Let C be the curve in $\mathbb{R}^2$ defined by,
C: $\dfrac{x^2}{9}+\dfrac{y^2}{4}=1, x\geq0,y\geq0$
Compute the line integral, in a direction so that the $y$-coordinate increases:
$\int_C (2x-3y)dx+(5x+6y)dy$.
I have come to the conclusion that the curve is an ellipse in the first quadrant, going from $(3,0)$ to $(0,2)$. By using Green's theorem, I got the integral \begin{align*} \int_C\int 8dA. \end{align*} I then used the Jacobian $dA=J(u,v)=6dudv$, with $x=3u$, $y=2v$ giving me $u^2+v^2=1$. Then switching to polar coordinates, I end up with the integral: \begin{align*} \int\limits_{0}^{\pi/2}\int\limits_{0}^{1} 48r drd\theta = 12\pi. \end{align*}
However the correct answer is $12\pi+3$. I am guessing it has something to do with the axis as this curve is not closed, though I am not sure.
Let's draw the first quarter of the ellipse. Let's call $C = C_1 + C_2 + C3$
It's great that you have found the surface integral of the area enclosed by $C$. However, by Green's theorem, the surface integral is the line integral of $C_1 + C_2 + C3$. In this case, you need to find out the line integral of $C_2$ and $C_3$ and subtract it from $12\pi$.
Since $C_2$ and $C_3$ are straight lines, the line integrals are pretty simple:
$C_2: r_2=<0,t>$ and $t$ goes from $2$ to $0$. Plugging the respective $x,y$ values into $F$ in the line integral yields $-12$.
Similarly, $C_3 : r_3=<t,0>$ where $t$ goes from $0$ to $3$. The integral yields $9$.
Therefore integral over the curve $C1 = C - C_2 - C_3 = 12\pi -(-12) -9 = 12\pi + 3$