Finding the area enclosed by a curve without usual 1-D explicit integration. I am trying to find the area enclosed by the curve
$$x^4 + y^4 = 4xy$$
in the first quadrant. Solving the roots of a quartic polynomial $y$ does not seem to be an efficient way to approach this. I am considering simplifying the problem converting to polar coordinates in the equation that defines the curve, or perhaps define the curve as a parametric function (in terms of some parameter $t$) and then use Green's Theorem. Any guidance regarding this idea or another technique is appreciated.
 A: Let's try polar coordinates. With $x=r\cos\phi, y=r\sin\phi$, the equation becomes
$$
r^4(\cos^4\phi+\sin^4\phi)=4r^2\cos\phi\sin\phi.
$$
From this we can solve (safe to ignore $r=0$)
$$
\begin{aligned}
\frac12r^2&=\frac{2\cos\phi\sin\phi}{\cos^4\phi+\sin^4\phi}=
\frac{\sin2\phi}{2\cos^4\phi-2\cos^2\phi+1}\\
&=\frac{2\sin2\phi}{4\cos^4\phi-4\cos^2\phi+2}=\frac{2\sin2\phi}{(2\cos^2\phi-1)^2+1}\\
&=\frac{2\sin2\phi}{1+\cos^22\phi}.
\end{aligned}
$$
From this last form we see that
$$
\int\frac 12 r^2\,d\phi=-\arctan(\cos 2\phi).
$$
This makes calculation of the area easy. We see that $\phi$ ranges over $[0,\pi/2]$, so
$$
A=\int_0^{\pi/2}\frac 12 r^2\,d\phi=-\arctan (\cos\pi)+\arctan(\cos 0)=\frac\pi2.
$$
Judging from the plot this feels reasonable.

A: If you look at the curve, that is produced by this equations it sprang to my mind that rotating it by 45° might be a good idea and it is:
$$(x,y) \mapsto \left(\frac{x-y}{\sqrt 2}, \frac{x+y}{\sqrt 2}\right)$$
gives the nice equation
$$ x^4 - 4x^2 + 6x^2y^2 + 4y^2 + y^4 = 0.$$
Here we can substitute $x^2 = a$ and $y^2 = b$ and get
$$ a^2 - 4a + 6ab + 4b + b^2 = 0.$$
This can be solved with standard methods and you will eventually get a parametrisation of your rotated curve as
$$y = \pm\sqrt{-3 x^2+2 \sqrt{2 x^4+4 x^2+1}-2}.$$
Now we can easily integrate over the plus term and get half the area you wanted, because the curce is symmetric:
$$ A/2 = \int_0^2 \sqrt{-3 x^2+2 \sqrt{2 x^4+4 x^2+1}-2} dx \approx 0.785398. $$
Unfortunately this integral cannot be solved analytically by my standard computer algebra tools. I can only approximate it.
