Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
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

3 Answers 3

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.

share|improve this answer

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 $$\iint_{\partial V} \mathbf{A} ~dS = \iiint_V \left(\nabla(\mathbf{\hat{n}} \cdot \mathbf{A})- \nabla \times (\mathbf{\hat{n}} \times \mathbf{A})\right) ~dV.$$

share|improve this answer
3  
$\mathrm{d}S$ 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

share|improve this answer
4  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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