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 $p=1$?
 A: Your volume is bounded by an outer sphere of radius 2 and an inner sphere of radius 1.
The normal vector $\hat n$ is always pointing outward of the volume. That means that it points out from the outer sphere ($+$) and it points inward for the inner sphere ($-$).
To calculate the flux, find the outward pointing vector's magnitude at the outer sphere and multiply it with the surface area of the outer sphere.
Do the same thing with the inner sphere, but count that one as negative.
The symbol $p$ is the radius of each of the spheres. It is $p=2$ for the outer sphere and $p=1$ for the inner sphere.
A: Hint: parameterise with spherical coordinates, and your normal will be the outward point normal, i.e. the radius of the sphere. (But there will be two outer surfaces)
A: 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.
A: Note: Used wrong dimension on area of surface. See correction below.
The surface is a "spherical shell" so one can intuit that the normal vector is $\frac{\pm (x, y, z)}{|(x,y,z)|}$ where the sign is positive and negative for the outer and inner surfaces respectively. Or, alternatively,parameterise $\sigma(u,v) = (\cos(u)\cos(v), \cos(u)\sin(v), \sin(u))$. Then $\sigma_{u} = (-\sin(u)\cos(v), -\sin(u)\sin(v), \cos(u))$ and $\sigma_{v} = (-\cos(u)\sin(v), \cos(u)\cos(v), 0)$ and:
$\sigma_{u} \times \sigma_{v} = (-\cos^2(u)\cos(v), -\cos^2(u)\sin(v), -\sin(u)\cos(u)) \\ = -\cos(u)\cdot ((\cos(u)\cos(v), \cos(u)\sin(v), \sin(u)).$
So our normal vector either points either radially inward or outward. Since the "hollowed out sphere" is a 3-manifold, the induced orientation of the boundary is given by n such that the frame $(-n, \sigma_{u}, \sigma{v})$ is right-handed. That is $\sigma_{u} \times \sigma_{v} \sim -n \sim ((\cos(u)\cos(v), \cos(u)\sin(v), \sin(u))$. So, for the positive orientation, it points outwards from the surface (or manifold depending on your preference).
Now, $F\cdot n = -\frac{x^2+y^2+z^2}{(x^2+y^2+z^2)^2} = -\frac{1}{x^2+y^2+z^2}$. So, we know the surface area of a spherical shell and changing sign for orientation we have that the integral evaluates to $-1/4\cdot (4\pi\cdot 2^2) + 1/1\cdot (4\pi\cdot 1^2) = -4\pi + 4\pi = 0$.
