What's the area enclosed by $x^4-x^2y^2+y^4=1$

Question

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

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?

Answer 1

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.

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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.

Answer 2

The courbe has in polar coordinates the equation $$r^4\left(1-\frac{3}{4}\sin^2(2 \theta)\right)=1.$$ Thus $0\leq r \leq \left(1-\frac{3}{4}\sin^2(2 \theta)\right)^{-1/4}.$