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?
Best Answer
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.