Evaluate the integral of $\vec{F}=(y(8x+1),18y^2)$ over the curve $\gamma$ along the ellipse $4x^2+9x^2=4,$ counterclockwise from the point $(-1,0)$ to $(-\sqrt{2}/2,-\sqrt{2}/3).$
I have that $P(x,y)=y(8x+1)$ and $Q(x,y)=18y^2.$ The formula I need to use is
$$I=\int_\gamma Pdx+Qdy.$$
The ellipse can be written as $(2x)^2+(3x)^2=4,$ so parameterizing by $(x,y)=\left(\frac{\cos{t}}{2},\frac{\sin{t}}{3}\right),$ so
$$\left\{ \begin{array}{rcr} dx & = & -\frac{\sin{t}}{2} \ dt \\ dy & = & \frac{\cos{t}}{3} \ dt\\ \end{array} \right.$$
Also, drawing a figure...
...reveals that $t\in[\pi, \pi + \arctan{2/3}],$ thus my integral becomes
$$I=\int_\pi^{\pi+\arctan{2/3}}\frac{\sin{t}}{3}\left(8\frac{\cos{t}}{2}+1\right)\cdot \left(-\frac{\sin{t}}{2}\right) + 18\left(\frac{\sin{t}}{3}\right)^2\cdot\frac{\cos{t}}{3} dt = \\ = -\frac{1}{6}\int_\pi^{\pi+\arctan{2/3}}\sin^2{t}=\frac{1}{26}-\frac{\arctan{2/3}}{12}.$$
This is not correct.