Verification of the Divergence Theorem

[haroldholt]

The question I was given asks to verify the divergence theorem by showing that both sides of the theorem show the same result. With the divergence theorem obviously being $\iint_S\mathbf{F}\cdot\mathbf{n}\,dS = \iiint_V \nabla\cdot\mathbf{F}\,dV$.

The vector field is $\mathbf{F}=\mathbf{i}+\mathbf{j}+\mathbf{k}$ and the surface through which it is flowing is the cone $z=\sqrt{x^2+y^2}$ below the plane $z = 1$.

Firstly, as far as I can tell, $\iiint_V \nabla\cdot\mathbf{F}\,dV = 0$ since $\nabla\cdot\mathbf{F} = 0$ which covers the right hand side of the divergence theorem.

So then I divided the surface into two, that is the top of the cone and the rest of the cone.

For the top of the cone I got the surface integral as $\frac{\pi}{2}$ since $\mathbf{F}\cdot\mathbf{n} = (\mathbf{i}+\mathbf{j}+\mathbf{k})\cdot(\mathbf{k}) = 1$ so then you're just left with the surface integral of 1 which by definition is the area of the surface. And since the radius is $\sqrt{\frac{1}{2}}$, the area is $\frac{\pi}{2}$.

Now for the other surface I found the unit normal to be $\frac{\cos{t}+\sin{t}-1}{\sqrt{2}}$ after parametrizing the surface as $\mathbf{r} = z\cos{t}\mathbf{i} + z\sin{t}\mathbf{j} + z\mathbf{k}$ thus finding $\iint_S\mathbf{F}\cdot\mathbf{n}\,dS = -\sqrt{2}\pi$.

Obviously I wanted the result to be $-\frac{\pi}{2}$ so that the total surface integral would be equal to zero which would verify the divergence theorem for this problem. But it's not and I can't figure out where I went wrong. Since $-\sqrt{2}\pi$ is fairly close to $-\frac{\pi}{2}$ I figure I'm on the right track at least with my working.

Hope this makes sense and thank you in advance for any assistance.

 

[latentcorpse]

on the upper surface, z=1 so you get $x^2+y^2=1$ is eqn of circle radius 1 and so your upper integral should be $\pi$.

how did you find the normal - you haven't specified a direction for the unit normal so at the moment it's just a scalar.

 

[haroldholt]

on the upper surface, z=1 so you get $x^2+y^2=1$ is eqn of circle radius 1 and so your upper integral should be $\pi$.

how did you find the normal - you haven't specified a direction for the unit normal so at the moment it's just a scalar.

Oh yes of course. Stupid mistake on my behalf. But if the integral of the top surface is $\pi$ I still don't get the correct answer.

I found the unit normal using $\frac{\partial{\mathbf{r}}}{\partial{t}}\times\frac{\partial{\mathbf{r}}}{\partial{z}}$ then dividing the result by the magnitude. There's supposed to be an i, j and k on the 3 different terms in the unit normal, my bad.