<s>[INST] What is the circumference (arc length) of $x^4 + x^2 + y^4 + y^2 = 2$? Consider
$$x^4 + x^2 + y^4 + y^2 = 2$$
It is a smooth non-intersecting circle like curve in the plane.
A bit like a Hyperellipse.
See https://en.m.wikipedia.org/wiki/Superellipse
What is the circumference (arc length) of $x^4 + x^2 + y^4 + y^2 = 2$ ?
Wolfram|Alpha plot
 [/INST] In polar coordinates: $$r^4\cos^4t+r^2\cos^2t+r^4\sin^4t+r^2\sin^2t=2$$
$$r^4\left(\cos^4t+\sin^4t\right)+r^2-2=0$$
$$r^2=\frac{-1+\sqrt{1+8\left(\cos^4t+\sin^4t\right)}}{2\left(\cos^4t+\sin^4t\right)}$$
Now the arc length for polar coordinates is $$\int_0^{2\pi}\sqrt{r^2+\left(\frac{dr}{dt}\right)^2}dt$$ So it appears you have a challenging integral to compute, and you may need to resort to an approximation technique.

Note $$4r^3\left(\cos^4t+\sin^4t\right)\frac{dr}{dt}+r^4\cdot4\left(-\cos^3t\sin t+\sin^3t\cos t\right)+2r\frac{dr}{dt}=0$$
$$\frac{dr}{dt}=\frac{2r^3\sin t\cos t\left(\cos^2t-\sin^2t\right)}{2r^2\left(\cos^4t+\sin^4t\right)+1}=\frac{2r^3\sin t\cos t\left(\cos^2t-\sin^2t\right)}{\sqrt{1+8\left(\cos^4t+\sin^4t\right)}}$$
$$\left(\frac{dr}{dt}\right)^2=\frac{4r^6\sin^2 t\cos^2 t\left(\cos^2t-\sin^2t\right)^2}{1+8\left(\cos^4t+\sin^4t\right)}$$
You can either use this to try to conquer the integral, or you can use this to help with an approximation technique like say Simpson's Rule (where it would be wise to integrate from $0$ to $\frac{\pi}{4}$ and octuple the result.)
Using Simpson, with $n=2$ over $[0,\pi/4]$ and then octuple: $$8\cdot\frac{\pi}{24}\left(1+4\sqrt{\frac{-1+\sqrt{7}}{3/2}+\frac1{28}\left(\frac{-1+\sqrt7}{3/2}\right)^3}+\sqrt{-1+\sqrt{5}}\right)\approx6.6923\ldots$$ which is within $0.16\%$ of Robert's Maple decimal output.
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