Help regarding the area and length of an implicit curve

Question

I want to know two things: 1.Given the equation of a plane curve ,how do we know whether it is closed or not ?To be specific ,consider the following $$(x+y)^{2}+(x-y)^{\left(\frac{2}{3}\right)}=1$$.I drew the above curve in desmos url{https://www.desmos.com/calculator/z39avcy2er}and found it be a nice closed curve. Second,how can I find the length and area of such a curve as the above example? Thanks for any hints/respences.

Answer 1

Starting from @Cye Waldman's answer $$L=2\sqrt{2}\int_0^1 \sqrt{1+\frac{1}{9(1-x^{2/3})x^{2/3}}}\ dx$$ we can make an exact calculation.

Let $x=\sin^3(t)$ to get $$L=\int_0^\frac \pi 2 \sin(t)\sqrt{8+18 \sin^2(2 t)} \, dt$$ This leads to awful elliptic integrals of the first and second kind.

Making the story short, the result is

$$\frac{i \left(\frac{9 }{\sqrt{2}}L-6\right)}{\sqrt{6 \left(3+\sqrt{13}\right)}}=$$ $$\sqrt{13} F\left(i \sinh ^{-1}\left(\sqrt{\frac{3}{2} \left(3+\sqrt{13}\right)}\right)|\frac{-11+3 \sqrt{13}}{2} \right)-3 E\left(i \sinh ^{-1}\left(\sqrt{\frac{3}{2} \left(3+\sqrt{13}\right)}\right)|\frac{-11+3 \sqrt{13}}{2} \right)$$ which makes $$L=4.1216964106052626439040756172731180548685478808226\cdots$$ which is not recognized by inverse symbolic calculators.

Answer 2

The curve $$(x-y)^{2/3}+(x+y)^2=1$$ rotated of $\pi/4$ with the transformation $$\left\{x\to \frac{x}{\sqrt{2}}-\frac{y}{\sqrt{2}},y\to \frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}\right\}$$ becomes $$2 x^2+\sqrt[3]{2} \,y^{2/3}=1$$ For the simmetry of the curve, the area is given by $$S=4 \int_0^{\frac{1}{\sqrt{2}}} \frac{\left(1-2 x^2\right)^{3/2}}{\sqrt{2}} \, dx=\frac{3 \pi }{8}$$ While the curve length leads to an elliptic integral $$L=4\int_0^{\frac{1}{\sqrt{2}}} \sqrt{-36 x^4+18 x^2+1} \, dx\approx 4.12$$

Edit.

I forgot to prove that this is a closed curve.

Solve the equation wrt $x$ $$x=\pm\frac{\sqrt{1-\sqrt[3]{2} y^{2/3}}}{\sqrt{2}}$$ in order to be real we must have $$\sqrt{1-\sqrt[3]{2} y^{2/3}}\ge 0 \to 0\leq y\leq \frac{1}{\sqrt{2}}$$ and so the limitations for $x$ are $$-\frac{1}{\sqrt{2}}\leq x\leq \frac{1}{\sqrt{2}}$$

Answer 3

I'm taking a slightly different approach to that of @Raffaele because I recognize the form as one for which we can obtain an analytic solution. We begin by simplifying the equation by rotating the axes by $\theta=-\pi/4$, thus

$$ x=x'\cos(\theta)-y'\sin(\theta)=\frac{\sqrt{2}}{2}(x'+y')\\ y=x'\sin(\theta)+y'\cos(\theta)=\frac{\sqrt{2}}{2}(-x'+y')\\ $$

I follows that

$$ x+y=\sqrt{2}y'\\ x-y=\sqrt{2}x'\\ (\sqrt{2}x')^{2/3}+(\sqrt{2}y')^2=1 $$

We recognize this to be in the form superconics that we developed (see, for example here). We can write the above equation more generally in the canonical form

$$ X^q+Y^{1/p}=1\\ \text{or}\\ f(X)=(1-|X|^q)^p $$

where $X=x'/a,\ Y=y'/b,\ a=b=1/\sqrt{2},\ q=2/3,\ p=1/2$ and $-1\le X\le 1$ and $Y\ge 0$.

The important point here is that we can solve for the area analytically. The area for the closed curve is shown as follows

$$ \begin{align} A&=2ab\int_{-1}^1 f(X)\ dX\\ &=2ab\int_{-1}^1 (1-|X|^q)^p\ dX\\ &=4ab\int_{0}^1 (1-X^q)^p\ dX\\ &=4ab\frac1qB(1/q,p+1,1)\\ &=4ab\frac{\Gamma(p+1)\Gamma(1+1/q)}{\Gamma(p+1+1/q)}\\ &=3\pi/8 \end{align} $$

where $B$ is the beta function and $\Gamma$ is the gamma function. The numerical result is for the parameters of the present problem.

No such good fortune for the arc length, however. Typically we resort to numerical methods. Thus we solved

$$ \begin{align} L&=4a\int_0^1 \sqrt{1+f'(X)^2}\ dX\\ &=2\sqrt{2}\int_0^1 \sqrt{1+\frac{1}{9(1-X^{2/3})X^{2/3}}}\ dX\\ &=4.1217... \end{align} $$