Proofs of Integral Theorems

Proof

Step 1 (Reduction to triply-regular regions). We first treat the case where $V$ is regular in all three coordinate directions, meaning that for each axis the upper and lower bounds are given by continuous functions of the remaining two coordinates. As an example, z-regularity means that there exist continuous functions $z_1(x,y) \le z_2(x,y)$ on a planar domain $D \subset \mathbb{R}^2$ such that \begin{equation}V = \{(x,y,z) : (x,y) \in D,\ z_1(x,y) \le z \le z_2(x,y)\} \label{eq:D-1}\end{equation} x-regularity and y-regularity are defined analogously. The general bounded closed region is handled by partitioning into finitely many triply-regular subregions (Step 6).

Step 2 (Fubini on the z-component). Apply Fubini's theorem to the volume integral of $\partial A_z/\partial z$.

\begin{equation}\displaystyle\iiint_V \dfrac{\partial A_z}{\partial z}\,dV = \displaystyle\iint_D \left( \displaystyle\int_{z_1(x,y)}^{z_2(x,y)} \dfrac{\partial A_z}{\partial z}\,dz \right) dA \label{eq:D-2}\end{equation}

The fundamental theorem of calculus evaluates the inner integral.

\begin{equation}\displaystyle\int_{z_1(x,y)}^{z_2(x,y)} \dfrac{\partial A_z}{\partial z}\,dz = A_z(x,y,z_2(x,y)) - A_z(x,y,z_1(x,y)) \label{eq:D-3}\end{equation}

Step 3 (Conversion to surface integrals on top and bottom). Decompose $\partial V$ into the top $S_+$ ($z = z_2$), the bottom $S_-$ ($z = z_1$), and the side $S_{\text{side}}$. The outward normal points upward on $S_+$ and downward on $S_-$, so the surface elements satisfy

\begin{equation}\hat{\mathbf{n}} \cdot \hat{\mathbf{e}}_z\,dS = +dA \quad (\text{on } S_+),\qquad \hat{\mathbf{n}} \cdot \hat{\mathbf{e}}_z\,dS = -dA \quad (\text{on } S_-) \label{eq:D-4}\end{equation}

Therefore the right-hand side of $\eqref{eq:D-3}$ equals

\begin{equation}\displaystyle\iint_{S_+} A_z\,(\hat{\mathbf{n}} \cdot \hat{\mathbf{e}}_z)\,dS + \displaystyle\iint_{S_-} A_z\,(\hat{\mathbf{n}} \cdot \hat{\mathbf{e}}_z)\,dS \label{eq:D-5}\end{equation}

Step 4 (The side contributes nothing). On $S_{\text{side}}$ the generators are parallel to the $z$-axis, so the outward normal lies in the horizontal plane and $\hat{\mathbf{n}} \cdot \hat{\mathbf{e}}_z = 0$. The side therefore contributes nothing to the surface integral of $A_z$.

\begin{equation}\displaystyle\iiint_V \dfrac{\partial A_z}{\partial z}\,dV = \displaystyle\iint_{\partial V} A_z\,(\hat{\mathbf{n}} \cdot \hat{\mathbf{e}}_z)\,dS \label{eq:D-6}\end{equation}

Step 5 (x and y components). By the triply-regular assumption from Step 1, $V$ is also x-regular ($x_1(y,z) \le x \le x_2(y,z)$) and y-regular, so repeating the argument of Steps 2-4 with the coordinate axes permuted yields

\begin{equation}\displaystyle\iiint_V \dfrac{\partial A_x}{\partial x}\,dV = \displaystyle\iint_{\partial V} A_x\,(\hat{\mathbf{n}} \cdot \hat{\mathbf{e}}_x)\,dS \label{eq:D-7}\end{equation}

\begin{equation}\displaystyle\iiint_V \dfrac{\partial A_y}{\partial y}\,dV = \displaystyle\iint_{\partial V} A_y\,(\hat{\mathbf{n}} \cdot \hat{\mathbf{e}}_y)\,dS \label{eq:D-8}\end{equation}

