3
$\begingroup$

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:

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

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

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Ted Shifrin May 31 '14 at 13:45
1
$\begingroup$

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:

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.