Simple Divergence Theorem

The (simple) Divergence Theorem is an integral theorem that relates the line integral of a vector field to the double integral of the divergence of the field.

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Definition §

Divergence Theorem ^theorem

Let $D$ be a simply connected region on the 2D plane bounded by a smooth (potentially piecewise) simple, closed curve (loop) $C$ that is positively orientated. Let $\hat{n}$ be the unit normal vector chosen such that it points ‘outwards’ from the region. Finally, let $\vec{F}=(P(x,y),Q(x,y))$ be a $C^1$ vector field on an open region containing $D$ (that is, the domain of the vector field is a superset of $D$)

Gauss’ Divergence Theorem states:

$${\int }_C\vec{F}\cdot \hat{n}\text{d}s={\iint }_D\nabla \cdot \vec{F}\text{d}x\text{d}y={\iint }_D\text{div}(F)\text{d}x\text{d}y={\iint }_D\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}\text{d}x\text{d}y$$

The divergence theorem is a special case of Green's Theorem

That is, we will prove that:

$${\int }_C\vec{F}\cdot \hat{n}\text{d}s={\iint }_D\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}\text{d}x\text{d}y$$

is a special case of Green’s Theorem, which is

$${\int }_C(P\mathrm{d}x+Q\mathrm{d}y)\mathrm{d}s={\iint }_D\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\mathrm{d}x\mathrm{d}y$$

First note that:

$$\hat{n}\cdot \mathrm{d}s=\frac{(\frac{\text{d}y}{\text{d}t},-\frac{\text{d}x}{\text{d}t})}{\Vert {\vec{c}}^{{}^{\prime }}(t)\Vert }\cdot \Vert {\vec{c}}^{{}^{\prime }}(t)\Vert \mathrm{d}t=(\frac{\text{d}y}{\text{d}t},-\frac{\text{d}x}{\text{d}t})\mathrm{d}t$$

Since $P,Q$ are arbitrary, we let $P=Q^{\ast }$ and $Q=-P^{\ast }$ Let’s start by showing both LHS are equivalent. Assume $C$ is traced by $\vec{c}(t)$ for $t\in [a,b]$

$$\begin{array}{rl}{\int }_C\vec{F}\cdot \hat{n}\text{d}s& ={\int }_0^{2\pi }(P(\vec{c}(t)),Q(\vec{c}(t)))\cdot \left(\frac{\text{d}y}{\text{d}t},-\frac{\text{d}x}{\text{d}t}\right)\text{d}t\\ & ={\int }_a^{b}P\text{d}y-Q\text{d}x\\ & ={\int }_a^b{P}^{\ast }\text{d}y+Q^{\ast }\text{d}x\end{array}$$

Which is exactly the left side of Green’s Theorem , then the right side of the Divergence Theorem becomes:

$$\begin{array}{r}{\iint }_D\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}\text{d}x\text{d}y={\iint }_D\frac{\partial Q^{\ast }}{\partial x}-\frac{\partial P^{\ast }}{\partial y}\text{d}x\text{d}y\end{array}$$

Which is the right side of Green’s Theorem. Hence:

$${\int }_C\vec{F}\cdot \hat{n}\text{d}s={\iint }_D\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}\text{d}x\text{d}y⟺{\int }_C(P^{\ast }\mathrm{d}x+Q^{\ast }\mathrm{d}y)\mathrm{d}s={\iint }_D\left(\frac{\partial Q^{\ast }}{\partial x}-\frac{\partial P^{\ast }}{\partial y}\right)\mathrm{d}x\mathrm{d}y$$