Adding $\eqref{eq:D-6}$, $\eqref{eq:D-7}$, $\eqref{eq:D-8}$,

\begin{equation}\displaystyle\iiint_V \left( \dfrac{\partial A_x}{\partial x} + \dfrac{\partial A_y}{\partial y} + \dfrac{\partial A_z}{\partial z} \right) dV = \displaystyle\iint_{\partial V} (A_x n_x + A_y n_y + A_z n_z)\,dS \label{eq:D-9}\end{equation}

The left-hand side is the volume integral of $\nabla \cdot \mathbf{A}$ and the right-hand side is the surface integral of $\mathbf{A} \cdot \hat{\mathbf{n}}$.

Step 6 (Extension to general regions). By the piecewise smoothness assumption, any bounded closed region $V$ can be partitioned into finitely many triply-regular subregions. On the common boundaries of adjacent subregions the outward normals point in opposite directions, so their surface integrals cancel and only the integral over the outer boundary $\partial V$ survives. Therefore

\begin{equation}\displaystyle\iiint_V (\nabla \cdot \mathbf{A})\,dV = \displaystyle\iint_{\partial V} \mathbf{A} \cdot \hat{\mathbf{n}}\,dS \label{eq:D-10}\end{equation}

holds for every bounded closed region $V$. $\square$

Proof

Step 1 (Reduction to a parametrization). We reduce to the case where $S$ is covered by a single smooth parametrization \begin{equation}\mathbf{r} : D \to \mathbb{R}^3,\quad (u,v) \mapsto \mathbf{r}(u,v),\quad D \subset \mathbb{R}^2 \label{eq:S-1}\end{equation} A general surface is handled by partitioning into finitely many parametric patches whose internal boundaries cancel.

Step 2 (Substitute the surface element). The oriented surface-element vector consistent with the orientation of $S$ is

\begin{equation}\hat{\mathbf{n}}\,dS = \left( \dfrac{\partial \mathbf{r}}{\partial u} \times \dfrac{\partial \mathbf{r}}{\partial v} \right) du\,dv \label{eq:S-2}\end{equation} so the surface integral becomes

\begin{equation}\displaystyle\iint_S (\nabla \times \mathbf{A}) \cdot \hat{\mathbf{n}}\,dS = \displaystyle\iint_D (\nabla \times \mathbf{A}) \cdot \left( \dfrac{\partial \mathbf{r}}{\partial u} \times \dfrac{\partial \mathbf{r}}{\partial v} \right) du\,dv \label{eq:S-3}\end{equation} Using the cyclic property of the scalar triple product, $\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = \mathbf{c} \cdot (\mathbf{a} \times \mathbf{b}) = \mathbf{b} \cdot (\mathbf{c} \times \mathbf{a})$, set $\mathbf{r}_u := \partial \mathbf{r}/\partial u$ and $\mathbf{r}_v := \partial \mathbf{r}/\partial v$.

Step 3 (Chain rule and Schwarz symmetry). Differentiate $\mathbf{A}(\mathbf{r}(u,v))$ via the chain rule.

\begin{equation}\dfrac{\partial}{\partial u}(\mathbf{A} \cdot \mathbf{r}_v) = \left( (\mathbf{r}_u \cdot \nabla)\mathbf{A} \right) \cdot \mathbf{r}_v + \mathbf{A} \cdot \mathbf{r}_{vu} \label{eq:S-4}\end{equation}

\begin{equation}\dfrac{\partial}{\partial v}(\mathbf{A} \cdot \mathbf{r}_u) = \left( (\mathbf{r}_v \cdot \nabla)\mathbf{A} \right) \cdot \mathbf{r}_u + \mathbf{A} \cdot \mathbf{r}_{uv} \label{eq:S-5}\end{equation}

