I want to solve the line integral
$$\oint_{C}\mathbf{F}\cdot\mathrm{d}\mathbf{r},$$
where
$$\mathbf{F}(x,y) = {\left(2\,y^2+e^{3\,x}\right)\,{\bf i}+3\,x\,y\,{\bf j}}$$
and $C$ is the curve tracing the x-axis and the circumference of the circles $x^2+y^2=4$ and $x^2+y^2=16$ in the upper half-plane. In addition the path C has an anticlockwise orientation (the x-axis is traced in the positive direction).
So
$$x = r\cos(t)\to x'=-r\sin(t)$$ $$y=r\sin(t)\to y' = r\cos(t)$$
and then
$$\oint_{C}\mathbf{F}\cdot\mathrm{d}\mathbf{r}= \int\mathbf F(r(t))\cdot r'(t)\,dt$$
so this gives me some random answer $(\frac{-(e^{24}+16e^6+144e^{12}-e^6-1)e^{-12}}{3})$. Don't know where is the mistake.
Notice that
$$\langle 2y^2 +e^{3x}, 3xy \rangle = \langle 2y^2 + e^{3x}, 4xy \rangle + \langle 0, -xy \rangle$$
where the vector field on the left is conservative. Since it's conservative, its contribution to the line integral will be $0$, so all we have to calculate is
$$\int_C -xy dy$$
On the pieces that connect the two arcs, $dy = 0$. All we need is to parametrize the two circles, leaving us with
$$\int_0^\pi -4^3\cos^2 t \sin t dt + \int_\pi^0 -2^3 \cos^2 t \sin t dt = \frac{64-8}{3}\cos^3 t \Bigr|_0^\pi = -\frac{112}{3}$$