suppose that vector field F = [3xz, 2xy, -yz^2]
then divergence of F is 3z + 2x - 2yz
but I wonder why doesn't consider partial derivative of 3xz w.r.t z, partial derivative of 2xy w.r.t x, and partial derivative of -yz^2 w.r.t y
For example, when I change x, 3xz and 2xy changes together, I think, in divergence formula, it should contain the variation of 2xy when change x.
When vector field F(v1, v2, v3) is given, I think divergence formula should be like this.
$$ divF = \frac{\partial v_1}{\partial x} + \frac{\partial v_1}{\partial y} + \frac{\partial v_1}{\partial z} + \frac{\partial v_2}{\partial x} + \frac{\partial v_2}{\partial y} + \frac{\partial v_2}{\partial z} + \frac{\partial v_3}{\partial x} + \frac{\partial v_3}{\partial y} + \frac{\partial v_3}{\partial z} $$
help me please
To me, divergence at a point signifies the flux of a field out of a small volume (for simplicity, a cube) centered at that point. More specifically, if have a vector field $F$, and we define $C(r,x)$ to be the cube with sides of length $r$, centered at $x$ and with sides aligned with the coordinate axes, then the divergence of $F$ at $x$ is $$ \nabla F(x)=\lim_{r\to 0}\frac{1}{r^3}\iint_{\partial C(r,x)}F\cdot \vec n\, dA $$ I'm not going to do all the details of the derivation here, but we can see what happens for a small cube: Take the two faces that are normal to the $x$-axis, and see what happens to the contribution for the integral there. The normal vectors are $[-1,0,0]$ and $[1,0,0]$ respectively, and that makes the contribution equal to $$ \frac1{r^3}\left(\iint_{\text{Face }1} F\cdot [-1,0,0]\,dA+\iint_{\text{Face }2} F\cdot [1,0,0]dA\right)\\ =\frac1{r^3}\left(\iint_{\text{Face }1} -F_x\,dA+\iint_{\text{Face }2} F_x\,dA\right) $$ where $F_x$ is the $x$-component of $F$. What are these integrals? They are each done over a face with area $r^2$, and if the cube is small, that means each of them is equal to the area times the value of the integrand at the center point of the face. We get, approximately $$ =\frac1{r^3}\left(-r^2F_x(x-[r/2,0,0])+r^2F_x(x+[r/2,0,0])\right)\\ =\frac{F_x(x+[r/2,0,0])-F_x(x-[r/2,0,0])}{r} $$ If we let $r\to 0$ here, we recognize this as $\frac{\partial F_x}{\partial x}$. Similarly, the contribution from the other two pairs of faces become $\frac{\partial F_y}{\partial y}$ and $\frac{\partial F_z}{\partial z}$. Sum them up, and you have the entire divergence.