2
$\begingroup$

Given a vector field F = <P(x,y), Q(x,y)>, the line integral over some path C, is $\int_C$ F ⋅ dr.

When expanded, this can be written as $\int_C$ P(x,y) dx + $\int_C$ Q(x,y) dy, which basically means that the line integral of this vector field is the sum of the line integral of its scalar field components, P and Q.

What I don't understand, however, is why we integrate P with respect to x and Q with respect to y. Since both P and Q are functions of (x,y), shouldn't we take the integral with respect to the arc length of the curve (ds) rather than dx or dy? I understand that this is how the dot product expands, but I don't get the intuition behind it. Could someone explain why it is written this way? Thanks in advance!

$\endgroup$

1 Answer 1

2
$\begingroup$

I would say is the natural way to integrate in two dimensions

$G(x)=\int dG(x)=\int \frac{dG(x)}{dx}dx=\int g(x)dx\quad$

$G(x,y)=\int dG(x,y) =\int (\frac{\partial G(x,y)}{\partial x}\,dx + \frac{\partial G(x,y)}{\partial y}\,dy)=\int (P(x,y)\,dx + Q(x,y)\,dy) =\int F\cdot dr\quad$

In physics is more usual to write it in terms of velocity and it is usually referred as work $\int F\cdot dr=\int F\cdot \frac{dr}{dt}dt=\int F\cdot vdt$

And the arc length can also be made to appear using the tangent unit vector as usually encountered in Frenet-Serret

$\int F\cdot dr=\int F\cdot \frac{dr}{dt}dt=\int F\cdot \frac{\frac{dr}{dt}}{||\frac{dr}{dt}||}||\frac{dr}{dt}||dt=\int F\cdot Tds$

Hope it helps

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .