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Suppose $F =(2x−4y)i +(x+3y)j$. Use Stokes' Theorem to make the following circulation calculations:

(a) Find the circulation of $F$ around the circle $C$ of radius $10$ centered at the origin in the $xy$-plane, oriented clockwise as viewed from the positive $z$-axis.

Circulation = $\int_CF\cdot dr$

Here is my work: Please tell me the correct answer and what I am doing wrong:

$$ \begin{align*} \int_C F\cdot dr&= \int\int_S \text{curl} F · dS & \text{by Stokes' Theorem} \\ &= \int\int\langle 0, 0, 5\rangle \cdot \langle 0, 0, 1\rangle~dA & \text{since the circle lies on }z = 0 \\ &= 5 * (\text{Area of }C) \\ &= 5 * 100π \\ &= 500π. \end{align*} $$

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  • $\begingroup$ Why the downvote? $\endgroup$ – Lays Apr 23 '13 at 1:10
  • $\begingroup$ @MichaelRametta To learn how to use LaTex, just right click on any LaTex, go to show math as and click on tex commands to give you an idea of how to use Latex. $\endgroup$ – Lays Apr 23 '13 at 1:11
  • $\begingroup$ So does anyone know what i am doing wrong? $\endgroup$ – Michael Rametta Apr 23 '13 at 1:15
  • $\begingroup$ For the rim, $x$ and $y$ are OK, but $z=2$. $\endgroup$ – André Nicolas Apr 23 '13 at 1:57
  • $\begingroup$ would part (b) be 8pi $\endgroup$ – Michael Rametta Apr 23 '13 at 2:29
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It's correct except for the small issue of orientation. That changes the sign of your answer. Officially, you need $\vec n = \langle 0,0,-1\rangle$.

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  • $\begingroup$ I see. Thanks a bunch A++ $\endgroup$ – Michael Rametta Apr 23 '13 at 3:41
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In your first equation, you take the circulation to be the line integral of the dot product of the field with the differential of the position vector- what you are calculating is the flux of the field across the circle. In order to get the circulation, you need to instead take the dot product of the field with the differential length element of the circle, which is orthogonal to the position vector, along the tangents of the circle.

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