Find the area bounded by the curve $x^4+y^4=x^2+y^2$ I am stuck with this problem which deals with evaluating an Area
The problem reads :

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

I tried factorizing the expression and expressing $y$ in terms of $x$, not able to proceed with that idea. Someone please help me out.
 A: Expressing the curve in polar coordinates, we have\begin{align}x^4+y^4=x^2+y^2&\iff\rho^4\cos^4\theta+\rho^4\sin^4\theta=\rho^2\\&\iff\rho=\frac1{\sqrt{\cos^4\theta+\sin^4\theta}}.\end{align}Hence, the area that you're after is\begin{align}\int_0^{2\pi}\int_0^{1/\sqrt{\cos^4\theta+\sin^4\theta}}\rho\,\mathrm d\rho\,\mathrm d\theta&=\int_0^{2\pi}\frac2{3+\cos(4\theta)}\,\mathrm d\theta\\&=\pi\sqrt2.\end{align}
A: Use polar co-ordinates $x=r \cos t, y=r \sin t$, then this curve is $$r^2=\frac{1}{\sin^4 t+ \cos^4 t}=\ $$, the area the area is
$$A=4 ~ \frac{1}{2}\int_{0}^{\pi/2} r^2 dt=2 \int_{0}^{\pi/2} \frac{1}{\cos^4 t+\sin^4 t}dt =2 \int_{0}^{\pi/2}\frac{\sec^4 t}{1+\tan^4 t} dt $$
Let $z=\tan t \implies dz=\sec^2 t dt$
$$=2\int_{0}^{\infty} \frac{1+z^2}{1+z^4} dz=2 \int_{0}^{\infty}\frac{1+1/z^2}{z^2+1/z^2}=2\int_{0}^{\infty} \frac{1+1/z^2}{(z-1/z)^2+2}$$ $$\implies A=2 \int_{-\infty}^{\infty} \frac{du}{u^2+2}=\sqrt{2} \pi.$$
A: It is possible to do this without using polar coordinates- but not easy.  I would start by noting that only even powers of x and y are involved so the graph of this is symmetric about the x and y axes.  That means that we can calculate the area for x and y positive and multiply by 4.
There is a little problem in that y is not a function of x even in the first quadrant.  Differentiating with respect to y gives $4x^3x'+ 4y^3= 2xx'+ 2y$, $(4x^3- 2x)x'= 2y- 4y^3$.  Setting x'= 0, for a maximum or minimum, $2y- 4y^3= 2y(1- 2y^2)= 0$. That has three roots, y= 0, $y= \frac{\sqrt{2}}{2}$, and $y= -\frac{\sqrt{2}}{2}$. y= 0 gives $x^4= x^2$ in the original equation which gives $x^2= 1$, $x= \pm 1$.
$y= \frac{\sqrt{2}}{2}$ gives $x^4- x^2= \frac{1}{4}- \frac{1}{2}= -\frac{1}{4}$.  Letting $z= x^2$, $z^2- z+ \frac{1}{4}= (z- \frac{1}{2})^2= 0$ so $z^2= x^2= \frac{1}{2}$.  Since we are in the first quadrant, $x= \frac{\sqrt{2}}{2}$.
For x between 0 and $\frac{\sqrt{2}}{2}$ we can solve y^4- y^2+ (x^4- x^2)= 0 using the quadratic formula: $y= \frac{1+ \sqrt{1- 4(x^4- x^2)}}{2}$.
$\frac{1}{2}\int_0^\frac{\sqrt{2}}{2} 1+ \sqrt{1- 4(x^4-x^2)}dx$.  But then we have to integrate from x= $\frac{\sqrt{2}}{2}$ to 1:  $\int_\frac{\sqrt{2}}{2}^1 \sqrt{1- 4(x^4- x^2)}dx$
