In the text we obtained the result of of $\nabla \cdot F$ in cartesian coordinates, by integrating over the surface of a small rectangular parallelepiped. As an example of the fact that this result is independent of the surface, rederive it using the wedge shaped surface as shown below
I've been stuck on this problem for quite a bit now, what's throwing me off is that tilted plane area. Anyhow , here is my work so far:
Reference centroid of prism as $(x,y,z)$ and let our vector field be $F(x,y,z) = <F_x,F_y,F_z>$
$$\phi_{S_1 + S_4} = \frac{ \Delta z \Delta y }{2}\left[ F_x(x+ \frac{\Delta x}{2},y,z) - F_x(x-\frac{\Delta x}{2},y,z) \right]$$
$$\phi_{S_3} = -\Delta z \Delta x \left[ F_y(x,y - \frac{\Delta y}{2},z) \right]$$
$$\phi_{S_5}= -\Delta x \Delta y \left[ F_z(x,y,z- \frac{\Delta z}{2}) \right]$$
Now the tricky part is the tilted area of $S_2$, I related it to the flat area of $S_5$ by the formula:
$$ dS' = \frac{dA}{\cos \theta}= \frac{\Delta y \Delta x}{\cos \theta}$$
Which is a formula previously discussed in the book, now for the value of the function on $S_2$, I think it should be the function plus the directional derivative in the direction to the face $\hat{n}$ from the cgeometric center of prism scaled by the distance to that face (l)
$$ \phi_{S_2} = \frac{ \Delta x \Delta y}{\cos \theta} \left[ F(x,y,z) + \nabla F \cdot l \hat{n} \right]$$
The real struggle here is on how to evaluate the gradient of $F$ and dot with the normal because, the calculation here is supposed to be done for an arbitary vector field. One idea I had was to consider the slant height as $\sqrt{\Delta z^2 + \Delta y^2}$ but I ditched that approach because I'd have the square of differentials in the end.
Here is a proof of divergence theorem over a small prism with vertices, $A \ (x, y, z), B \ (x+\Delta x, y, z), C \ (x + \frac{\Delta x}{2}, y, z + \Delta z), \ D (x, y + \Delta y, z)$, $E \ (x+\Delta x, y + \Delta y, z)$ and $F \ (x + \frac{\Delta x}{2}, y + \Delta y, z + \Delta z) $
Vector field $ \vec F = (P, Q, R)$
Outward unit normal vector to surface $ABC$ is $(0, -1, 0)$ and to surface $DEF$ is $(0, 1, 0)$
If centroid of $ABC$ is $(x_g, y, z_g)$ then of $DEF$ is $(x_g, y + \Delta y, z_g)$.
Outward flux through surface $ABC = -Q (x_g, y, z_g) \ dA = - \frac{1}{2} \Delta x \ \Delta z \ Q (x_g, y, z_g)$
Similarly outward flux through surface $DEF = \frac{1}{2} \Delta x \ \Delta z \ Q (x_g, y + \Delta y, z_g)$
Net outward flux through $ABC$ and $DEF$ $= \frac{1}{2} \Delta x \ \Delta y \ \Delta z \ \frac{\Delta Q}{\Delta y} = \frac{\partial Q}{\partial y} \Delta V \tag1$
Similarly outward flux through surface $ABED$,
$ = - \Delta x \ \Delta y \ R (x + \frac{\Delta x}{2}, y + \frac{\Delta y}{2}, z) \tag 2$
To find outward normal vector to surface $ADFC$, we take cross product of $\vec {AC}$ and $\vec {AD}$ in positive z-direction.
$\vec{AC} = (\frac{\Delta x}{2}, 0, \Delta z), \vec{AD} = (0, \Delta y, 0)$
$\vec N = (- \Delta y \ \Delta z, 0, \frac{\Delta x \ \Delta y}{2})$
Outward flux through surface $ADFC$ $= - \Delta y \ \Delta z \ P(x + \frac{\Delta x}{4}, y + \frac{\Delta y}{2}, z + \frac{\Delta z}{2}) + \frac{\Delta x \ \Delta y}{2} \ R(x + \frac{\Delta x}{4}, y + \frac{\Delta y}{2}, z + \frac{\Delta z}{2})$
Similarly, outward flux through surface $BCFE$ $= \Delta y \ \Delta z \ P(x + \frac{3 \Delta x}{4}, y + \frac{\Delta y}{2}, z + \frac{\Delta z}{2}) + \frac{\Delta x \ \Delta y}{2} \ R(x + \frac{3 \Delta x}{4}, y + \frac{\Delta y}{2}, z + \frac{\Delta z}{2})$
Adding, net flux through $ADFC$ and $BCFE$ can be approximated as,
$ = \Delta x \ \Delta y \ \Delta z \ (\frac{3}{4} - \frac{1}{2}) \frac{\partial P}{\partial x} + \Delta x \ \Delta y \ R(x + \frac{\Delta x}{2}, y + \frac{\Delta y}{2}, z + \frac{\Delta z}{2})$
$ = \frac{\partial P}{\partial x} \ \Delta V + \Delta x \ \Delta y \ R(x + \frac{\Delta x}{2}, y + \frac{\Delta y}{2}, z + \frac{\Delta z}{2}) \tag3$
Adding $(2)$ and $(3)$, we get net flux through $ABED, ADFC$ and $BCFE$
$ = \frac{\partial P}{\partial x} \ \Delta V + \Delta x \ \Delta y \ \frac{\Delta z}{2} \ \frac{\partial R}{\partial z} = \frac{\partial P}{\partial x} \ \Delta V + \frac{\partial R}{\partial z} \ \Delta V \tag4$
Finally adding $(1)$ and $(4)$,
Net outward flux through the prism $ = (\nabla \cdot \vec F) \ \Delta V$