Suppose A is a vector satisfying curlA = aA, a a function of position.

Is it possible to turn the volume integral of A^2 into a surface integral? I know I have to write A^2 as the divergence of something, but what?

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- Oct 7th 2010, 01:44 PMthefrogVolume integral to a surface integral ... divergence theorem
Suppose A is a vector satisfying curlA = aA, a a function of position.

Is it possible to turn the volume integral of A^2 into a surface integral? I know I have to write A^2 as the divergence of something, but what? - Oct 7th 2010, 05:31 PMAckbeet
A couple of questions for clarification.

1. What quantities here are vector quantities, and which ones are scalar?

2. Which multiplication do you mean when you write $\displaystyle \mathbf{A}^{\!2}?$ Is that $\displaystyle \mathbf{A}\cdot\mathbf{A}?$ Or is it $\displaystyle \mathbf{A}\times\mathbf{A}?$

3. I'm also curious as to what physical quantity $\displaystyle \mathbf{A}$ represents. Is that the vector magnetic potential? - Oct 7th 2010, 05:35 PMthefrog
Hi Ackbeet,

1/. a is scalar, A is a vector

2/. A^2 = A dot A

3/. Yes, it is a vector magnetic potential - Oct 7th 2010, 06:11 PMAckbeet
Ok, so re-writing your problem statement, I have the following:

Assume $\displaystyle \nabla\times\mathbf{A}=a\mathbf{A},$ where $\displaystyle \mathbf{A}$ is the vector magnetic potential defined by $\displaystyle \mathbf{B}=\nabla\times\mathbf{A},$ where $\displaystyle \mathbf{B}$ is the magnetic field, and the scalar function $\displaystyle a$ depends only on the coordinates.

We wish to show that we can convert

$\displaystyle \displaystyle \iiint_{V}(\mathbf{A}\cdot\mathbf{A})\,dV$ into a surface integral via the divergence theorem.

The divergence theorem states that if $\displaystyle \mathbf{F}$ is a continuously differentiable vector field defined on a neighborhood of $\displaystyle V,$ then

$\displaystyle \displaystyle\iiint_{V}(\nabla\cdot \mathbf{F})\,dV=\iint_{S}\mathbf{F}\cdot\mathbf{n} \,dS.$

Here's a question: do you know if $\displaystyle a$ is ever zero? Because, if not, you might be able to manipulate the equation

$\displaystyle \displaystyle \frac{\nabla\times\mathbf{A}}{a}=\mathbf{A}$ to simplify

$\displaystyle \displaystyle\mathbf{A}\cdot\mathbf{A}=\left(\frac {\nabla\times\mathbf{A}}{a}\right)\cdot\left(\frac {\nabla\times\mathbf{A}}{a}\right)=\frac{1}{a^{2}} \left(\nabla\times\mathbf{A}\right)\cdot\left(\nab la\times\mathbf{A}\right).$

I have another question: are you told that you can perform this transformation? Or are you merely trying to simplify a particular volume integral? If the second, can you provide a little more context?