Suppose we have a circuit in the xy-plane, with a random geometry (it just needs to be closed). I now wanna calculate the magnetic moment of this setup:

$\vec{m} = \frac{1}{2} \int_V d^3 x' \vec{x}' \times \vec{j}(\vec{x}')$ Now assuming the current is constant we get $I d \vec{x}' = \vec{j}(\vec{x}')$ which turns the integral into:

$\vec{m} = \frac{1}{2} \int_V d^3 x' \vec{x}' \times (I d\vec{x}')$ Now the next step in the computation I'm trying to follow is immediately writing down $\vec{m} = \frac{1}{2} I \int_{\delta A} \vec{x}' \times d\vec{x}'$

I probably flunked an important lesson in Calc, but I can't seem to understand what's happening here. How does the integral over the volume/surface transform into one over its edge? We didn't use Stokes, because then the $\times$ would go away.

Anyone care to help?

Greetings!

Imagine we have a circuit on the xy-plane, with a random geometry (it just need to be closed). I want to calculate the magnetic moment of this setup:

$$\vec{m} = \frac{1}{2} \int_V d^3 x' \vec{x}' \times \vec{j}(\vec{x}')$$

Assuming that current is constant we get $I d \vec{x}' = \vec{j}(\vec{x}')$ which turns the integral into:

$$\vec{m} = \frac{1}{2} \int_V d^3 x' \vec{x}' \times (I d\vec{x}')$$

Next step in the computation, I'm trying to follow, is immediately writing down: $$\vec{m} = \frac{1}{2} I \int_{\delta A} \vec{x}' \times d\vec{x}'$$

I probably flunked an important lesson and can't understand what is happening here. We don't use Stokes, because then the "$\times$" would go away.

My question is: how does the integral over the volume/surface transform into one over its edge?