If I want to prove that for any scalar field $f:\;\mathbb{R}^3\to\mathbb{R}:$ $$\int_V \boldsymbol{\nabla} f\;\mathrm{d}V=\int_{\partial V} f\;\mathrm{d}\mathbf{S}$$ Can I apply the divergence theorem to $\mathbf{a}_1=(f,0,0),\;\mathbf{a}_2=(0,f,0),\;\mathbf{a}_3=(0,0,f)$ and then stack the equalities into a single vector? So using: $$\int_V \frac{\partial f}{\partial x_i}\mathrm{d}V=\int_{\partial V}f\cdot n_i\;\mathrm{d}S$$ ($n_i$ is the $i$th component in the outward normal $\mathbf{n}$) can I deduce:

$$\int_V \left(\frac{\partial f}{\partial x_1},\frac{\partial f}{\partial x_2},\frac{\partial f}{\partial x_3}\right)\mathrm{d}V=\int_{\partial V}f\Big(n_1,n_2,n_3\Big)\mathrm{d}S$$

$$\Rightarrow \int_V \boldsymbol{\nabla} f\;\mathrm{d}V=\int_{\partial V} f\;\mathrm{d}\mathbf{S}?$$

  • $\begingroup$ One way to show the identity is to apply the divergence theorem on $\vec{k}f$, where $\vec{k}$ is a constant vector, and noting that $\nabla\cdot\vec{k}f=\vec{k}\cdot \nabla f$, since $\vec{k}$ is constant. $\endgroup$
    – Sarastro
    May 31 '14 at 13:33
  • $\begingroup$ Your function $f$ is not a scalar field, it is a vector field (since the codomain is $\Bbb{R}^3$ rather than $\Bbb{R}$). What you are trying to show is that $$\int_{V} (\nabla \cdot f )dV = \int_{\partial V} (f \cdot {\bf n}) dS.$$ $\endgroup$
    – Tom
    May 31 '14 at 13:37
  • $\begingroup$ @Sarastro Thanks, I thought of this but felt uneasy about $\mathbf{k}\cdot \left(\int_V \nabla f \;\mathrm{d}V\right)=\mathbf{k}\cdot \left(\int_S f\mathrm{d}\mathbf{S}\right)\Rightarrow$ the result. Does the implication hold because $\mathbf{k}$ is arbitrary? $\endgroup$
    – custodia
    May 31 '14 at 13:38
  • $\begingroup$ @Tom Sorry the $\mathbb{R}^3$ was a typo. It is a scalar field $\endgroup$
    – custodia
    May 31 '14 at 13:41
  • $\begingroup$ Your proof is fine, and the alternative of dotting with an arbitrary vector is fine, too. $\endgroup$ May 31 '14 at 13:45

Let $\mathbf c$ be some constant vector. Taking the divergence of $\mathbf c f$,

$$\nabla\cdot(\mathbf c f)=\mathbf c \cdot \nabla f+f\nabla\cdot\mathbf c\\ =\mathbf c \cdot \nabla f,$$

since the divergence of a constant vector is zero.

Applying the divergence theorem to $\mathbf c f$,

$$\begin{align} \int_V \mathbf c \cdot \nabla f \,\mathrm{d} V&=\int_{\partial V}\mathbf c f\cdot\mathrm{d} \mathbf{S}\\ \implies \mathbf c \cdot \int_V \nabla f \,\mathrm{d} V&=\mathbf c \cdot \int_{\partial V}f\,\mathrm{d} \mathbf{S}\\ \implies 0&=\mathbf{c}\cdot\left(\int_V \nabla f \,\mathrm{d} V - \int_{\partial V}f\,\mathrm{d} \mathbf{S}\right). \end{align}$$

Since the previous equation holds for any constant vector $\mathbf{c}$, it follows that:

$$\int_V \nabla f \,\mathrm{d} V = \int_{\partial V}f\,\mathrm{d} \mathbf{S}.$$


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