Skip to main content

Subsection 1.4.4 Gauss’s Divergence Theorem

The surface integral of the normal component of function \(\vec{F}\) over the boundary of a closed surface \(S\) of any shape is equal to the volume integral of the divergence of \(\vec{F}\) taken throughout the enclosed volume V, i.e.,
\begin{equation*} \mathop{\vcenter{\huge\unicode{x222F}\,}}_{S} \vec{F} \cdot \hat{n} \,dS =\iiint_{V}(\vec{\nabla}\cdot \vec{F})\,dV \end{equation*}

Proof.

From right - hand side of the above equation, we have -
\begin{equation} \iiint\limits_{V}(\vec{\nabla}\cdot \vec{F})\,dV=\iiint\limits_{V}\left[\frac{\partial F_{x}}{\partial x}+\frac{\partial F_{y}}{\partial y} +\frac{\partial F_{z}}{\partial z}\right] \,dV\tag{1.4.1} \end{equation}
Since the theorem is valid for any shape. We chose parallelopiped of volume V enclosed by the surface S, as showh in the figure Figure 1.4.4.
Figure 1.4.4.
From eqn. (1.4.1) let us first evaluate -
\begin{equation*} \iiint\limits_{V}\frac{\partial F_{z}}{\partial z}\,dx\,dy\,dz \quad \text{from z' to z''.} \end{equation*}
That is,
\begin{equation*} \iiint\limits_{z'}^{z"}\frac{\delta F_{z}}{\delta z}\,dx\,dy\,dz \end{equation*}
[value of \(\vec{F}\) along z axis is \(F_{z}(x,y,z)\)]
\begin{equation*} \therefore \hspace{2pt} \iint\limits_{S}[F_{z}(x,y,z")-F_{z}(x,y,z')\,dx\,dy] \end{equation*}
\begin{equation} =\iint\limits_{S_{2}}F_{z}(x,y,z")\,dx\,dy-\iint\limits_{S_{1}}F_{z}(x,y,z')\,dx\,dy\tag{1.4.2} \end{equation}
But, \(\hat{n_{1}}\cdot\hat{k}\,dS_{1}=-\,dx\,dy\) (bottom surface ) and \(\hat{n_{2}}\cdot\hat{k}dS_{2}=\,dx\,dy\) (top).
Hence from eqn. (1.4.2), we get-
\begin{equation*} \iiint\limits_{z'}^{z"}\frac{\partial F_{z}}{\partial z}\,dx\,dy\,dz =\iint\limits_{S_{2}}F_{z}(x,y,z")\hat{n_{2}}\cdot\hat{k}\,dS_{2}+\iint\limits_{S_{1}}F_{z}(x,y,z')\hat{n_{1}}\cdot\hat{k}dS_{1} \end{equation*}
\begin{equation} \therefore\quad \iiint\limits_{z'}^{z"}\frac{\partial F_{z}}{\partial z}\,dx\,dy\,dz = \mathop{\vcenter{\huge\unicode{x222F}\,}}_{S} F_{z}\hat{k} \cdot \hat{n} \,dS \tag{1.4.3} \end{equation}
where surface integrals of the other sides are zero because \(\hat{k}\) is perpendicular to the surfaces \(\,dS_{3},\,dS_{4},\,dS_{5},\,dS_{6}\text{.}\) Similarly, we can show that
\begin{equation} \iiint\limits_{V}\frac{\partial F_{x}}{\partial x}dxdydz = \mathop{\vcenter{\huge\unicode{x222F}\,}}\limits_{S} F_{x}\hat{i} \cdot \hat{n} \,dS \tag{1.4.4} \end{equation}
and
\begin{equation} \iiint\limits_{V}\frac{\partial F_{y}}{\partial y}\,dx\,dy\,dz = \mathop{\vcenter{\huge\unicode{x222F}\,}}\limits_{S} F_{y}\hat{j} \cdot \hat{n} \,dS \tag{1.4.5} \end{equation}
Adding eqns. (1.4.3)–(1.4.5), we get-
\begin{equation*} \iiint\limits_{V}(\vec{\nabla}\cdot \vec{F})\,dV = \mathop{\vcenter{\huge\unicode{x222F}\,}}\limits_{S} (F_{x}\hat{i}+F_{y}\hat{j}+F_{z}\hat{k}) \cdot \hat{n} \,dS = \mathop{\vcenter{\huge\unicode{x222F}\,}}\limits_{S} \vec{F}\cdot\vec{\,dS} \end{equation*}