<s>[INST] Which is the normal vector?? Apply the divergence theorem over the region $1 \leq x^2+y^2+z^2 \leq 4$ for the vector field $\overrightarrow{F}=-\frac{\hat{i}x+\hat{j}y+\hat{k}z}{p^3}$, where $p=(x^2+y^2+z^2)^\frac{1}{2}$.
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The divergence theorem is the following:
$$\iiint_D \nabla{\overrightarrow{F}}  \cdot dV=\iint_S \overrightarrow{F} \cdot \hat{n} d \sigma$$
How can I calculate the integral: $$\iint_S \overrightarrow{F} \cdot \hat{n} d \sigma$$ ??
Which is the normal vector $\hat{n}$??
EDIT:
In my notes it is:
$f=x^2+y^2+z^2$
$\displaystyle{\hat{n}=\pm \frac{\bigtriangledown f}{|\bigtriangledown f|}=\pm \frac{\hat{i}x+\hat{j}y+\hat{k}z}{p}}$
$"+": p=2$
$"-" \text{ for } p=1$
I haven't understood the signs.. Could you explain me why it's $\pm$? And also why is it $"+"$ when $p=2$ and $"-"$ when $p=1$?
 [/INST] First you need two surfaces to bind your region, the two surfaces are $x^2+y^2+z^2 = 4$ and $x^2+y^2+z^2=1$.First calculate the surface with the outer surface as 
$$\int_{-2}^2\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} \vec F(x, y, \sqrt{4-x^2-y^2})\cdot \frac{\nabla(x^2+y^2+z^2)}{|\nabla(x^2+y^2+z^2)\cdot \hat k|}dydx - \int_{-2}^2\int_{-\sqrt{4-x^2}}^{-\sqrt{4-x^2}} \vec F(x, y, -\sqrt{4-x^2-y^2})\cdot \frac{\nabla(x^2+y^2+z^2)}{|\nabla(x^2+y^2+z^2)\cdot (-\hat k)|}dydx$$
Note that the first integral is for upper surface of sphere and second integral is for lower surface. Calculate the gradient first in those two integrals, and replace $z = \sqrt{4-x^2-y^2}$ on upper surface of sphere and $z = -\sqrt{4-x^2-y^2}$ on lower surface.
For inner surface, the expression is similar, just note the direction of normal surface. It's inward. And it's better to use spherical coordinate as Elya suggested on other answer.
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