# Gauss-divergence theorem for volume integral of a gradient field

I need to make sure that the derivation in the book I am using is mathematically correct. The problem is about finding the volume integral of the gradient field. The author directly uses the Gauss-divergence theorem to relate the volume integral of gradient of a scalar to the surface integral of the flux through the surface surrounding this volume, i.e.

$$\int_{CV}^{ } \nabla \phi dV=\int_{\delta CV}^{ } \phi d\mathbf{S}$$

The book page is available via this link: http://imgh.us/Esx.jpg

Is that true? is there any mathematical derivation available for Gauss-divergence theorem (or similar theorem) when we consider gradient instead of divergence?

Does that has any physical significance as in case of divergence?

The statement is true. It is typically proved using following properties of vectors.

Two vectors $$\vec{p}, \vec{q} \in \mathbb{R}^n$$ equals to each other if and only if for all vectors $$\vec{r} \in \mathbb{R}^n$$, $$\vec{r}\cdot \vec{p} = \vec{r}\cdot \vec{q}$$.

Back to our original identity. For any constant vector $$\vec{k}$$, we have

$$\vec{k} \cdot \left(\int_{CV}\nabla\phi dV\right) = \int_{CV} \nabla\cdot(\phi \vec{k}) dV \stackrel{\color{blue}{\verb/div. theorem/}}{=} \int_{\partial CV} \phi \vec{k} \cdot dS = \vec{k} \cdot \left(\int_{\partial CV} \phi dS\right)$$

The first equality holds because $$\vec{k}\cdot\nabla\phi = \nabla\cdot(\phi \vec{k}) - \phi(\nabla\cdot \vec{k})$$ Additionally, since $$\vec{k}$$ is a constant vector, $$\nabla\cdot\vec{k} = 0$$. Hence, $$\vec{k}\cdot\nabla\phi = \nabla\cdot(\phi\vec{k})$$.

Since this is true for all constant vector $$k$$, the two vectors defined by the integrals equal to each other. i.e.

$$\int_{CV}\nabla\phi dV = \int_{\partial CV} \phi dS$$

$\newcommand{\bbx}[1]{\bbox[8px,border:1px groove navy]{{#1}}\ } \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{\mrm{CV}}\nabla\phi\,\dd V & = \sum_{i}\hat{x}_{i}\int_{\mrm{CV}}\partiald{\phi}{x_{i}}\,\dd V = \sum_{i}\hat{x}_{i}\int_{\mrm{CV}}\nabla\cdot\pars{\phi\,\hat{x}_{i}}\,\dd V = \sum_{i}\hat{x}_{i}\int_{\mrm{\partial CV}}\phi\,\hat{x}_{i}\cdot\dd\vec{S} \\[5mm] & = \sum_{i}\hat{x}_{i}\int_{\mrm{\partial CV}}\phi\,\pars{\dd\vec{S}}_{i} = \int_{\mrm{\partial CV}}\phi\,\sum_{i}\pars{\dd\vec{S}}_{i}\hat{x}_{i} =\ \bbx{\int_{\mrm{\partial CV}}\phi\,\dd\vec{S}} \end{align}

One interesting application of this identity is the Archimedes Principle derivation ( the force magnitude over a body in a fluid is equal to the weight of the mass of fluid displaced by the body ):

$$\left\{\begin{array}{rl} \ds{P_{\mrm{atm.}}:} & \mbox{Atmospheric Pressure.} \\[1mm] \ds{\rho:} & \mbox{Fluid Density.} \\[1mm] \ds{g:} & \mbox{Gravity Acceleration}\ds{\ \approx 9.8\ \mrm{m \over sec^{2}}.} \\[1mm] \ds{z:} & \mbox{Depth.} \\[1mm] \ds{m_{\mrm{fluid.}}:} & \ds{\rho V_{\mrm{body}} = \rho\int_{\mrm{CV}}\,\dd V} \end{array}\right.$$

$$\int_{\mrm{\partial CV}}\pars{P_{\mrm{atm.}} + \rho gz}\pars{-\dd\vec{S}} = -\int_{\mrm{CV}}\nabla\pars{P_{\mrm{atm.}} + \rho gz}\,\dd V = -\int_{\mrm{CV}}\rho g\,\hat{z}\,\dd V = -m_{\mrm{fluid}}\, g\,\hat{z}$$