Verify Gauss divergence theorem of $F=xi+yj+zk$ over the sphere $x^{2}+y^{2}+z^{2}=a^{2}$ Verify Gauss divergence theorem of $F=xi+yj+zk$ over the sphere $x^{2}+y^{2}+z^{2}=a^{2}$ 
When I evaluate taking normal $N=k$, I get the answer $2\pi a^{3}$.
But when I take normal to the surface by taking out gradient  of $f(x,y,z)= x^{2}+y^{2}+z^{2}$ I am able to verify theorem.
But my course instructor told we can take normal $N=k$ when whenever we are able to translate system in $xy$ plane . 
And if plane is given then we have to find gradient otherwise not.
Please clarify!!!
 A: Let $S_a$ be your sphere, and $B_a$ the enclosed ball. For each point ${\bf r}\in S_a$ the outwards unit normal ${\bf n}$ is given by ${\bf n}={{\bf r}\over a}$. Furthermore ${\bf F}({\bf r})={\bf r}$. Since ${\bf r}\cdot{\bf r}=a^2$ on $S_a$ it follows that
$$\int_{S_a}{\bf F}\cdot{\bf n}\>{\rm d}\omega=\int_{S_a}{\bf r}\cdot{{\bf r}\over a}\>{\rm d}\omega=a\int_{S_a}{\rm d}\omega=4\pi a^3\ .$$
On the other hand, ${\rm div}({\bf F})\equiv3$, and therefore
$$\int_{B_a}{\rm div}({\bf F})\>{\rm dvol}=3\,{\rm vol}(B_a)=4\pi a^3\ .$$
A: Parameterize your sphere using spherical coordinates as $\psi : [0,\pi] \times [0,2\pi] \to \mathbb{R}^3$ given by
$$\psi(\theta, \phi) = a(\sin\theta\cos\phi, \sin\theta\sin\phi, \cos\theta)$$
Then your normal is given by
$$\mathbf{n}(\theta, \phi) = \frac{\partial\psi}{\partial\theta} \times \frac{\partial\psi}{\partial\phi} =  a^2(\sin^2\theta\cos\phi, \sin^2\theta\sin\phi, \sin\theta\cos\theta) = (a\sin\theta)\,\psi(\theta,\phi)$$
so $$\int\limits_{S(0,a)} \mathbf{F}\cdot d\mathbf{A} = \int\limits_{[0,\pi] \times [0,2\pi]} \mathbf{F}(\psi(\theta, \phi))\cdot \,\mathbf{n}(\theta, \phi)\,d\theta\,d\phi $$
Now notice that $\mathbf{F}$ is actually the identity function so the integral equals
$$\int\limits_{[0,\pi] \times [0,2\pi]} (a\cos\theta)\psi(\theta, \phi)\cdot \,\psi(\theta, \phi)\,d\theta\,d\phi = \int_{\theta=0}^\pi \int_{\phi=0}^{2\pi} a^3\sin\theta\,d\theta\,d\phi$$
which is $4\pi a^3$.
On the other hand, $\operatorname{div} \mathbf{F} = \operatorname{Tr} \nabla\mathbf{F} = \operatorname{Tr} \mathbf{F} = 3$ so 
$$\int_{B(0,a)} \operatorname{div} \mathbf{F} \,dV = 3 \operatorname{vol}\big(B(0,a)\big) = 4\pi a^3$$
