2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ By using Green's Theorem, you can turn this into a trivial integral and simultaneously avoid parameterizing the curve. $\endgroup$ – Travis Jan 16 '15 at 0:33
4
$\begingroup$

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)

enter image description here

Jokes apart, $\iint_R \,{\rm d}A = \frac{3}{4}\pi r^2$, and no calculus is needed for this part.

$\endgroup$
  • $\begingroup$ Is a requirement of Green's theorem that the velocity field vector must not be conservative? In this case, how do we show that the vector is not conservative? Or does that not matter? $\endgroup$ – inggumnator Jan 16 '15 at 4:42
  • $\begingroup$ This is not important :) $\endgroup$ – Ivo Terek Jan 16 '15 at 7:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.