Calculate the circulation of vector field

[Jonsson]

Hello there,

I've got a vector field which you can see here: Sketch of the vector field . It is: \vec{v} = \cos(x)\,\sin(y)\vec{i}-\sin(x)\,\cos(y)\vec{j}

Say I want to find the circulation around the square formed by -\frac{\pi}{2} \, \leq x \leq \, \frac{\pi}{2} and -\frac{\pi}{2} \, \leq y \leq \, \frac{\pi}{2}. I think that I should do this by finding the circulation along one of the sides, and multiply by 4 (I can tell from the vector field that the circulation along one of the sides is going to be equal to every other side.

This is where I become unconfident. \int_{y = -\frac{\pi}{2}}^{y = \frac{\pi}{2}}\vec{v}\,\mathrm{d}\vec{r} Please correct me if I am wrong. Any criticism is appreciated:

My \mathrm{d}\vec{r} = \mathrm{d}y\vec{j} along the y-axis, for the first side: (\Delta y = -\frac{\pi}{2}, x = -\frac{\pi}{2}).

\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos(x)\,\sin(y) \, \mathrm{d}y -\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin(x)\,\cos(y) \, \mathrm{d}y

\cos(-\frac{\pi}{2}) = 0, so the first integral is equal to zero.

-\sin(-\frac{\pi}{2})\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos(y) \, \mathrm{d}y = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos(y) \, \mathrm{d}y = [\sin(y)]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \sin(\frac{\pi}{2}) - \sin(-\frac{\pi}{2}) = 2

First question:
Is the above correct?

Second question:
Can you think of any additional criticism?

Thank you for your time.

Kind regards,
Marius

 

[kuruman]

You are correct if you multiply your result by 4 to get the value of the line integral $I=8$.

You can check the result using Stokes' theorem $$\oint \vec{v}\cdot d\vec l=\int_S \vec {\nabla}\times \vec v \cdot \hat n~dS$$Here, $\vec {\nabla}\times \vec v=-2\cos(x)\cos(y)~\hat z$. We have already chosen clockwise line integration therefore $\hat n=-\hat z$.Thus, $$\int_S \vec {\nabla}\times \vec v \cdot \hat n~dS=+2\int_S\cos(x)\cos(y)dx~dy=2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos(x)dx\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos(y)dy=8.$$