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Calculate the circulation of vector field

  1. Feb 26, 2013 #1
    Hello there,

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

    Say I want to find the circulation around the square formed by [itex]-\frac{\pi}{2} \, \leq x \leq \, \frac{\pi}{2}[/itex] and [itex]-\frac{\pi}{2} \, \leq y \leq \, \frac{\pi}{2}[/itex]. 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. [itex]\int_{y = -\frac{\pi}{2}}^{y = \frac{\pi}{2}}\vec{v}\,\mathrm{d}\vec{r}[/itex] Please correct me if I am wrong. Any criticism is appreciated:

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

    [itex]\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[/itex]

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

    [itex]-\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[/itex]

    First question:
    Is the above correct?

    Second question:
    Can you think of any additional criticism?

    Thank you for your time.

    Kind regards,
    Marius
     
  2. jcsd
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