If $\mathbf{r}$ is $C^2$ then Schwarz's theorem gives $\mathbf{r}_{uv} = \mathbf{r}_{vu}$, and subtracting $\eqref{eq:S-5}$ from $\eqref{eq:S-4}$,

\begin{equation}\dfrac{\partial}{\partial u}(\mathbf{A} \cdot \mathbf{r}_v) - \dfrac{\partial}{\partial v}(\mathbf{A} \cdot \mathbf{r}_u) = \left( (\mathbf{r}_u \cdot \nabla)\mathbf{A} \right) \cdot \mathbf{r}_v - \left( (\mathbf{r}_v \cdot \nabla)\mathbf{A} \right) \cdot \mathbf{r}_u \label{eq:S-6}\end{equation}

The right-hand side equals $(\nabla \times \mathbf{A}) \cdot (\mathbf{r}_u \times \mathbf{r}_v)$ by direct component computation with the Levi-Civita symbol. Indeed, the $i$-th component contribution

\begin{equation}\left[ (\mathbf{r}_u \cdot \nabla)\mathbf{A} \cdot \mathbf{r}_v - (\mathbf{r}_v \cdot \nabla)\mathbf{A} \cdot \mathbf{r}_u \right]_i = \varepsilon_{ijk}\,(\partial_j A_k)\,(\mathbf{r}_u \times \mathbf{r}_v)_i \label{eq:S-7}\end{equation} sums to $(\nabla \times \mathbf{A}) \cdot (\mathbf{r}_u \times \mathbf{r}_v)$. Hence

\begin{equation}(\nabla \times \mathbf{A}) \cdot (\mathbf{r}_u \times \mathbf{r}_v) = \dfrac{\partial}{\partial u}(\mathbf{A} \cdot \mathbf{r}_v) - \dfrac{\partial}{\partial v}(\mathbf{A} \cdot \mathbf{r}_u) \label{eq:S-8}\end{equation}

Step 4 (Apply planar Green's theorem). Substituting $\eqref{eq:S-8}$ into $\eqref{eq:S-3}$,

\begin{equation}\displaystyle\iint_S (\nabla \times \mathbf{A}) \cdot \hat{\mathbf{n}}\,dS = \displaystyle\iint_D \left[ \dfrac{\partial}{\partial u}(\mathbf{A} \cdot \mathbf{r}_v) - \dfrac{\partial}{\partial v}(\mathbf{A} \cdot \mathbf{r}_u) \right] du\,dv \label{eq:S-9}\end{equation} The integrand is exactly $\partial Q/\partial u - \partial P/\partial v$ with $P = \mathbf{A} \cdot \mathbf{r}_u$ and $Q = \mathbf{A} \cdot \mathbf{r}_v$, so planar Green's theorem (G) applies directly.

\begin{equation}\displaystyle\iint_D \left( \dfrac{\partial Q}{\partial u} - \dfrac{\partial P}{\partial v} \right) du\,dv = \displaystyle\oint_{\partial D} (P\,du + Q\,dv) \label{eq:S-10}\end{equation}

Step 5 (Boundary line integrals match). When $\partial D$ winds counterclockwise around $D$, its image $\mathbf{r}(\partial D)$ traverses $\partial S$ in the orientation prescribed by the right-hand rule with $\hat{\mathbf{n}} = (\mathbf{r}_u \times \mathbf{r}_v)/|\mathbf{r}_u \times \mathbf{r}_v|$. On the boundary $d\mathbf{r} = \mathbf{r}_u\,du + \mathbf{r}_v\,dv$, so

\begin{equation}\mathbf{A} \cdot d\mathbf{r} = (\mathbf{A} \cdot \mathbf{r}_u)\,du + (\mathbf{A} \cdot \mathbf{r}_v)\,dv = P\,du + Q\,dv \label{eq:S-11}\end{equation}

and therefore

\begin{equation}\displaystyle\oint_{\partial D} (P\,du + Q\,dv) = \displaystyle\oint_{\partial S} \mathbf{A} \cdot d\boldsymbol{\ell} \label{eq:S-12}\end{equation}

Chaining $\eqref{eq:S-9}$, $\eqref{eq:S-10}$, $\eqref{eq:S-12}$ proves (S). $\square$

