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$$\overrightarrow{F}=-y\hat{i}+x\hat{j}$$ $$C:r=a \cos{t}\hat{i}+a \sin{t}\hat{j}, o \leq t \leq 2 \pi$$ $$R:x^2+y^2 \leq a^2 $$

Green's Theorem: $$\oint_C{F}dr=\int \int_R{\bigtriangledown \times \overrightarrow{F} \cdot \hat{n}}dA$$

$$\bigtriangledown \times \overrightarrow{F}=2 \hat{k}$$ But how can I find $ \int \int_R{ \bigtriangledown \times \overrightarrow{F} \cdot \hat{n}}dA $ ??

Mary Star
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    Can you find $\hat n$? It is the outward normal to the surface you are integrating over. – user88595 Mar 05 '14 at 15:21
  • @user88595 Could you give me a hint how to find this? – Mary Star Mar 05 '14 at 15:22
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    What is $R$? You have defined it as $x^2 + y^2 \le a^2$. On which plane does it lie? From there you should be able to see what the normal vector is. – user88595 Mar 05 '14 at 15:26
  • @user88595 $R$ lies on the plane $xy$, right? How can find the normal vector? The normal vector should be perpendicular to the plane, shouldn't it? So should it be on the z-axis? – Mary Star Mar 05 '14 at 19:24
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    That's correct. – user88595 Mar 06 '14 at 09:14
  • @user88595 Ok...So we have now $ \int \int_R{ \bigtriangledown \times \overrightarrow{F} \cdot \hat{n}}dA= \int \int_R{ \bigtriangledown \times \overrightarrow{F} \cdot \hat{k}}dA =\int \int_R{2}dA$..To find this do I have to say that this double integral is the area of the circle which equals to $ \pi r^2$ (r: radius)? Or is there an other way? – Mary Star Mar 06 '14 at 22:29
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    Well you can just say it is the area of the circle to avoid calculations. Having said that you can also do it by calculating it. Here's a post showing how to calculate it http://math.stackexchange.com/questions/187987/calculus-proof-for-the-area-of-a-circle – user88595 Mar 07 '14 at 10:02
  • @user88595 Ok! In the link the integrals are of the form $\int \int_R dxdy$..Is this the same as the integral $\int \int_R{ \nabla \times \overrightarrow{F} \cdot \hat{n}}dA$ ?? – Mary Star Mar 07 '14 at 10:09
  • $$\int\int dxdy = \int\int dA$$ – user88595 Mar 07 '14 at 11:37
  • Aha! Ok!! Thank you for your help!! – Mary Star Mar 08 '14 at 00:51

1 Answers1

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First of all Greens Theorm implies : $$\oint_C{F}dr=\int \int_R{\nabla \times \overrightarrow{F}\cdot \hat{k} }dA$$ with $\hat{k}$ is pointing the $z$ direction and the R.H.S is just the normal double integral of $\nabla \times \overrightarrow{F}$ over region $R$ which is the area inside the circle centered at the origin and of radius $a$.

Since $\nabla \times \overrightarrow{F}=2 \hat{k}$, we get
$$\oint_C{F}dr=2\int \int_R{ }dA= 2\pi a^2 $$

user88595
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ketan
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  • So you mean that $\hat{n}=\hat{k}$?? $\hat{n}$ should be perpendicular to the plane which is the $xy$,that's why $\hat{n}$ is on $z$.Is this correct? – Mary Star Mar 05 '14 at 21:43
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    yeap, greens theorm is stokes theorm for plane, so n=k – ketan Mar 06 '14 at 03:36
  • A ok! How did you calculate the last double integral? Is this the area of the circle which equals to $ \pi r^2$ (r: radius)? Or is there an other way? – Mary Star Mar 06 '14 at 22:31