Area enclosed by a biquadratic function

Question

Find the area enclosed by the curve $x^4+y^4=2xy$.

I've asked my teacher also about it, and he just said to put x= rcos(t) and y= rsin(t), first of all, can we do so?, Even after doing so I didn't got anything.

Answer 1

Hint:

Clearly, $xy\ge0$

and the required area $=2\cdot$ area in the first quadrant

$$\dfrac{x^3}y+\dfrac{y^3}x=2$$

For $x,y>0$ WLOG $\dfrac{x^3}y=\sqrt2\cos^2t,\dfrac{y^3}x=\sqrt2\sin^2t$

$$y^8=4\cos^2t\sin^6t\implies y=\sqrt[4]{2\cos t\sin^3t}$$

and similarly, $x=\sqrt[4]{2\cos^3t\sin t}\implies dx=\sqrt[4]2\left(\dfrac{\cos^{1+3/4}t}{4\sin^{3/4}t}-\dfrac{3\sin^{1+1/4}t}{4\cos^{1/4}t}\right)$

We need $$2\int_0^{\pi/2}y(t)\ d x(t)$$