# Volume integral turned in to surface + line integral?

1. May 5, 2014

### AntiElephant

Hi, I have a book that makes the equality.

$\vec{B}dV = (\vec{e_1}B_1 + \vec{e_2}B_2 + \vec{e_1}B_2)dx_1 dx_2 dx_3 \\[1ex] = dx_1 \vec{e}_1(B_1 dx_2 dx_3 ) + dx_2 \vec{e}_2(B_2 dx_1 dx_3 ) + dx_3 \vec{e}_3 (B_3 dx_1 dx_2) = (\vec{B}\cdot d\vec{S}) d\vec{l}.$

I'm a bit confused as to how it makes that last equality. In a very general sense, the surface element is given by;

$d\vec{S} = (dx_2dx_3,dx_1dx_3,dx_1dx_2)$

right? What I need is a way of represententing $d\vec{l} = (dx_1,dx_2,dx_3)$ as being multiplied component-wise by the 3 summation terms of $\vec{B} \cdot d\vec{S}$, but as far as I can tell the notation $(\vec{B}\cdot d\vec{S})d\vec{l}$ doesn't seem to do that?

If this is not possible, it might be cause it's specific to my situation. I'm looking at the integral of $B$ over the volume a plasma flux rope - which is defined as the volume encompassed by a fixed selection of magnetic field lines.

2. May 5, 2014

### xiavatar

I think its just a shorthand that they are using. I can't seem to use the formal rules of vectors to transform the right side into the left. What book is this from?

3. May 5, 2014

### AntiElephant

Lectures in magnetohydrodynamics by D.Schnack. Think it's possible to access it online. Chapter 12 when it talks about magnetic helicity (page 73).

If it's not possible in general to write it like that, it might because it's relevant to the situation.

Maybe the volume element is chosen to be a line element $dl$ in the direction of the field, mutiplying a surface element $dS$ which is perpendicular to the line and field direction (covering the poloial cross section). In this coordinate system $B_2 = B_3 = 0$ always. And

$\vec{B} dV = dx_1 \vec{e_1} (B_1 dx_2 dx_3) = (\vec{B} \cdot d\vec{S})\cdot d\vec{l}$

I'm not entirely sure how correct this is.

Last edited: May 5, 2014