How do the derivatives [the divergence] in the volume give the flux [vectors] out of the boundary?
It doesn't appear to do that; when ever I think of these derivatives, particularly derivatives like constant values, they don't total up to any value which I calculate on the boundary. What is the step I have forgotten?
For example: When ever $F = \langle x, y, 0 \rangle$ inside the rectangle between the axis and the planes where $x=2,y=5,z=2$:
$\dfrac{\partial F_{x}}{\partial x}$ = $1$.
$\dfrac{\partial F_{y}}{\partial y}$ = $1$.
$\dfrac{\partial F_{z}}{\partial z}$ = $0$.
$\displaystyle \iiint_{V} div(F)dV = \int_{z=0}^{z=2} \int_{y=0}^{y=5} \int_{x=0}^{x=2} 2 \; dxdydz$
$\displaystyle \iiint_{V} div(F)dV = \int_{z=0}^{z=2} \int_{y=0}^{y=5} 4 \; dydz$
$\displaystyle \iiint_{V} div(F)dV = \int_{z=0}^{z=2} 20 \; dz$
$\displaystyle \iiint_{V} div(F)dV = 40 = \iint_{S} F \cdot n \; dS$
That is as far as I can get to, will someone write out the surface integral [and thereby teach how to dot with the normal vector] because that would be so helpful – the divergence's scalar and the values along the boundary do not seem to be equal when I thought about them..
Best Answer
There are 6 faces, and the flux has to be calculated separately through each. I will show a couple:
(1) One face is in the $xy$-plane. The flux through that surface is zero, since the normal vector is in the negative $\hat{k}$ direction, and the vector field has a zero component in that way.
(2) Take the face in the plane $x=2$. For this face, the outward normal is in the positive $\hat{i}$ direction. This yields the integral $$\int F \cdot d\hat{n}=\int_0^2 \int_0^5 ⟨x,y,0⟩ \cdot <1,0,0>dydz$$ Since $x=2$ in this plane, we simply get $$\int_0^2 \int_0^5 2dydz=20$$ Can you do the others?