Proof (a) From Stokes' theorem

Regard $\mathbf{A} := (P, Q, 0)$ as a vector field on the $xy$-plane, take $S := D \times \{0\}$ (the flat surface at $z=0$) and $\hat{\mathbf{n}} := \hat{\mathbf{e}}_z$. Then

\begin{equation}\nabla \times \mathbf{A} = \left( -\dfrac{\partial Q}{\partial z},\ \dfrac{\partial P}{\partial z},\ \dfrac{\partial Q}{\partial x} - \dfrac{\partial P}{\partial y} \right) = \left( 0,\ 0,\ \dfrac{\partial Q}{\partial x} - \dfrac{\partial P}{\partial y} \right) \label{eq:G-1}\end{equation} (the first two components vanish because $P, Q$ are independent of $z$). Hence $(\nabla \times \mathbf{A}) \cdot \hat{\mathbf{n}} = \partial Q/\partial x - \partial P/\partial y$. The boundary $\partial S$ is $C$ traversed counterclockwise, and with $d\boldsymbol{\ell} = (dx, dy, 0)$ we have $\mathbf{A} \cdot d\boldsymbol{\ell} = P\,dx + Q\,dy$. Applying Stokes' theorem (S) immediately yields (G). $\square$

Proof (b) Direct argument

Step 1 (y-regular decomposition). Suppose first that $D$ is y-regular, i.e.

\begin{equation}D = \{(x,y) : a \le x \le b,\ g_1(x) \le y \le g_2(x)\} \label{eq:G-2}\end{equation} and process the double integral of $\partial P/\partial y$ by Fubini.

\begin{equation}\displaystyle\iint_D \dfrac{\partial P}{\partial y}\,dA = \displaystyle\int_a^b \displaystyle\int_{g_1(x)}^{g_2(x)} \dfrac{\partial P}{\partial y}\,dy\,dx = \displaystyle\int_a^b \big[ P(x, g_2(x)) - P(x, g_1(x)) \big]\,dx \label{eq:G-3}\end{equation}

Step 2 (Convert to a boundary line integral). Splitting $C = \partial D$ counterclockwise into bottom $C_1: y=g_1(x)$, $a \to b$; right $C_2: x=b$, $y$ increasing; top $C_3: y=g_2(x)$, $b \to a$; left $C_4: x=a$, $y$ decreasing. On $C_2$ and $C_4$ we have $dx = 0$, so

\begin{equation}\displaystyle\oint_C P\,dx = \displaystyle\int_a^b P(x, g_1(x))\,dx + \displaystyle\int_b^a P(x, g_2(x))\,dx = \displaystyle\int_a^b \big[ P(x, g_1(x)) - P(x, g_2(x)) \big]\,dx \label{eq:G-4}\end{equation}

Comparing $\eqref{eq:G-3}$ and $\eqref{eq:G-4}$,

\begin{equation}-\displaystyle\iint_D \dfrac{\partial P}{\partial y}\,dA = \displaystyle\oint_C P\,dx \label{eq:G-5}\end{equation}

Step 3 (x-regular decomposition for $Q$). Assume $D$ is also x-regular, $D = \{(x,y) : c \le y \le d,\ h_1(y) \le x \le h_2(y)\}$. The analogous computation gives

\begin{equation}\displaystyle\iint_D \dfrac{\partial Q}{\partial x}\,dA = \displaystyle\int_c^d \big[ Q(h_2(y), y) - Q(h_1(y), y) \big]\,dy = \displaystyle\oint_C Q\,dy \label{eq:G-6}\end{equation}

Step 4 (Combine). Adding $\eqref{eq:G-5}$ and $\eqref{eq:G-6}$,

\begin{equation}\displaystyle\oint_C (P\,dx + Q\,dy) = \displaystyle\iint_D \left( \dfrac{\partial Q}{\partial x} - \dfrac{\partial P}{\partial y} \right) dA \label{eq:G-7}\end{equation}

A general $D$ can be expressed as a finite union of subregions that are simultaneously y- and x-regular; the internal-boundary line integrals cancel because adjacent regions traverse them in opposite directions. This proves (G). $\square$

Proof of (G1)

