1
$\begingroup$

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 $D\vec A$?

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

$\vec{dA}=\vec{n}dA$, 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 $\phi$.

$\endgroup$

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.