$\int_\gamma\textbf{u}\ d\textbf{r}$, wtih $\gamma$ a curve along an ellipse. Let $$\textbf{u}(x,y)=(y(8x+1),2y^2)\\ \gamma: 4x^2+y^2=1, \ \text{from}\ (-\frac{1}{4},\frac{\sqrt{3}}{2})\ \text{to} \ (\frac{\sqrt2}{4},\frac{\sqrt2}{2})$$
and calculate the integral $\int_\gamma\textbf{u}\ d\textbf{r}$.
So my take is to use Green's formula to get: $$\int_\gamma y(8x+1) dx+2y^2dy=\iint_D \frac{\partial}{\partial x}(2y^2)-\frac{\partial}{\partial y}(y(8x+1))\ dxdy=\iint_D -8x-1\ dxdy $$
I can substitute to polar coordinates; $0\leq r\leq 1, t_1=(-\frac{\pi}{2}-\arctan{(\frac{1}{2\sqrt3})})\leq \theta \leq \arctan{(2)}=t_2$) and get $$x=\frac{1}{2}r\cos(\theta)\\y=r\sin(\theta)\\J=\frac{1}{2}r$$
When I integrate $$\int_{t_1}^{t_2}\int_1^0 2r^2cos(\theta)+\frac{r}{2} drd\theta$$
This gives me the wrong evaluation, but I am not sure why. Is it because I have not defined the the path of the curve back to its starting point and thus not getting a well defined shape or something else?

You cannot use Green's Theorem as the path is not a closed curve. So you will have to do line integral directly.
$\vec F = (8xy+y,2y^2)$
$\gamma: 4x^2+y^2=1, \ \text{from}\ \text{point A} (-\frac{1}{4},\frac{\sqrt{3}}{2})\ \text{to} \ \text{B} (\frac{\sqrt2}{4},\frac{\sqrt2}{2})$.
Parametrize the ellipse as $ \ \gamma(t) = (\frac{1}{2} \cos t, \sin t)$. Based on values of $x, y$ coordinates, point A is represented by $t = \frac{2\pi}{3}$ and point B by $t = \frac{\pi}{4}$.
$\gamma'(t) = (-\frac{1}{2} \sin t, \cos t)$
$\vec F(\gamma(t)) = (4 \sin t \cos t + \sin t, 2 \sin^2t)$
$\vec F(\gamma(t)) \cdot \gamma'(t) = - \frac{1}{2} \sin^2t$.
Now integrate going from point A to point B.
