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I encountered the following problem in a practice Advanced Calc exam, and I have an issue.

Suppose $\phi:\mathbb{R}^3\to\mathbb{R}$ is a strictly positive function satisfying $|\nabla\phi|^2=4\phi$ and $\Delta(\phi^2)=20\phi$. Evaluate $$\int_S\frac{\partial\phi}{\partial n}ds,$$ where $S$ is the surface of the unit sphere centered at the origin, $\cfrac{\partial\phi}{\partial n}$ is the directional derivative of the unit outward notmal to $S$, and $ds$ is the surface measure of $S$.

I cannot for the life of me recall what the $\Delta$ means in this context. Can someone help me out?

Note that I am not (at present) looking for any hints or help in evaluating the integral, though I will update this post later if I'm still stymied even after having this notation issue cleared up.

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Maybe it is the Laplacian $$\Delta = \sum\limits_{k = 1}^n {\frac{{{\partial ^2}}}{{\partial x_k^2}}} $$ –  Pedro Tamaroff Jun 7 '12 at 17:25
    
For $$\Bbb R^2$$ is is $$\Delta f = {\nabla ^2}f = \frac{{{\partial ^2}f}}{{\partial {x^2}}} + \frac{{{\partial ^2}f}}{{\partial {y^2}}}$$ For $$\Bbb R^3$$ is is $$\Delta f = {\nabla ^2}f = \frac{{{\partial ^2}f}}{{\partial {x^2}}} + \frac{{{\partial ^2}f}}{{\partial {y^2}}} + \frac{{{\partial ^2}f}}{{\partial {z^2}}}$$ –  Pedro Tamaroff Jun 7 '12 at 17:27
    
@PeterTamaroff: I followed the initial comment, but thanks for making sure! –  Cameron Buie Jun 7 '12 at 17:33
    
@PeterTamaroff: That turned out to be exactly what was needed. If you'd like to post it as an answer, I'll accept it, and we can put this thing to bed. –  Cameron Buie Jun 11 '12 at 15:07

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It is probably the Laplacian Operator

$$\Delta = \sum\limits_{k = 1}^n {\frac{{{\partial ^2}}}{{\partial x_k^2}}}$$

For two dimesions, you'd get

$$\Delta f = {\nabla ^2}f = \frac{{{\partial ^2}f}}{{\partial {x^2}}} + \frac{{{\partial ^2}f}}{{\partial {y^2}}}$$

For three,

$$\Delta f = {\nabla ^2}f = \frac{{{\partial ^2}f}}{{\partial {x^2}}} + \frac{{{\partial ^2}f}}{{\partial {y^2}}} + \frac{{{\partial ^2}f}}{{\partial {z^2}}}$$ and so on.

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