A remark on the divergence theorem

The divergence theorem states that for a compact domain {D} in {\mathbb R^3} with piecewise smooth boundary {\partial D}, then for a smooth vector field {\textbf{F}} on {D}, we have

\displaystyle \begin{array}{rl} \displaystyle \iint_{\partial D}\textbf{F}\cdot \textbf{n}\,dS=\iiint_D \nabla \cdot\textbf{F}\,dV,\end{array}

where {\textbf{n}} is the unit outward normal and {\nabla \cdot\textbf{F}} is the divergence of {\textbf{F}}.

In most textbooks, the divergence theorem is proved by the following strategy:

  1. Prove that

    \displaystyle \begin{array}{rl} \displaystyle  \iint _{\partial D}(0, 0, R)\cdot \textbf{n} \;dS= \iiint _D\frac{\partial R}{\partial z}\;dV \ \ \ \ \ (1)\end{array}

    where {D} is a so called type {1} domain, i.e. {D=\{(x, y, z): (x, y)\in \Omega, \underline f(x, y)\le z\le \bar f(x, y)\}} for some region {\Omega\subset \mathbb R^2} and {\bar f, \underline f\in C^1(R)}. i.e. {D} is bounded by the graphs of two functions with variables {x, y}. This can be quite easily done using the fundamental theorem of calculus.

  2. Similarly prove

    \displaystyle \begin{array}{rl} \displaystyle  \iint _{\partial D}(P, 0, 0)\cdot \textbf{n} \;dS= \iiint _D\frac{\partial P}{\partial x}\;dV \ \ \ \ \ (2)\end{array}


    \displaystyle \begin{array}{rl} \displaystyle  \iint _{\partial D}(0, Q, 0)\cdot \textbf{n} \;dS= \iiint _D\frac{\partial Q}{\partial y}\;dV \ \ \ \ \ (3)\end{array}

    for the so called type {2} domain and type {3} domain respectively.

  3. Argue that for sufficiently nice {D}, it can be cut into finitely many pieces {D_i}, each of which are of both type {1}, {2}, and {3} (the so called type {4} domain). Thus we can compute the divergence integral {\displaystyle \iiint_D \nabla \cdot \textbf{F}\,dV} by adding the integrals of the pieces and apply the previous result (using linearity). Argue that the orientation of the common face of these {D_i} (if any) is opposite to each other and thus a cancellation argument will give the surface integral {\displaystyle \iint _{\partial D}\textbf{F}\cdot \textbf{n}\;dS}.

I am slightly disappointed by this approach because even for the so called type {1} domain, we still cannot prove the divergence theorem very directly (say, by reducing to a double integral on some planar domain {\Omega} and apply Green’s theorem), but have to further cut it into even smaller type {4} domains.

In this note we modify the strategy above by proving (1), (2), (3) all hold on a type {1} domain. In particular, I will show that (2) and (3) can be proved for a type {1} domain by using Green’s theorem.


  1. We first prove (1). Let {\textbf{F}=(0,0, R)}, {D=\{(x, y, z): (x, y)\in \Omega, \underline f(x, y)\le z\le \bar f(x, y)\}}.

    Let {S_{\textrm{top}}=\{z=\bar f(x,y)\}} and {S_{\textrm{bottom}}=\{z=\underline f(x,y)\}} be the top and bottom face respectively.

    As an outward normal of {S_{\textrm{top}}} and {S_{\textrm{bottom}}} is given by {(-\bar f_x, -\bar f_y, 1)} and {({\underline f\,}_x, {\underline f\,}_y, -1)} respectively, it is easy to see that on the lateral face {S_{\textrm{lateral}}}, {\textbf{F}\cdot n=0} and so

