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?
 A: $\newcommand{\bbx}[1]{\bbox[8px,border:1px groove navy]{{#1}}\ }
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 \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}}}
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\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
\\[5mm] & =
\sum_{i}\hat{x}_{i}\int_{\mrm{\partial CV}}\phi\,\hat{x}_{i}\cdot\dd\vec{S}
=
\sum_{i}\hat{x}_{i}\int_{\mrm{\partial CV}}\phi\,\pars{\dd\vec{S}}_{i}
\\[5mm] & =
\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 ):

$$
\left\{\begin{array}{rl}
\ds{P_{\mrm{atm.}}:} & \mbox{Atmospheric Pressure.}
\\[1mm]
\ds{\rho:} & \mbox{Fluid Density.}
\\[1mm]
\ds{g:} & \mbox{Gravity Acceleration}\ds{\ \approx 9.8\ \mrm{m \over sec^{2}}.}
\\[1mm]
\ds{z:} & \mbox{Depth.}
\\[1mm]
\ds{m_{\mrm{fluid.}}:} & \ds{\rho V_{\mrm{body}} = \rho\int_{\mrm{CV}}\,\dd V}
\end{array}\right.
$$
\begin{align}
\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 \\[2mm] & =
-\int_{\mrm{CV}}\rho g\,\hat{z}\,\dd V = -m_{\mrm{fluid}}\, g\,\hat{z}
\end{align}
A: 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
$$\vec{k} \cdot \left(\int_{CV}\nabla\phi dV\right) = \int_{CV} \nabla\cdot(\phi \vec{k}) dV
\stackrel{\color{blue}{\verb/div. theorem/}}{=} \int_{\partial CV} \phi \vec{k} \cdot d\vec{S} = \vec{k} \cdot \left(\int_{\partial CV} \phi d\vec{S}\right)$$
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.
$$\int_{CV}\nabla\phi dV = \int_{\partial CV} \phi d\vec{S}$$
