## line integral of a vector field over a square curve

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

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

a) by direct integration

b) by Stokes' theorem

2. Relevant equations

Stokes' theorem: $\oint$ V $\cdot$ dr = ∫∫ (∇ x V)$\cdot$n d$\sigma$

3. The attempt at a solution

I know I have to split up the sides of the square. I get confused when $\vec{r}$ is involved. I know the limits are at a/2.. not sure where to go after that.
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 For the direct integration, just write it out. What is $d\vec{r} \cdot \vec{v}$? 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.
 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: $\oint$ d$\vec{r}$$\cdot$$\vec{v}$ from -a/2 to a/2 $\vec{v}$ = (y,0,0) $\vec{r}$ = (x, a/2, 0) d$\vec{r}$ = (1, 0, 0) d$\vec{r}$$\cdot$$\vec{v}$ = y but aren't the y limits a/2 to a/2, making the integral zero?

## line integral of a vector field over a square curve

wait hold on $\vec{r}$ = (d$\vec{x}$, 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?

 Quote by marineric but that's (a/2*x) from -a/2 to a/2 ... which equals zero?
Are you sure it equals zero? :)
 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)?
 You got it
 for the stokes' theorem part, would it just be the ∇$\times$$\vec{v}$ times the area of the square, which is a^2?

 Quote by marineric for the stokes' theorem part, would it just be the ∇$\times$$\vec{v}$ times the area of the square, which is a^2?
Yep, although it's not immediately clear what's the correct sign.