# Line integral of a vector field over a square curve

## Homework Statement

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

## Homework Equations

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

## 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?

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?

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?

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.