Gauss-divergence theorem for volume integral of a gradient field

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 ):

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

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