# 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.

• $div F =3$ you should find $4\pi a^3$. – hamam_Abdallah Jan 11 at 19:11
• You can only use $\vec{k}$ as the normal vector to your surface when the surface is parallel to the $xy-$plane. – pwerth Jan 11 at 19:12
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 $$\int_{B_a}{\rm div}({\bf F})\>{\rm dvol}=3\,{\rm vol}(B_a)=4\pi a^3\ .$$
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 $$\int_{B(0,a)} \operatorname{div} \mathbf{F} \,dV = 3 \operatorname{vol}\big(B(0,a)\big) = 4\pi a^3$$