Divergence Theorem To Calculate Surface Integral

Question

$M=${$(x,y,z):x^2+y^2+z^2=16$,$\sqrt {x^2+y^2}\leq z$}

We are asked to find the surface area of this surface. This is my way:

$\partial M=${$(x,y,z):x^2+y^2+z^2=16$,$\sqrt {x^2+y^2}= z$} so the latter condition is $x^2+y^2=z^2$.

If we place this equation in the first condition we receive $2x^2+2y^2=16$, and this is very easy to parameterize with $$r(t)=(\sqrt 8 \cos(t),\sqrt 8 \sin(t), \sqrt 8), \quad 0\leq t <2\pi.$$

Now according to the divergence theorem, $$\int_Mdxdydz=\int_{\partial M}F(r(t))\cdot r'(t)dt$$ when $F=(P,Q,R)$ so that $\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}=1$. I chose $F=(x,0,0)$.

So calculating $$\int_0^{2\pi}(\sqrt 8 \cos(t),0,0)\cdot(-\sqrt8 \sin(t),\sqrt8 \cos(t),0)=0.$$

Is my method of parameterization wrong or is it the way I applied the diveregence theorem?

Correct me if I'm wrong, but seeing as if I had decided $F=(0,y,0)$ or $F=(0,0,z)$ the integral would still be $0$, that means that I used the divergence theorem right?

Answer 1

One could use the Divergence theorem, $\int _{M}(F\cdot n)dS=\int _{B}(\nabla\cdot F)dV$, with $F=n$, where B is the volume enclosed by M, to find the surface area of the whole sphere. As $\nabla\cdot n=\frac{2}{r}$, $dV=r^2\sin\theta dr d\theta d\phi$, the volume integral for the $\phi\in[0,2\pi]$, $\theta\in[0,2\pi]$, $r\in[0,R]$ equals $4\pi R^2$.

According to http://en.wikipedia.org/wiki/Divergence_theorem#Mathematical_statement, M has to be a closed surface. Even if you try to close the cap with the disk $x^2+y^2\leq8$, it would still fail to satisfy the other condition: F has to be a continuously differentiable vector field.

So I think you should calculate the integral directly in spherical coordinates with the area element $R\sin\theta d\theta d\phi$, and $\theta\in[0,\frac{\pi}{4}]$.