Line integral of a vector field over a square curve

[marineric]

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)\cdotn 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.

 

[clamtrox]

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.

 

[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:
\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?

 

[marineric]

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?

 

[clamtrox]

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

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 ∇\times\vec{v} times the area of the square, which is a^2?

 

[clamtrox]

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.