# Converting a volume-integral to a surface integral using Gauss' theorem

I have the following integral: $$\int_V ((\vec\nabla.\vec A)(\vec\nabla.\vec B)+\vec B.\vec\nabla(\vec\nabla.\vec A))dV$$

Using Gauss' theorem I can convert this into a surface integral. However, I need to rewrite the integrandum into something of the form $\vec\nabla.\vec V$ in order to apply Gauss.

In short:

$$(\vec\nabla.\vec A)(\vec\nabla.\vec B)+\vec B.\vec\nabla(\vec\nabla.\vec A)=\vec\nabla.\vec V$$

Where I need to find $\vec V$ so that the identity is correct.

I've tried using index notation but since the result is a scalar that doesn't make much sense so I was a bit hesitant to try it. I decided to give it a shot and got the following:
$$(\partial_iA_i)(\partial_jB_j)+B_k\partial_l\partial_lA_k$$
Which leads me nowhere.

I've also tried putting $(\vec\nabla.\vec A)$ in front (applying distributivity) but that's not correct because, while $(\vec\nabla.\vec A)$ does appear in the second term, there is still a $\vec\nabla$ that works in on it.

I honestly don't know what else I can try.

Write everything in index notation. Your integrand is $$\frac{\partial A_i}{\partial x_i}\frac{\partial B_j}{\partial x_j} + B_j \hat{e}_j \cdot \left ( \hat{e}_m \frac{\partial}{\partial x_m} \left ( \frac{\partial A_p}{\partial x_p} \right)\right)$$ which reduces, on simplification, to $$\frac{\partial A_i}{\partial x_i}\frac{\partial B_j}{\partial x_j} + B_j \frac{\partial^2 A_p}{\partial x_j \partial x_p}$$

Now consider $$\frac{\partial}{\partial x_p} \left( B_p \frac{\partial A_j}{\partial x_j} \right) \\ = \frac{\partial B_p}{\partial x_p}\frac{\partial A_j}{\partial x_j} + B_p \frac{\partial^2 A_j}{\partial x_p \partial x_j} \\ = \frac{\partial A_i}{\partial x_i} \frac{\partial B_j}{\partial x_j} + B_j \frac{\partial^2 A_p}{\partial x_j \partial x_p}$$ where we have interchanged the dummy indices j and p in the last step, and renamed the dummy indices in the first term. This is precisely your integrand above.

So your integrand can be written as $$\int_V \frac{\partial}{\partial x_p} \left( B_p \frac{\partial A_j}{\partial x_j} \right) dV$$ which is now in a suitable form for application of the divergence theorem.

ETA: If you want everything back in vector form, $$\frac{\partial}{\partial x_p} \left( B_p \frac{\partial A_j}{\partial x_j} \right) =\nabla \cdot (B (\nabla \cdot A))$$

The key is that you have both $B \cdot \nabla$ and $\nabla \cdot B$. This is a tell-tale sign of the product rule being applied. You can use "overdot notation" to write the expression as a total divergence. First, recognize that

$$(B \cdot \nabla) (\nabla \cdot A) + (\nabla \cdot B)(\nabla \cdot A)= (\dot \nabla \cdot B)(\nabla \cdot \dot A) + (\dot \nabla \cdot \dot B)(\nabla \cdot A)$$

The dots denote what is to be differentiated; in $\dot \nabla \cdot B$, $B$ is "held constant", and as such, this is equivalent to $B \cdot\nabla$.

The overdot notation makes it easy to recognize that the product rule has been used to expand, and easy to see how the product rule should be undone:

$$(\dot \nabla \cdot B)(\nabla \cdot \dot A) + (\dot \nabla \cdot \dot B)(\nabla \cdot A) = \dot \nabla \cdot [\dot B (\nabla \cdot A) + B(\nabla \cdot \dot A)] = \nabla \cdot [B (\nabla \cdot A)]$$

• I'm not sure what you mean when you say that it's "held constant", but I'd still like to understand your solution since it seems like a simple way to keep track of things Dec 18, 2014 at 14:36
• I mean that it's not differentiated. For instance, $(\dot \nabla \cdot B)(\nabla \cdot \dot A) = \dot \partial_i B^i (\partial_j \dot A^j)$ in index notation. Because $B^i$ is not being differentiated, you can rearrange this to $B^i \partial_i (\partial_j A^j)$. The overdot notation just gives you the same flexibility you'd have in index notation: the flexibility to differentiate only what should be differentiated. But overdot notation lets you do so with familiar dot and cross products instead of having to resort to index notation. Dec 18, 2014 at 16:45