3 \phi \cos^2 \theta+b\sin^3 \phi \sin^2 \theta + c\cos^2\phi \sin \phi)\ d\phi d\theta = \Big(2\pi a+2\pi b +\frac{2}{3}c\Big).$$ Both the answers are different. I know that I made some mistake in latter part. Please help.

Check your last evaluation. Note that
$$\int_{\phi=0}^{\pi}\int_{\theta=0}^{2\pi} \sin^3 \phi \cos^2 \theta d\theta d\phi=\int_{\phi=0}^{\pi} \sin^3 \phi d\phi\int_{\theta=0}^{2\pi}\cos^2 \theta d\theta = \frac{4}{3}\cdot \pi$$
(the same result if you replace $\cos^2 \theta$ with $\sin^2 \theta$) and 
$$\int_{\phi=0}^{\pi}\int_{\theta=0}^{2\pi} \cos^2 \phi \sin \phi d\theta d\phi=2\pi\int_{\phi=0}^{\pi} \cos^2 \phi \sin \phi d\phi = 2\pi\cdot\frac{2}{3}.$$
