Applied Divergence Theorem

I'm struggling immensely with this problem, I can't seem to evaluate and use the Divergence Theorem correctly. I've approached this problem in both spherical and cylindrical coordinates and can't solve it. The problem states;

A non-zero scalar field $\psi$ is such that $||\vec{\nabla}\psi||^2$ = 3$\psi$ and $\vec{\nabla}\cdot(\psi\vec{\nabla}\psi)$ = 10$\psi$. Evaluate $$\iint_{S}\vec{\nabla}\psi \cdot \hat{n}dS$$ when S is the surface of the region in the first octant bounded by z = $\sqrt{x^2+y^2}$, z = $\sqrt{1-x^2-y^2}$, y = x and y = $\sqrt{3}x$. Also we are required to solve this problem using $\underline{both}$ spherical and cylindrical coordinates.

• Are you sure you wrote down the integral correctly? It seems like the information given about $\psi$ is not useful at all. – Chee Han Nov 27 '16 at 22:54
• @CheeHan Please see my "HINT" in the posted solution. – Mark Viola Nov 27 '16 at 22:54
• @Dr.MV I wrote down exactly the same thing, but it doesn't correspond to the integral at all since if you were to apply Divergence theorem, you get the integral over $D$ of the Laplacian of $\psi$. (Ahhhhhh I see, nvm, good catch!) – Chee Han Nov 27 '16 at 22:56

HINT:

Using the product rule for the divergence, we can write

\begin{align} \nabla \cdot \left(\psi \nabla \psi\right)&=\psi \nabla ^2(\phi)+\nabla (\psi)\cdot \nabla (\psi)\\\\ &=\psi \nabla ^2(\phi)+\left|\nabla (\psi)\right|^2 \tag 1 \end{align}

SPOILER ALERT: Scroll over the highlighted area to reveal the solution.

Solving $(1)$ for $\nabla^2(\psi)$, we find that $$\nabla^2(\psi)=\frac{\nabla \cdot (\psi \nabla (\psi))-\left|\nabla (\psi)\right|^2}{\psi}=7$$Using the Divergence Theorem, the result is $7\times \,\,\text{Volume enclosed}$.

Now, solve for $\nabla^2 \cdot (\psi)$, apply the given relationships, and integrate a constant over the volume of interest.

• Sorry I asked such a stupid question, good hint (: – Chee Han Nov 27 '16 at 22:57
• @CheeHan Thank you! Much appreciated. -Mark – Mark Viola Nov 27 '16 at 23:00
• Thank you very much Dr. MV! However after I have the Laplacian of $\psi$ = 7 how would I proceed to implementing the Divergence Theorem and evaluating the triple integral with the integrand div($\psi$) over the volume of interest? I'm just not seeing it :( – Nelly Nov 28 '16 at 1:04
• You're welcome. My pleasure. The divergence of the gradient is the Laplacian. And the Laplacian is $7$. Now integrate $7$ over the volume of interest. – Mark Viola Nov 28 '16 at 1:07
• Oh my, I just realized what I was misinterpreting. Thank you very much for your detailed explanation and your patience! – Nelly Nov 28 '16 at 1:43