The question is:
$$
\int_\gamma\frac{-y^2dx+2xy\ dy}{x^2+y^4}
\quad \gamma:r(t)=(t,2t^2-2), \quad -1\leq t\leq 1
$$
I have tried to solve it like this:
since $Q_x=P_y$ it's potential vector field but the singularity is at origin, first i thought that i could evaluate it from $(-1,0)$ to $(1,0)$ along $x$ axis but the singularity is the problem, how should i proceed in this case?
Any suggestion would be great Thanks
Best Answer
I will present two options to avoid doing the integral directly.
$\textbf{Option 1}$: Fundamental Theorem of Line Integrals
Since the vector field is conservative on any domain that doesn't contain the origin, we can find a potential function
$$f(x,y) = -\arctan\left(\frac{x}{y^2}\right)$$
There are many options for the potential function here, but this one is continuous on the lower half plane $y<0$. Being careful how we take limits, we approach the endpoints of the integral $(1,0)$ and $(-1,0)$ from below
$$I = \lim_{(x,y)\to(1,0^-)}-\arctan\left(\frac{x}{y^2}\right)+\lim_{(x,y)\to(-1,0^-)}\arctan\left(\frac{x}{y^2}\right) = -\frac{\pi}{2}-\frac{\pi}{2}=-\pi$$
$\textbf{Option 2}$: Green's Theorem/Partial Path Independence
As long we have two paths that (when taken together) do not enclose the origin, then we have path independence. This can be proven via Green's theorem. In this case we will automatically have path independence if we restrict our attention to paths only in the lower half plane. In this case consider the curve
$$x^2+y^4=1 \hspace{20 pt} y\leq 0$$
This curve is chosen because
(a) it simplifies the denominator to a constant and
(b) it contains the start and end points of the original curve.
We can parametrize this curve by
$$\begin{cases}x(t) = \cos t \\ y(t) = -\sqrt{-\sin t} \end{cases} \hspace{24 pt} t\in[-\pi,0] \hspace{18 pt}\implies \hspace{18 pt} \begin{cases}x'(t) = -\sin t \\ y'(t) = \frac{\cos t}{2\sqrt{-\sin t}} \end{cases}$$
Plugging into the integral gives us
$$\int_{-\pi}^0-\sin^2t-\cos^2t\:dt = -\pi$$