<s>[INST] Find the area bounded by $ \ x^4+y^4=4(x^2+y^2) \ $ . Find the area bounded by $ \ x^4+y^4=4(x^2+y^2) \ $ . 
Answer:
The graph is above : 
         
Since the region is symmetrical , 
$  Area =4  \times \int_{0}^{2} \int_{0}^{\sqrt{2+\sqrt{4x^2-x^4+4}}} dxdy $
Am I right ? Is there any help ? 
 [/INST] According to your own sketch, the integral you wrote would not cover the area $|y| > 2$, never mind that the order of integration does not match the limits of integration.  So the setup is problematic in multiple ways.
The suggestion to use polar coordinates is worth exploring.  Let $(x,y) = (r \cos \theta, r \sin \theta)$ so that we have $$x^2 + y^2 = r^2,$$ and $$x^4 + y^4 = (x^2 + y^2)^2 - 2 x^2 y^2 = r^4 - 2 r^4 \sin^2 \theta \cos^2 \theta = r^4 (1  - \tfrac{1}{2} \sin^2 2\theta).$$  Thus we have $$0 = (x^4 + y^4) - 4(x^2 + y^2) = r^4 (1  - \tfrac{1}{2} \sin^2 2\theta) - 4r^2,$$ and eliminating the trivial solution $r = 0$, we get $$r^2 = 4(1  - \tfrac{1}{2} \sin^2 2\theta)^{-1}.$$  In the first quadrant, $0 \le \theta \le \pi/2$ and we can write $$\frac{A}{4} = \int_{\theta=0}^{\pi/2} \frac{1}{2}r^2 \, d\theta =  2 \int_{\theta=0}^{\pi/2}  \frac{1}{1-\frac{1}{2}\sin^2 2\theta} \, d\theta = 8\int_{\theta=0}^{\pi/2} \frac{d\theta}{3+\cos 4\theta},$$ and the rest is quite straightforward with the standard trigonometric substitutions.
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