Step 1 (Apply the divergence theorem to a product field). Apply the divergence theorem (D) to $\mathbf{A} := f\,\nabla g$.

\begin{equation}\displaystyle\iiint_V \nabla \cdot (f\,\nabla g)\,dV = \displaystyle\iint_{\partial V} (f\,\nabla g) \cdot \hat{\mathbf{n}}\,dS \label{eq:GI-1}\end{equation}

Step 2 (Expand the divergence). Use the Leibniz rule for the divergence of a scalar-vector product, $\nabla \cdot (f\mathbf{B}) = f\,\nabla \cdot \mathbf{B} + \nabla f \cdot \mathbf{B}$, with $\mathbf{B} = \nabla g$.

\begin{equation}\nabla \cdot (f\,\nabla g) = f\,(\nabla \cdot \nabla g) + \nabla f \cdot \nabla g = f\,\nabla^2 g + \nabla f \cdot \nabla g \label{eq:GI-2}\end{equation} where we used $\nabla \cdot \nabla = \nabla^2$ (the Laplacian).

Step 3 (Assemble the identity). Substituting $\eqref{eq:GI-2}$ into $\eqref{eq:GI-1}$,

\begin{equation}\displaystyle\iiint_V (f\,\nabla^2 g + \nabla f \cdot \nabla g)\,dV = \displaystyle\iint_{\partial V} f\,(\nabla g \cdot \hat{\mathbf{n}})\,dS \label{eq:GI-3}\end{equation} which is the first identity (G1). $\square$

Proof of (G2)

Step 1 (Swap the roles of $f$ and $g$). Exchanging $f$ and $g$ in $\eqref{eq:GI-3}$,

\begin{equation}\displaystyle\iiint_V (g\,\nabla^2 f + \nabla g \cdot \nabla f)\,dV = \displaystyle\iint_{\partial V} g\,(\nabla f \cdot \hat{\mathbf{n}})\,dS \label{eq:GI-4}\end{equation}

Step 2 (Subtract). Subtract $\eqref{eq:GI-4}$ from $\eqref{eq:GI-3}$. Because $\nabla f \cdot \nabla g = \nabla g \cdot \nabla f$, the cross term in the volume integral cancels.

\begin{equation}\displaystyle\iiint_V (f\,\nabla^2 g - g\,\nabla^2 f)\,dV = \displaystyle\iint_{\partial V} \big[ f\,(\nabla g \cdot \hat{\mathbf{n}}) - g\,(\nabla f \cdot \hat{\mathbf{n}}) \big]\,dS \label{eq:GI-5}\end{equation}

The right-hand side combines into $(f\,\nabla g - g\,\nabla f) \cdot \hat{\mathbf{n}}$.

\begin{equation}\displaystyle\iiint_V (f\,\nabla^2 g - g\,\nabla^2 f)\,dV = \displaystyle\iint_{\partial V} (f\,\nabla g - g\,\nabla f) \cdot \hat{\mathbf{n}}\,dS \label{eq:GI-6}\end{equation} which is the second identity (G2). $\square$

Proof

Step 1 (Construct Newton-potential solutions). Define the scalar potential $\phi$ and vector potential $\mathbf{A}$ as convolutions with the Newton kernel (the fundamental solution $-1/(4\pi |\mathbf{r}-\mathbf{r}'|)$).

\begin{equation}\phi(\mathbf{r}) := \dfrac{1}{4\pi}\,\displaystyle\iiint_{\mathbb{R}^3} \dfrac{\nabla' \cdot \mathbf{F}(\mathbf{r}')}{|\mathbf{r} - \mathbf{r}'|}\,dV' \label{eq:H-1}\end{equation}

\begin{equation}\mathbf{A}(\mathbf{r}) := \dfrac{1}{4\pi}\,\displaystyle\iiint_{\mathbb{R}^3} \dfrac{\nabla' \times \mathbf{F}(\mathbf{r}')}{|\mathbf{r} - \mathbf{r}'|}\,dV' \label{eq:H-2}\end{equation} where $\nabla'$ denotes differentiation with respect to $\mathbf{r}'$. The decay assumptions make these integrals absolutely convergent.

