$\int_C (\alpha x, -\alpha y) . dr = 0$ where C is the unit circle

Question

Circulation is given by

$$\int_C u . dr$$

I want to show that the circulation around the unit circle is $0$ for $u = (\alpha x, \alpha y)$. Ie.

$$\int_C (\alpha x, -\alpha y) . dr = 0$$

How would I go about this? (My exam is in 4 hours time and I just noticed that this comes up now and again as an optional question. We never covered it in class and though it looks very straightforward I don't have time to spend on it now so hints are no good in this case, I would really appreciate an actual answer.)

Answer 1

Parametrize the curve $c(t)=(\cos t,\sin t), t\in [0,2\pi].$ Thus, given $u=(\alpha x,-\alpha y),$ we have

$$\int_C (\alpha x, -\alpha y) \cdot dr =\int_0^{2\pi} (\alpha(\cos t,-\sin t)\cdot c'(t)) dt \\= \alpha\int_0^{2\pi} ((\cos t,-\sin t)\cdot (-\sin t,\cos t))dt \\=\alpha\int_0^{2\pi} (-\cos t\sin t-\sin t\cos t)dt \\=-\alpha \int_0^{2\pi}\sin(2t)dt=0. $$