    \displaystyle \begin{array}{rl} \displaystyle    \iint_{\partial D} \textbf{F}\cdot \textbf{n} \;dS =& \displaystyle \left(\iint_{S_{ \textrm{top} }}+\iint _{S_{\textrm{bottom}}}\right)\textbf{F}\cdot \textbf{n} \;dS\\ =& \displaystyle \iint_\Omega \left[R(x, y, \bar f(x, y))-R(x, y, \underline f(x, y))\right] dA\\ =& \displaystyle \iint_\Omega \left(\int_{\underline f(x,y)}^{\bar f(x,y)}\frac{\partial R}{\partial z}dz\right) dxdy\\ =& \displaystyle \iiint_D \frac{\partial R}{\partial z}dV.   \end{array}

  2. Now we prove the remaining cases. For simplicity let us prove only (2), as (3) is similar. With the same setting as in the previous part except that {\textbf{F}=(P, 0, 0)}, we compute

    \displaystyle \begin{array}{rl} \displaystyle   \left(\iint_{S_{ \textrm{top} }}+\iint _{S_{\textrm{bottom}}}\right)\textbf{F}\cdot \textbf{n} \;dS =\iint_\Omega \left[-P(x, y, \bar f (x, y))\bar f_x +P(x, y, \underline f (x, y)) {\underline f\,} _x\right] dA  \ \ \ \ \ (4)\end{array}

    On the other hand, let {\partial \Omega} be parametrized by {\textbf{x}(t)=(x(t), y(t))} for {t\in I} (with positive orientation), then the lateral face {S_{\textrm{lateral}}} is given by

    \displaystyle \begin{array}{rl} \displaystyle S_{\textrm{lateral}}=\{(\textbf{x}(t), z): t\in I, \underline f(\textbf{x}(t))\le z\le \bar f(\textbf{x}(t))\}\end{array}

    and an outward normal is then given by {(y', -x', 0)}. From this we compute

    \displaystyle \begin{array}{rl} \displaystyle   \iint _{S_{\textrm{lateral}}} \textbf{F}\cdot \textbf{n} \;dS =\int_{\partial \Omega}\left(\int_{\underline f(\textbf{x}(t))}^{\bar f(\textbf{x}(t))}P(x, y, z)dz\right)dy. \end{array}

    Define {\phi(x, y): \Omega\rightarrow \mathbb R} by

    \displaystyle \begin{array}{rl} \displaystyle \phi(x, y)= \int_{\underline f(x,y)}^{\bar f(x, y)}P(x, y, z)dz. \end{array}

    Then by Green’s theorem,

    \displaystyle \begin{array}{rl} \displaystyle   \iint _{S_{\textrm{lateral}}} \textbf{F}\cdot \textbf{n} \;dS =\int_{\partial \Omega}\phi \;dy =\iint_{\Omega}\frac{\partial \phi}{\partial x} \;dA.  \ \ \ \ \ (5)\end{array}

    We compute

    \displaystyle \begin{array}{rl} \displaystyle  \frac{\partial \phi}{\partial x}= P(x, y, \bar f(x, y))\bar f_x -P(x, y, \underline f(x, y)){\underline f\,}_x +\int_{\underline f(x,y)}^{\bar f (x, y)}\frac{\partial P}{\partial x}dz. \ \ \ \ \ (6)\end{array}

    Exercise 1 Show that

    \displaystyle \begin{array}{rl} \displaystyle \frac{d}{dx}\left(\int_{a(x)}^{b(x)}c(x, t)dt\right)=c(x, b(x))b'(x)-c(x, a(x))a'(x)+ \int_{a(x)}^{b(x)}\frac{\partial c}{\partial x}(x, t)dt. \end{array}

    Hint: let {\psi(u, v, x)=\int_{u}^{v}c(x, t)dt} and use the chain rule and the fundamental theorem of calculus.

    Adding (4) and (5) together and comparing with (6), we obtain

    \displaystyle \begin{array}{rl} \displaystyle    \iint_{\partial D}\textbf{F}\cdot \textbf{n}\;dS =\iint_\Omega \left(\int_{\underline f(x,y)}^{\bar f(x, y)}\frac{\partial P}{\partial x}dz\right)dx\;dy =\iiint _D\frac{\partial P}{\partial x}dV.   \end{array}


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