Assume a vector field $\boldsymbol{V}=M(x,y)\boldsymbol{i}+N(x,y)\boldsymbol{j}$ The divergence is then: $$\frac{\partial M}{\partial x}+\frac{\partial N}{\partial y}$$

My book(Thomas's calculus) explains this using a square in the x-y plane whose area goes to zero which is similar to this answer How would one arrive at the formulas for divergence and curl? Following the same argument, why is the change in the x For N (i.e $N(x+\Delta x,y))$) as we move along the x-axis not included, similarly for the change in y for M.

i.e why are these not included in the derivation of the divergence formula?

$$(-N(x,y)+N(x+\Delta x,y))\Delta x$$ and $$(-M(x,y)+M(x,y+\Delta y))\Delta y$$

  • $\begingroup$ Because you're finding the flux outwards across the rectangle. So when you're moving parallel to the $x$-axis, you want the component of $\mathbf V$ normal to that, which means the $\mathbf j$ component, etc. $\endgroup$ – Ted Shifrin Jan 24 '18 at 1:40
  • $\begingroup$ Sorry I made a mistake in my question. I edited the question. $\endgroup$ – NegativeTension Jan 24 '18 at 1:55

So, as you go around the little rectangle, the flux is (approximately) the sum of the $\mathbf V\cdot\mathbf n \Delta s$ contributions. Starting with the bottom edge and proceeding counterclockwise, we have \begin{align*} \text{FLUX} &\approx\mathbf V\cdot (-\mathbf j) \Delta x + \mathbf V\cdot\mathbf i \Delta y + \mathbf V\cdot\mathbf j \Delta x + \mathbf V\cdot(-\mathbf i)\Delta y\\ &= -N(x,y)\Delta x + M(x+\Delta x,y)\Delta y + N(x,y+\Delta y)\Delta x + M(x,y)\Delta y \\&= \left(M(x+\Delta x,y)-M(x,y)\right)\Delta y + \left(N(x,y+\Delta y)-N(x,y)\right)\Delta x \\ &\approx \left(\frac{\partial M}{\partial x} + \frac{\partial N}{\partial y}\right)\Delta x\Delta y. \end{align*}

The terms to which you refer don't show up because they would be higher order in $\Delta x$. As we move along the bottom edge, of course $x$ is varying, but this variation will disappear in the limit. By the intermediate value theorem for integrals, the bottom edge will actually be $-N(\xi,y)\Delta x$ for some $\xi$ satisfying $x\le\xi\le x+\Delta x$. We then approximate this as $$-N(\xi,y)\Delta x \approx -\big(N(x,y)+\left(\frac{\partial N}{\partial x}\right)(\xi-x)\big)\Delta x.$$ Since $\xi-x \le \Delta x$, this term will contribute something bounded by $(\Delta x)^2$. When we combine this with the corresponding term from the upper edge, we'll have something that looks like $$\frac{\partial}{\partial y}\left(\frac{\partial N}{\partial x}\right) (\Delta x)^2\Delta y,$$ and such terms will disappear in the limit when we convert to a double integral.


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