Why line integral of f(x.y)=(x.y) is not zero along the circle? I am asked to determine whether f(x.y)=(x.y) is gradient or not.
It is clear that there exists a function g whose derivative with respect to x and y is equal to first and second component of f.
So, f is gradient. 
If f is gradient, then on closed path , its line integral must be zero.
I considered a circle with radius $1$ at the origin from $0$ to $2\pi$
This is closed path. 
And I did integrate f along the circle and ended up getting $2 \pi$.
Why is it different? 

$$\int_C f(x,y)d\vec x=\int_0^{2\pi}f(\cos \theta,\sin\theta)\cdot d(\cos\theta,\sin \theta)=\int_0^{2\pi}(\cos\theta,\sin\theta)\cdot (-\sin\theta,\cos\theta)d\theta=\int_0^{2\pi}\underbrace{(-\cos\theta\sin\theta+\cos\theta\sin\theta)}_{=0}d\theta=0.$$