Step 2 (Laplacian of the fundamental solution). A basic property of the Newton kernel is

\begin{equation}\nabla^2 \left( \dfrac{1}{|\mathbf{r} - \mathbf{r}'|} \right) = -4\pi\,\delta^3(\mathbf{r} - \mathbf{r}') \label{eq:H-3}\end{equation} in the distributional sense ($\delta^3$ is the 3-dimensional Dirac delta).

Step 3 (Compute $-\nabla \phi + \nabla \times \mathbf{A}$). Applying the Laplacian to $\phi$ and $\mathbf{A}$,

\begin{equation}\nabla^2 \phi(\mathbf{r}) = \dfrac{1}{4\pi}\,\displaystyle\iiint \nabla^2 \left( \dfrac{1}{|\mathbf{r}-\mathbf{r}'|} \right) (\nabla' \cdot \mathbf{F})\,dV' = -\,\nabla \cdot \mathbf{F}(\mathbf{r}) \label{eq:H-4}\end{equation}

\begin{equation}\nabla^2 \mathbf{A}(\mathbf{r}) = -\,\nabla \times \mathbf{F}(\mathbf{r}) \label{eq:H-5}\end{equation}

Step 4 (Vector identity). The identity

\begin{equation}\nabla \times (\nabla \times \mathbf{X}) = \nabla(\nabla \cdot \mathbf{X}) - \nabla^2 \mathbf{X} \label{eq:H-6}\end{equation} applied with $\mathbf{X} = \mathbf{A}$ together with $\nabla \cdot \mathbf{A} = 0$ (which follows from $\eqref{eq:H-2}$ by integration by parts using the decay assumption and $\nabla \cdot (\nabla \times \mathbf{F}) = 0$) gives

\begin{equation}\nabla \times (\nabla \times \mathbf{A}) = -\,\nabla^2 \mathbf{A} = \nabla \times \mathbf{F} \label{eq:H-7}\end{equation}

It remains to verify that $-\nabla \phi + \nabla \times \mathbf{A}$ has the same divergence and curl as $\mathbf{F}$.

\begin{equation}\nabla \cdot (-\nabla \phi + \nabla \times \mathbf{A}) = -\nabla^2 \phi + 0 = \nabla \cdot \mathbf{F} \label{eq:H-8}\end{equation}

\begin{equation}\nabla \times (-\nabla \phi + \nabla \times \mathbf{A}) = 0 + \nabla \times \mathbf{F} = \nabla \times \mathbf{F} \label{eq:H-9}\end{equation}

Step 5 (Difference is harmonic and decays, hence vanishes). Set $\mathbf{H} := \mathbf{F} - (-\nabla \phi + \nabla \times \mathbf{A})$. From $\eqref{eq:H-8}$ and $\eqref{eq:H-9}$, $\nabla \cdot \mathbf{H} = 0$ and $\nabla \times \mathbf{H} = \mathbf{0}$. The latter gives $\mathbf{H} = -\nabla \psi$, and the former forces $\nabla^2 \psi = 0$ (Laplace's equation). The decay of $\mathbf{F}$, $\nabla \phi$, and $\nabla \times \mathbf{A}$ implies that $\mathbf{H}$, and hence $\nabla \psi$, vanishes at infinity. Liouville's theorem (a bounded harmonic function is constant) then forces $\psi$ to be constant, so

\begin{equation}\mathbf{H} = \mathbf{0},\qquad \mathbf{F} = -\nabla \phi + \nabla \times \mathbf{A} \label{eq:H-10}\end{equation}

Step 6 (Uniqueness). If $\mathbf{F} = -\nabla \phi_1 + \nabla \times \mathbf{A}_1 = -\nabla \phi_2 + \nabla \times \mathbf{A}_2$ are two such decompositions, the difference is harmonic and decays at infinity, so by Step 5 it vanishes. Therefore $\nabla \phi_1 = \nabla \phi_2$ (up to a constant) and $\nabla \times \mathbf{A}_1 = \nabla \times \mathbf{A}_2$. $\square$

