# Prove $\oint_{\partial V} (\mathbf{\hat{n}} \times \mathbf{A}) \; \mathrm{d}S = \int_V (\nabla \times \mathbf{A}) \; \mathrm{d}V$

I need help proving the following vector calculus identity:

$$\oint_{\partial V} (\mathbf{\hat{n}} \times \mathbf{A}) \; \mathrm{d}S = \int_V (\nabla \times \mathbf{A}) \; \mathrm{d}V$$

the identity is also found on this link, under "Surface–volume integrals".

Thank you very much

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It looks like a corollary to the Divergence Theorem, like applying it to the cross product of a vector field and a nonconstant vector. Do you know that theorem? –  mixedmath Jul 8 '12 at 4:42

This alternate version of Divergence theorem can be proved using Gauss-Green formula: $$\int_{\Omega} \partial_{x_i} u\, dx = \int_{\partial \Omega} u n_i dS \,,$$ where $\boldsymbol{n} = (n_1,\ldots,n_k)$ is the unit outward normal vector to a smooth domain $\Omega\subset \mathbb{R}^k$.

Then let's consider this special case in $\mathbb{R}^3$. Let $\mathbf{A} = (A_1,A_2,A_3)$. The first component of $\nabla\times\mathbf{A}$ is $\partial_{x_2} A_3 - \partial_{x_3} A_2$, then: $$\int_{V} (\partial_{x_2} A_3 - \partial_{x_3} A_2)dV = \int_{\partial V} (n_2 A_3 - n_3 A_2)dS,$$ in which the right hand side is exactly the first component for $\mathbf{n}\times \mathbf{A}$. The second and the third components are proved in the same way.

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Since $\mathbf{\hat{n}}$ and $\mathbf{dS}$ are parallel, $\mathbf{dS} \times \mathbf{\hat{n}} = 0$, so \begin{align*} \mathbf{\hat{n}} \times (\mathbf{A} \times \mathbf{dS}) & = - \mathbf{A} \times (\mathbf{dS} \times \mathbf{\hat{n}}) - \mathbf{dS} \times (\mathbf{\hat{n}} \times \mathbf{A}) \\ & = - \mathbf{dS} \times (\mathbf{\hat{n}} \times \mathbf{A}) \\ & = (\mathbf{\hat{n}} \times \mathbf{A}) \times \mathbf{dS} \end{align*} by the the Jacobi Identity and divergence theorem. $$\mathbf{A} ~dS = (\mathbf{\hat{n}} \times \mathbf{A}) \times \mathbf{dS} + \mathbf{\hat{n}} \cdot \mathbf{A} ~\mathbf{dS}.$$ Then use the corollaries to show that $$\iint_{\partial V} \mathbf{A} ~dS = \iiint_V \left(\nabla(\mathbf{\hat{n}} \cdot \mathbf{A})- \nabla \times (\mathbf{\hat{n}} \times \mathbf{A})\right) ~dV.$$

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$\mathrm{d}S$ is not a vector –  alqubaisi Jul 8 '12 at 6:33
ecb is not saying $\text{d}S$ is a vector. However, $\mathbf{\text{d}S} = \mathbf{\widehat{n}}\,\text{d}S$ is a vector. –  Jesse Madnick Nov 15 '12 at 2:22

here this guy explains it nicely http://www.youtube.com/watch?v=1qLb0B40YnA

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While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. –  Julian Kuelshammer Feb 1 '13 at 11:48