Find the integral on this curve. (Vector field containing one point that is problematic). 
$\vec \ell(t)=(\sin(t), \sin(t)\cos(t)-1)$
$t\in[0,2\pi]$. 
Find $\displaystyle\int_\ell \Big(\frac{-y}{\sqrt{2x^2+2y^2}}\Big) \, dx + \Big(\frac{x}{\sqrt{2x^2+2y^2}}\Big) \, dy$.

My attempt: 
First thing that crossed my mind when I saw this question is I don't want to do the integral with the given parametrization.. so I checked if $\vec F=(\frac{-y}{\sqrt{2x^2+2y^2}},\frac{x}{\sqrt{2x^2+2y^2}})$ could be conservative.
I found out that $Q_x=P_y$, in all $\mathbb{R}^2\setminus\{(0,0)\}.$
But I know that now, for any curve I take that rounds $(0,0)$ whatever value of the integral I get, will be the value of any other integral over a curve that rounds $(0,0).$
So I went up and $\vec r(t)=(\cos(t), \sin(t))$.
$$\int_r\vec F\cdot d\vec r=\int_0^{2\pi}\frac{1}{2}(\cos^2t+\sin^2t)=\pi.$$
So I thought my normal integral would be the same. 
Plot twist: The answer was actually $0$. 

Suspections: 

*

*I made a mistake calculating the integral. (I checked alot of times so I don't think this is the case). 

*I did all this for nothing because $\vec l$ doesn't round the point $(0,0)$.. ($99\%$). 
Questions: 

*

*How do I check if the given curve rounds the point $(0,0)$?

*Knowing that my curve doesn't round the point, is there any way to use the fact that the field is conservative there, other than finding the potential function?

Note: Final answer is $0$. 
Any feedback is appreciated, thanks in advance!
 A: Given how the question reads, the line integral of the given vector field over the given curve is zero.
$\vec \ell(t)=(\sin(t), \sin(t)\cos(t)-1), t\in[0,2\pi]$
$\vec F = \displaystyle \Big(\frac{-y}{\sqrt{2x^2+2y^2}}, \frac{x}{\sqrt{2x^2+2y^2}}\Big)$
Here is the path from $t: 0 \to 2\pi$.

The loop to the right of y-axis is clockwise whereas the loop to the left of y-axis is anti-clockwise.
Checking the numerator of the vector field and the symmetry of the curve, it can be shown that the line integral over the part of the curve in the fourth quadrant will cancel out the integral over the part of the curve in first quadrant. Same with the line integral in second and fourth quadrant. However I will show this using Green's theorem.
$Q_x = \dfrac{y^2}{\sqrt2 (x^2+y^2)^{3/2}}, P_y = - \dfrac{x^2}{\sqrt2 (x^2+y^2)^{3/2}}$
$Q_x - P_y = \dfrac{1}{\sqrt{2x^2+2y^2}}$
For the first loop (for $x \geq 0$) between $0 \leq t \leq \pi$, we are in the clockwise direction.
So the line integral is $\displaystyle \iint_{D_{12}} - \dfrac{1}{\sqrt{2x^2+2y^2}} \ dA$
For the second loop, it is $\displaystyle \iint_{D_{34}} \dfrac{1}{\sqrt{2x^2+2y^2}} \ dA$
The integrand is an even function but with opposite signs in both integrals and regions are symmetric about y-axis. The two integrals will cancel each other out.

Some additional details - in cartesian coordinates, the curve is
$(y+1)^2 = x^2 - x^4, \ \ -1 \leq x \leq 1$. So the first integral can be written as,
$\displaystyle \int_0^1 \int_{-1 - x \sqrt{1-x^2}}^{-1 + x \sqrt{1-x^2}} - \dfrac{1}{\sqrt{2x^2+2y^2}} dy \ dx$
And the second integral can be written as,
$\displaystyle \int_{-1}^0 \int_{-1 + x \sqrt{1-x^2}}^{-1 - x \sqrt{1-x^2}} \dfrac{1}{\sqrt{2x^2+2y^2}} dy \ dx$
(Substituting $t = -x$ in the second integral gives you first integral but with opposite sign.)
