# compute $\int_\gamma\frac{-y^2dx+2xy\ dy}{x^2+y^4}$

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

• The singularity is not a problem since you have path which avoids it. Feb 3, 2021 at 17:10
• You just need another curve that has the same start and end points, but when taken together with the original curve doesn't enclose the origin. Try the curve $x^2+y^4=1$ with the restriction $y\leq 0$. How would you parametrize it? Feb 3, 2021 at 18:03

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

$$\int_{-\pi}^0-\sin^2t-\cos^2t\:dt = -\pi$$

• Thank you @Ninad Munshi, I have some questions because I am not a professional and I'm a beginner, how did you find the potential I had big trouble when I wanted to integrate with regard to x or y? also did not understand why you did limit of 0 from the negative side? why did you parameterize the curve in that way? how should i know that this is a curve to begin with? also i mean x^2+y^4=1 is that a curve? Feb 4, 2021 at 4:39
• @simon when you are starting, I will recommend to go with option 2 that Ninad has mentioned. $x^2+y^4 = 1, y \leq 0$ is a curve between $(-1,0)$ and $(1, 0)$ as you can see in the sketch. The choice of the curve is based on the vector field and the idea is to simplify the integrand taking advantage of the fact that line integral is path independent as the vector field is conservative. Feb 4, 2021 at 6:22
• As far as option $1$, either you know from experience looking at the denominator as to what could be the potential function and then go from there, otherwise there is no easier way than to integrate and find the potential function. Feb 4, 2021 at 6:27
• @simon and by the way, $x^{2n} + y^{2n} = 1$ are all closed curves between $-1 \leq x \leq 1$, like $x^2+y^2 = 1$ is. If one of them is odd, then I think it is open curve like $x^2 + y = 1$ (a parabola), $x^2 + y^3 = 1$ etc. Feb 4, 2021 at 6:52
• Thank you @MathLover Feb 4, 2021 at 15:06

$$\int_\gamma\frac{-y^2dx+2xy\ dy}{x^2+y^4}=\int_\gamma\frac{-y^2}{x^2+y^4}\,dx+\int_\gamma\frac{2xy}{x^2+y^4}\,dy$$ $$\gamma(t)=(t,2 t^2-2);\;t\in[-1,1];\;\gamma'(t)=(1, 4 t)$$ Substitute $$\int_{-1}^1 -\frac{\left(2 t^2-2\right)^2}{\left(2 t^2-2\right)^4+t^2}\cdot 1\,dt+\int_{-1}^1 \frac{2 t \left(2 t^2-2\right)}{\left(2 t^2-2\right)^4+t^2}\cdot(4t)\,dt=$$ $$=\int_{-1}^1 \frac{4 \left(3 t^4-2 t^2-1\right)}{16 t^8-64 t^6+96 t^4-63 t^2+16}\,dt$$ Integrating is quite hard

denominator $$16 t^8-64 t^6+96 t^4-63 t^2+16=(16 t^8-64 t^6+96 t^4-64 t^2+16)+t^2=16 \left(t^2-1\right)^4+t^2$$

numerator can be factored $$4 \left(3 t^4-2 t^2-1\right)=4\left(t^2-1\right) \left(3 t^2+1\right)$$

and integral can be written as $$\int_{-1}^1\frac{4\left(t^2-1\right) \left(3 t^2+1\right)}{16 \left(t^2-1\right)^4+t^2}\,dt$$ Now divide numerator and denominator by $$16 \left(t^2-1\right)^4$$ $$\int_{-1}^1\frac{\frac{4\left(t^2-1\right) \left(3 t^2+1\right)}{16 \left(t^2-1\right)^4}}{1+\frac{t^2}{16 \left(t^2-1\right)^4}}\,dt=\int_{-1}^1\frac{\frac{ 3 t^2+1}{4 \left(t^2-1\right)^3}}{1+\frac{t^2}{16 \left(t^2-1\right)^4}}\,dt$$ Now substitute $$\frac{t}{4\left(t^2-1\right)^2}=u$$

extremes become $$t=-1\to u=-\infty;\;t=1\to u=+\infty$$

while differentiating we get $$\frac{3 t^2+1}{4 \left(t^2-1\right)^3}\,dt=-du$$

Therefore the integral becomes $$-\int_{-\infty}^{\infty} \frac{du}{1+u^2}=-\left[\arctan u\right]_{-\infty}^{\infty}=-\pi$$

You may also note that by changing variable $$y\to{y'}=y^2$$ you get: $$\int_\gamma\frac{-y^2dx+2xy\ dy}{x^2+y^4}=\int\frac{-y'dx+x\ dy'}{x^2+y'^2}$$. We know that the integrand is a potential vector field, so the integral depends only on the starting and finishing points and does not depend on the way - we just have to avoid the singularity at $$(0,0)$$.

The starting point for $$(x,y')$$ is $$(-1,0)$$ and finishing $$(1,0)$$. Now we can introduce a parametrization $$x=\cos(t)$$ and $$y'=\sin(t)$$, where $$t$$ is changing from $$\pi$$ to $$0$$ - to meet the requirements for the starting and finishing points.

After that the integration is trivial: $$\int_{(-1.0)}^{(1,0)}\frac{-y'dx+x\ dy}{x^2+y'^2}=\int_{\pi}^{0}\frac{-\sin(s)(-\sin(s))ds+\cos(s)\cos(s) ds}{\cos^2(s)+\sin^2(s)}=\int_\pi^0ds=-\pi$$