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I have found the following integral to be zero, but i don't think its correct. $$ C = \oint_K d\mathbf r\cdot \mathbf A $$ Where $\mathbf A = \frac 1 2 \mathbf n \times \mathbf r$ and $\mathbf n \cdot \mathbf n=1$.

Taking $K$ as a circle with radius $R$ and $\mathbf n$ is the normal to the plane where the circle lives.

Any kind of help is appreciated . The problem is that when i found $A$ it came out to be zero. i found normal as $\nabla f$ where $f=x^2+y^2-1$ .

Progress : I applied stokes theorem, and also $\nabla \times \mathbf A =2\mathbf n$ , then i get $C=\pi R^2$ . am i right ?? how do i do it using line integral ?

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2 Answers 2

In the following I assume the path $K$ is counter-clockwise with respect to the orientation $\mathbf n$ of the plane.
Let $$ K\equiv \gamma(t) := R(\cos(t), \sin(t), 0)\quad t\in[0, 2\pi] $$ we have $$ \mathbf A(\gamma(t)) = \frac 1 2 \mathbf n\times \gamma(t) = \frac R 2 (0, 0, 1)\times (\cos(t), \sin(t), 0) = \frac R 2 (-\sin(t), \cos(t), 0) $$ and so $$ \begin{align} \oint_K d\mathbf r \cdot \mathbf A &= \int_0^{2\pi} \mathbf A(\gamma(t))\cdot \dot\gamma(t)dt \\ &= \frac {R^2} 2 \int_0^{2\pi} (\sin(t)^2 + \cos(t)^2)dt\\ &= \pi R^2 \end{align} $$

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First note that we stay on the plane, because ${\bf n}\times{\bf r}$ is just $\bf r$ rotated by $+90^\circ$.

You can parametrize the integral, writing $${\bf r}=(R\cos t,R\sin t)$$ Then $d{\bf r}=(-R\sin t,R\cos t)dt\ $ which is thus looks in the same direction as $\bf A$,so ${\bf A}\cdot d{\bf r}= \displaystyle\frac12R^2$, so $$C=\int_{t=0}^{2\pi} \frac12R^2 = R^2\pi$$

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what about if i choose $K$ to be in some space rather than as in $xy$ plane ?? would it not change under co-ordinate transformation ? –  Theorem Oct 22 '12 at 23:51

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