Evaluate $$I = \int_{C}\mathbf{F}\cdot d\mathbf{R},$$ where $\mathbf{F} = \left\langle ye^{xy},xe^{xy}\right\rangle$ and $C$ is the curve parameterized by $\mathbf{R} = \left\langle \cos{t},\sin{t}\right\rangle$ for $0\leq t\leq 2\pi$.
a) $I = \pi$
b) $I=2\pi$
c) $I = 0$
d) $I = 1$
e) $I = \frac{1}{2}$
Okay, so I am confused on where to go with solving this problem. I tried to plug in $\cos(t)$ for $x$ and $\sin(t)$ in for $y$ and ended up. I then took the dot product of that and $d\mathbf{R}$, but I am not getting an integral that could be easily integrated. The dot product I end up with is $$-\sin^2(t)e^{\cos(t)\sin(t)}+\cos^2(t)e^{\cos(t)\sin(t)}.$$
Maybe I a missing a step? Perhaps an identity I am not seeing?
Thanks.
Start working from the vector field. you can notice that its conservative by checking the symmetric properties of the jacobi-matrix. That is $\frac{\partial F_1}{\partial y} = \frac{\partial F_2}{\partial x}$. Since it's conservative that means that the integral can be evaluated by a potential $\phi$ but since the start- and endpoint is the same point the value of the integral is zero (i.e $\phi(A)-\phi(A)=0$).
But you can also calculate the integral $\int_0^{2\pi} e^{\frac{1}{2}\sin(2t)}\cos(2t)dt$ Which turn out to be $0$