# Finding the line integral of a gradient

This is a problem from my last years mid-term exam:

The problem goes as following: $C$ is a circle paratamized by $r(t)=\cos t\mathbf{i} +\sin t\mathbf{j}$, where $t\in[0,2\pi]$ what is $\int_c{\nabla\phi\cdot dr}$ where $\phi(x,y)=e^{\sin{(xy)}}$

I know that this line integral is path dependent $\frac{\partial \phi}{\partial x}\neq \frac{\partial \phi}{\partial y}$ but i'm not sure on what to do next.

do i just plug in $\phi(r(t)\cdot dr$ and then solve it like that?

• Here $F$ is $\nabla\phi$ not just $\phi$..it is path independent – Believer Mar 10 '18 at 18:57

At $t=0,r(t)=(1,0)$and at $t=2\pi, r(t)=(1,0)$
$\displaystyle\int_c \nabla\phi.dr =[\phi]_{(1,0)}^{(1,0)}=0.$