# Calculate the circulation of vector field

1. Feb 26, 2013

### 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?