Integrating curve over a line segment

I'm having some trouble understanding this question: what is $\int_C xdy - ydx$, where C is the curve composed of a straight line segment ￼￼￼￼￼￼C from $(−2, 0)$ to $(0, 0)$, a straight line segment from $(0, 0)$ to $(0, −2)$, and the part of the circle of radius 2, centered at the origin, traversed counterclockwise starting from $(0,−2)$ and ending at $(−2, 0)$. I tried breaking it down into 2 curves, and integrating separately but I'm not able to get the answer of $6\pi$. Anyone knows what's the correct way to solve this problem?

• By using Green's Theorem, you can turn this into a trivial integral and simultaneously avoid parameterizing the curve. – Travis Jan 16 '15 at 0:33

Recall Green-Stokes's theorem: if a closed curve $C$ encloses a region $R$, going counterclockwise, then $$\int_C P \,{\rm d}x + Q\,{\rm d}y = \iint_R \left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\,{\rm d}A,$$ for $\cal C^1$ functions $P$ and $Q$.
So, using Green-Stokes, we have: $$\int_C -y \,{\rm d}x + x \,{\rm d}y = \iint_R \frac{\partial x}{\partial x} - \left(\frac{\partial(-y)}{\partial y}\right) \,{\rm d}A = 2 \iint_R \,{\rm d}A = 2 \,\frac{3}{4}\,\pi \,2^2 = 6\pi,$$ where $R$ is the interior of this pacman (its eye is just me having fun, it is simply connected) Jokes apart, $\iint_R \,{\rm d}A = \frac{3}{4}\pi r^2$, and no calculus is needed for this part.