Calculating the following integral . 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 ? 
 A: 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
$$
\begin{align}
\oint_K d\mathbf r \cdot \mathbf A &= \int_0^{2\pi} \mathbf A(\gamma(t))\cdot \dot\gamma(t)dt \\
&= \frac {R^2} 2 \int_0^{2\pi} (\sin(t)^2 + \cos(t)^2)dt\\
&= \pi R^2 
\end{align}
$$
A: 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
$$C=\int_{t=0}^{2\pi} \frac12R^2 = R^2\pi$$
A: Invoking Stokes theorem over a circle of radius $R$ leads to the following calculation: 
$$\eqalign{
\oint{\underline F}\cdot{\underline {dr}}
 &= \int(\nabla\times{\underline F})\cdot{\underline {da}} \cr
 &= \int\Big(\nabla\times\tfrac{1}{2}({\underline n}\times{\underline r})\Big)\cdot{\underline n\,{da}} \cr
 &= \int \Big({\underline n}\Big)\cdot{\underline n\,{da}} \cr
 &= \int da = \pi R^2 \cr
}$$
No need for any trig functions, you just have to know the area of a circle. This seems to be the direction in which you were headed with your "progress" update except you stated the curl incorrectly:  $\,\nabla\times{\underline A}={\underline n}$.
