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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!!!

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    $\begingroup$ $div F =3$ you should find $4\pi a^3$. $\endgroup$ – hamam_Abdallah Jan 11 at 19:11
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    $\begingroup$ You can only use $\vec{k}$ as the normal vector to your surface when the surface is parallel to the $xy-$plane. $\endgroup$ – pwerth Jan 11 at 19:12
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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 $$\int_{B_a}{\rm div}({\bf F})\>{\rm dvol}=3\,{\rm vol}(B_a)=4\pi a^3\ .$$

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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 $$\int_{B(0,a)} \operatorname{div} \mathbf{F} \,dV = 3 \operatorname{vol}\big(B(0,a)\big) = 4\pi a^3$$

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