line integral of a vector field over a square curve

marineric

1. The problem statement, all variables and given/known data

Please evaluate the line integral [itex]\oint[/itex] dr[itex]\cdot[/itex][itex]\vec{v}[/itex], where [itex]\vec{v}[/itex] = (y, 0, 0) along the curve C that is a square in the xy-plane of side length a center at [itex]\vec{r}[/itex] = 0

a) by direct integration

b) by Stokes' theorem


2. Relevant equations

Stokes' theorem: [itex]\oint[/itex] V [itex]\cdot[/itex] dr = ∫∫ (∇ x V)[itex]\cdot[/itex]n d[itex]\sigma[/itex]


3. The attempt at a solution

I know I have to split up the sides of the square. I get confused when [itex]\vec{r}[/itex] is involved. I know the limits are at a/2.. not sure where to go after that.

clamtrox

For the direct integration, just write it out. What is [itex] d\vec{r} \cdot \vec{v} [/itex]? Since you're integrating over a square, you can just write the integral as a sum of four one-dimensional integrals wrt. x and y.

marineric

Ok well I'll try the top line of the square and you can tell me what I'm doing wrong.

From left to right:
[itex]\oint[/itex] d[itex]\vec{r}[/itex][itex]\cdot[/itex][itex]\vec{v}[/itex] from -a/2 to a/2

[itex]\vec{v}[/itex] = (y,0,0)

[itex]\vec{r}[/itex] = (x, a/2, 0)

d[itex]\vec{r}[/itex] = (1, 0, 0)

d[itex]\vec{r}[/itex][itex]\cdot[/itex][itex]\vec{v}[/itex] = y

but aren't the y limits a/2 to a/2, making the integral zero?

marineric

wait hold on [itex]\vec{r}[/itex] = (d[itex]\vec{x}[/itex], 0, 0) ??

then it would be the integral from -a/2 to a/2 of y*dx where y = a/2?

but that's (a/2*x) from -a/2 to a/2 ... which equals zero?

clamtrox

Are you sure it equals zero? :)

marineric

oh, right, minus sign.. so it equals a^2/2 for the top line, and i just do it similarly for all the other sides? do i have to do it in the same order (clockwise)?

clamtrox

You got it

marineric

for the stokes' theorem part, would it just be the ∇[itex]\times[/itex][itex]\vec{v}[/itex] times the area of the square, which is a^2?

clamtrox

Yep, although it's not immediately clear what's the correct sign.