# Circulation using Green's Theorem

I have a problem where I'm asked to find the circulation along boundary of the half annulus with bounds $\{(r,\theta)|1\le r\le 3, 0\le \theta \le \pi\}$ in the force field $F =\ <-y, x>$. I found the answer to be $4\pi$ but my teacher said the answer is $16\pi$ and I can't figure out how she got that. Here's my steps...

Use Green's Theorem: $\int\int\limits_R\ \frac{\partial\ g}{\partial\ x}-\frac{\partial\ f}{\partial\ y}dA$

For the partial which would be the inside of the integral, I got 2, and using the bounds given I set up my integral as $\int\limits_0^\pi\int\limits_1^32\ dr\ d\theta$.

$=\int\limits_0^\pi\ 2r\bigg|_1^3d\theta$

$=\int\limits_0^\pi4\ d\theta$

$=4\bigg|_0^\pi\ =4\pi$

I double checked so I must not be using the formula correctly. Could anyone point out what I'm missing?

The infinitesimal area element in polar coordinates is given by $$\mathrm{d}A=r\mathrm{d}r\mathrm{d}\theta$$as you can prove calculating Jacobian determinant of this substitution.
• so I left out an $r$. I see, thanks! – Sabien Dec 9 '14 at 7:42
• But wait, when I calculate this with the extra $r$, I get only $8\pi$, not $16\pi$... Am I still missing something? – Sabien Dec 9 '14 at 7:44
• I get $8\pi$ too and I think this is the right answer because the integral reduces to twice the area of the half anulus (namely the area of the entire anulus) and even with simple geometric arguments ($A= \pi(R_{2}^2-R_{1}^2), \, R_{2}>R_{1}$) we still obtain $8 \pi$. – Ivan Dec 9 '14 at 8:23