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What's the area enclosed in $x^4-x^2y^2+y^4=n$?

x^4+y^4-(xy)^2=1

The image above has $n=1$.

We can convert to polar coordinates of course. $r^4[\cos^4(\theta)-\cos^2(\theta)\sin^2(\theta)+\sin^4(\theta)]=n$ Not exactly sure how that might help us.

I can approximate the area by creating some random points between $0$ and $2$ and asking how often they fall in the space enclosed by $x^4-x^2y^2+y^4=1$.

This leads me to an area of $\approx 4.3$. Can I get an exact value for this?

Mason
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    You can get an integral out since its a quadratic in $y^2$ and hence you can describe part of the boundary as a 4th root. This + symmetries gets you something you could plug into wolfram or matlab but idk if it could be exact – Calvin Khor Oct 07 '18 at 21:15
  • Suggestion: Try $u=x^2,\ v=y^2$. Then convert to polar coordinates. – herb steinberg Oct 07 '18 at 21:17
  • https://www.wolframalpha.com/input/?i=solve+for+:+x%5E4-x%5E2y%5E2%2By%5E4%3Dn – Mason Oct 07 '18 at 21:51
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    @Mason the more useful branch is $y=\frac{\sqrt{2}}{2}\sqrt{x^2+\sqrt{4n-3x^4}}$, In addition to the symmetries $x\mapsto -x$ and $y\mapsto -y$, there is also the symmetries $y=-x$, $y=x$ so you just need to find the area bound by $y\ge 0$, $x=0$ and $x=y$, after subtracting off the triangle you're left with a wavy bit given by the above and this is the thing i meant you can integrate – Calvin Khor Oct 07 '18 at 22:08
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    https://www.desmos.com/calculator/logbtfxngw – Mason Oct 07 '18 at 22:11
  • @Mason it should be $f(x) - 1$ in the integral since you're "lowering the graph by 1". This gives for $n=1$ the integral 0.313031294999 and then you add 4 when you add back the 8 triangles / 4 squares – Calvin Khor Oct 07 '18 at 22:14
  • I could be wrong but I think you might have a slightly more complicated approach actually. I am taking advantage of some symmetries to make the job easier. https://www.desmos.com/calculator/yoaeaid6c9 – Mason Oct 07 '18 at 22:17
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    oh sure that works too I just wanted to toss out the square because i knew that was a 4 – Calvin Khor Oct 07 '18 at 22:18
  • I wrote out some of my motivations – Mason Oct 07 '18 at 23:00
  • https://www.desmos.com/calculator/7bemizkxjf – Mason Oct 08 '18 at 00:03

2 Answers2

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You are right on the spot with the polar form. You need an area in polar coordinates, which is (quite obviously, from circle geometry):

$$A=\frac12\int_0^{2\pi} r^2\,d\phi$$

You have

$$r^4=\frac{1}{\cos^4 \phi-\cos^2\phi\sin^2\phi+\sin^4\phi}=\frac{1}{1-3\cos^2\phi\sin^2\phi}=\frac{1}{1-\frac{3}{4}\sin^2 2\phi}$$ $$r^2=\frac{1}{\sqrt{1-\frac{3}{4}\sin^2 2\phi}}$$

and finally

$$A=\frac12 \int_0^{2\pi} \frac{1}{\sqrt{1-\frac{3}{4}\sin^2 2\phi}}d\phi$$

This integral does not have a closed form in terms of algebraic and trigonometric functions, but can be put into Wolfram Alpha to calculate numerically (you get about 4.31). WA will express it with with a complete elliptic integral (one of the special functions that are not part of the "standard" set), but that's probably not that useful.

Maybe an explanation of the trig simplification:

$$1^2=(\cos^2\phi+\sin^2\phi)^2=\cos^4\phi+\sin^4\phi+2\cos^2\phi\sin^2\phi$$ From where both fourth power terms can be extracted with ease.

orion
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  • https://www.desmos.com/calculator/tjsbx6gplq – Mason Oct 07 '18 at 22:02
  • https://www.wolframalpha.com/input/?i=%5Cfrac%7B1%7D%7B2%7D%5Cint_0%5E%7B2%5Cpi%7D+1%2F%7B%5Csqrt%7B1-%5Cfrac%7B3%7D%7B4%7D%5Csin%5E2%5Cleft(2x%5Cright)%7D%7Ddx – Mason Oct 07 '18 at 22:04
  • Thanks! Fantastic. Is the case that for $n>1$ we just multiply $A$ by $n^2$ compute the area? – Mason Oct 07 '18 at 22:07
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    You multiply by $\sqrt{n}$ (we needed to take a square root from $r^4$ to $r^2$). – orion Oct 07 '18 at 22:10
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The courbe has in polar coordinates the equation $$r^4\left(1-\frac{3}{4}\sin^2(2 \theta)\right)=1.$$ Thus $0\leq r \leq \left(1-\frac{3}{4}\sin^2(2 \theta)\right)^{-1/4}.$

user376343
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