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1. The problem statement, all variables and given/known data

I am having trouble with part iii) of the following problem:

Verify the divergence theorem for the function $\vec{u} = (xy,- y^2, + z)$ and the surface enclosed by the three parts:

(i) $z = 0, x^2+y^2 < 1$,

(ii) $x^2+y^2 = 1, 0 \le z \le 1$ and

(iii) $(a^2-1)x^2+(a^2-1)y^2+z^2 = a^2 , 1 \le z \le a, a > 1$.

2. Relevant equations

Divergence theorem:

$\int_V\vec{ \nabla}\cdot\vec{u}~dV=\oint_{\partial v}\vec{u}\cdot\vec{dS}$

3. The attempt at a solution

The boundaries in iii) describe part of an ellipsoid. Rewrite the equation as follows: $\frac{x^2}{(\frac{a}{\sqrt{a^2-1}})^2}+\frac{y^2}{(\frac{a}{\sqrt{a^2-1}})^2}+\frac{z^2}{a^2}=1$

LHS$ = \int_0^{2\pi}\int_1^a\int_0^{\frac{z^2-a^2}{1-a^2}}\vec{ \nabla}\cdot\vec{u}~\rho~ d\rho ~dz ~d\theta$ $~~~~ = \int_0^{2\pi}\int_1^a\int_0^{\frac{z^2-a^2}{1-a^2}}\rho - \rho^2\sin{\theta}~ d\rho ~dz ~d\theta$ $~~~~=\frac{4\pi}{1-a^2}$

I think that is correct... maybe someone can check my setup.

I'm stuck with the RHS.. I tried to use cylindrical coordinates and it got ugly quickly, I tried to use generalised spherical coordinates but it got ugly equally as fast. I expect the process to be somewhat ugly, but mine is so much I feel for certain i'm doing something wrong. I'll give a brief overview of my strategy:

For cylindrical coordinates I tried to parameterize it as follows: $\vec{r}(\theta,z)=(\frac{z^2-a^2}{1-a^2}\cos{\theta},\frac{z^2-a^2}{1-a^2}\sin{\theta},z)$. For starters im not even sure if this is right. I then get the normal vector to this by computing: $\vec{n}=\frac{\partial \vec{r}}{\partial \theta}\times\frac{\partial \vec{r}}{\partial z}$. Computing this and normalising i get: $\hat{n}=\frac{(\cos{\theta},\sin{\theta},\frac{-2z}{1-a^2})}{\sqrt{1+\frac{4z^2}{(1-a^2)}}}$.Plugging all of this into the flux integral looks extremely messy and I don't think I will be able to do the integral. So I abort and resort to spherical coordinates...

For spherical coordinates I tried to parameterize it as follows: $\vec{r}(\theta,\phi)= (\frac{a}{\sqrt{a^2-1}}\sin{\theta}\cos{\phi},\frac{a}{\sqrt{a^2-1}}\sin{\theta}\sin{\phi},a\cos{\theta})$ and the rest of the story goes like the cylindrical one..Any help would be greatly appreciated!

NormalsNotFar
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  • Notice $\vec{\nabla}\cdot u = 1-y$ and the integration region is symmetric wrt $y$, the $-y$ part doesn't contribute in final result. the integral reduce to the computation of the volume of the region. Since the cross section for fixed $z$ are circles, the 3d integral reduce to an integral in $z$. The trick to do all these integral is being lazy and don't integrate unless necessary. – achille hui Jan 19 '16 at 21:02

1 Answers1

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You will want to use "spherical coordinates" in the following sense: for an ellipsoid of radii $a,b,c$, we have the standard parameterization $$\mathbf{r}(\theta,\phi)=(a\cos\theta\sin\phi,b\sin\theta\sin\phi,c\cos\phi),$$ for $0\leq\theta\leq2\pi$ and $0\leq\phi\leq\pi$.

To restrict this parameterization to the upper cap you are after, we simply need to find the angle $\phi$ for which the parameterization hits the edge: here's a picture (notice the $xy-$projection is a circle)

enter image description here

The appropriate $\phi$ will occur when $z=1$, which after a little algebra turns out to be $\phi=\frac{\pi}{4}$! Thus it will be a relatively simple spherical integral using these coordinates.

Of course for the bottom flat part of this cap you may use regular ole cylindrical with outward unit normal $-\mathbf{k}$.

charlestoncrabb
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  • Hi charlestoncrabb, thank you for your response!

    From what I can find, the density correction (Jacobian) for these coordinates is $abcr^2\sin{\theta}$. So what do I sub for the $r^2$?

     – NormalsNotFar Jan 19 '16 at 20:59