Evaluating an integral with divergence theorem 
Evaluate the integral  $\int_S(x^2+y^2)dS$ where S is the unit sphere in $\mathbb{R}^3$

I'm being tripped up when the question asks me to evaluate this integral with the divergence theorem because I keep getting $2\pi/3$ but I should get $8\pi/3$ since I got that for the integral.
 A: Take $\mathbf{F}=(x,y,0)$,  then, since  $\mathbf{n}=(x,y,z)$ on the sphere $S$, by the divergence theorem it follows that
$$\int_S(x^2+y^2)dS=\int_S(\mathbf{F}\cdot\mathbf{n})dS=\int_V\text{div}(\mathbf{F})dV=\int_V 2dV=2\cdot \frac{4\pi}{3}.$$
A: other approach
we have
$$x^2+y^2=(x,y,0)\cdot(x,y,z)$$
$$div(x,y,0)=1+1=\color{red}{2}.$$
Using spherical coordinates as
$$x=\sin(f)\cos(t)$$
$$y=\sin(f)\sin(t)$$
$$z=\cos(f)$$
with
$$0\le t\le 2\pi \; , \; 0\le f\le \pi$$
the integral becomes
$$\int_0^1\int_0^{2\pi}\int_0^\pi\color{red}{2}r^2\sin(f)drdtdf=\frac{8\pi}{3}$$
A: The divergence theorem states that:
$\iiint_D$div(F)$dV$=$\iint_S$((F)*n))dS, where n is a an outward pointing normal vector to S (which forms the (closed) boundary of the region D).
We thus require that (F)*n=$x^2+y^2$.
The outward normal vector n is given by $(x,y,z)$ (Why?).
Hence, if F=($F_1$,$F_2$,$F_3$), then we have that:
(F)*n=$(F_1)x$+$(F_2)y$+$(F_3)z$=$x^2+y^2$.
So F=$(x,y,0)$
And div(F)=2 (Why?)
So $\iint_S$($x^2+y^2$)$ds$= $\iint_S$((F)*n))dS= $2$ $\iiint_D$dV=2($\frac{4}{3}$$\pi$)=($\frac{8}{3}$$\pi$)
A: I just thought I'd mention you can use symmetry to do this without any theorems. Let $I$ denote the integral in question. By symmetry, one has
$$I = \int_S(x^2+y^2)dS =  \int_S(y^2+z^2)dS = \int_S(x^2+z^2)dS$$
Adding these together, one sees that 
$$3I = 2\int_S(x^2+y^2 + z^2)dS$$
Since $x^2 + y^2 + z^2 = 1$ on the unit sphere, the integral is the area of the unit sphere, or $4\pi$. Thus $I = {8\pi \over 3}$.
