I'm trying to resolve this line integral who has a vector field with a singularity so with Green's theorem, using that the curl = 1 it's very difficult.
My curve is $$C(t) = (3/2\cos(t) - \cos^2(t) , 3/2 \sin(t)- \cos(t)\sin(t))$$ where $$0 \le t \le 2\pi$$ and my vector field is
$$ \int_C \left(\frac{6y}{4x^2 + 9y^2} -y \right)dx + \left(\frac{-6x}{4x^2 + 9y^2} \right)dy $$
Please this is a typical exam problem and I can't figure out how to solve it.
Thanks in advance.
Hint:
You can check that this curve $C$ can be represented by $r=\frac{3}{2}-\cos t$. It is a heart shape with the tip on the left at $(-\frac{5}{2})$.
The ellipse $4x^2+9y^2=1$ is inside it. Let this be $C_1$. Then we can use Green's theorem in the region $D$ enclosed by $C$ and $C_1$. So $$\int_C \left(\frac{6y}{4x^2 + 9y^2} -y \right)dx + \left(\frac{-6x}{4x^2 + 9y^2} \right)dy=\int_{C_1}\left(\frac{6y}{4x^2 + 9y^2} -y \right)dx + \left(\frac{-6x}{4x^2 + 9y^2} \right)+\iint_D Q_x-P_y dxdy.$$
Since $Q_x-P_y=1$, the last integral is the area of $D$, which can be found by subtracting from the area of the region enclosed by the curve $C$ the area of the ellipse:
$$\iint_D dxdy =\int_0^{2\pi} \int_0^{3/2-\cos t}rdrdt-\pi \left(\frac{1}{2}\right)\left(\frac{1}{3}\right).$$
The integral on $C_1$ can be done directly.