Well, I hope this is a stokes problem. Im honestly a bit lost on this so please help me out!

Suppose I have a simple closed curve, C, in the plane w/ counterclockwise direction. I need to calculate $\int_C F\bullet dr$ in terms of the area inside the curve. With $F=\frac x2j$.

$$\int_C F\bullet dr=\int_Scurl\vec F\bullet D\vec A$$

I calculated $curl\vec F$ to be: $$\nabla \times\vec F= \frac 12\vec k$$

How do I find $D\vec A$?

  • $\begingroup$ Can you give more details on "simple closed curve, $C$, in the plane w/ counterclockwise direction"? $dA$ is calculated from the information about which plane it's on, and is it any arbitrary $C$ on the plane? $\endgroup$ – sillyme Dec 6 '13 at 1:37
  • $\begingroup$ I believe that yes, C is any arbitrary curve on the plane since no other details were given about the curve. $\endgroup$ – Student Dec 6 '13 at 1:39

$\vec{dA}=\vec{n}dA$, where $\vec{n}$ is the outward (unit) normal to the surface $S$ and $dA$ is the surface measure on $S$, given by $|\phi_{u}\times\phi_{v}|dudv$ after pulling back the integral to the parameter domain. $\phi(u,v)$ is of course the parameterization of $S$. Since $\vec{n}=\frac{\phi_{u}\times\phi_{v}}{|\phi_{u}\times\phi_{v}|}$, we see that the terms in absolute values cancel and so the result is (assuming your computation of the curl is correct)

$$\int_{S}\text{curl}\;F\cdot\vec{n}\;dA=\int\int_{U}\frac{1}{2}(\phi_{u}\times\phi_{v})\cdot\vec{k}\;dudv$$ where $U$ is the parameter domain of $\phi$.


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