Proof of (D-grad)

Step 1 (Multiply by an arbitrary constant vector $\mathbf{c}$ and take the divergence). For any constant vector $\mathbf{c} \in \mathbb{R}^3$, compute the divergence of $\mathbf{B} := f\,\mathbf{c}$. Since $\mathbf{c}$ is constant, $\nabla \cdot \mathbf{c} = 0$, so

\begin{equation}\nabla \cdot (f\,\mathbf{c}) = \mathbf{c} \cdot \nabla f + f\,(\nabla \cdot \mathbf{c}) = \mathbf{c} \cdot \nabla f \label{eq:DG-1}\end{equation}

Step 2 (Apply the divergence theorem). Apply (D) to $\mathbf{B} = f\,\mathbf{c}$.

\begin{equation}\displaystyle\iiint_V \mathbf{c} \cdot \nabla f\,dV = \displaystyle\iint_{\partial V} f\,(\mathbf{c} \cdot \hat{\mathbf{n}})\,dS \label{eq:DG-2}\end{equation} The constant $\mathbf{c}$ can be pulled outside the integrals.

\begin{equation}\mathbf{c} \cdot \displaystyle\iiint_V \nabla f\,dV = \mathbf{c} \cdot \displaystyle\iint_{\partial V} f\,\hat{\mathbf{n}}\,dS \label{eq:DG-3}\end{equation}

Step 3 (Arbitrariness of $\mathbf{c}$). Since $\eqref{eq:DG-3}$ holds for every $\mathbf{c} \in \mathbb{R}^3$, the two vectors must be equal.

\begin{equation}\displaystyle\iiint_V \nabla f\,dV = \displaystyle\iint_{\partial V} f\,\hat{\mathbf{n}}\,dS \label{eq:DG-4}\end{equation} $\square$

Proof of (D-curl)

Step 1 (Divergence of $\mathbf{A} \times \mathbf{c}$). For an arbitrary constant vector $\mathbf{c}$, use the divergence-of-cross-product identity $\nabla \cdot (\mathbf{A} \times \mathbf{c}) = (\nabla \times \mathbf{A}) \cdot \mathbf{c} - \mathbf{A} \cdot (\nabla \times \mathbf{c})$. Since $\nabla \times \mathbf{c} = \mathbf{0}$,

\begin{equation}\nabla \cdot (\mathbf{A} \times \mathbf{c}) = (\nabla \times \mathbf{A}) \cdot \mathbf{c} = \mathbf{c} \cdot (\nabla \times \mathbf{A}) \label{eq:DC-1}\end{equation}

Step 2 (Divergence theorem and scalar triple product). Apply (D) to $\mathbf{B} := \mathbf{A} \times \mathbf{c}$ and use the cyclic identity $(\mathbf{A} \times \mathbf{c}) \cdot \hat{\mathbf{n}} = \mathbf{c} \cdot (\hat{\mathbf{n}} \times \mathbf{A})$ on the surface integral.

\begin{equation}\displaystyle\iiint_V \mathbf{c} \cdot (\nabla \times \mathbf{A})\,dV = \displaystyle\iint_{\partial V} \mathbf{c} \cdot (\hat{\mathbf{n}} \times \mathbf{A})\,dS \label{eq:DC-2}\end{equation} Pull the constant $\mathbf{c}$ out.

\begin{equation}\mathbf{c} \cdot \displaystyle\iiint_V (\nabla \times \mathbf{A})\,dV = \mathbf{c} \cdot \displaystyle\iint_{\partial V} (\hat{\mathbf{n}} \times \mathbf{A})\,dS \label{eq:DC-3}\end{equation}

Step 3 (Arbitrariness of $\mathbf{c}$). Since $\eqref{eq:DC-3}$ holds for every $\mathbf{c}$,

\begin{equation}\displaystyle\iiint_V \nabla \times \mathbf{A}\,dV = \displaystyle\iint_{\partial V} \hat{\mathbf{n}} \times \mathbf{A}\,dS \label{eq:DC-4}\end{equation} $\square$