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

Easy estimation: $2\pi = 6.28\ldots \le s \le 8$.
$$
u = x^2 + 1/2 \Rightarrow u^2 = x^4 + x^2 + 1/4 \\
v = y^2 + 1/2 \Rightarrow v^2 = y^4 + y^2 + 1/4
$$
Then 
$$
r^2 = (\sqrt{2+1/2})^2 = u^2 + v^2 = x^4 + x^2 + y^4 + y^2 + 1/2 \iff \\
2 = x^4 + x^2 + y^4 + y^2 \quad (*)
$$
So the transformed curve is a circle in the $u$-$v$-plane with radius $r = \sqrt{5/2}$ and has the known arc length $S = \pi \sqrt{10}$.
I looked for a way to transform this arc length into the wanted arc length $s$, but had no success. 
For the upper half of the curve we get:
\begin{align}
v &= \sqrt{r^2 - u^2} \iff \\
y^2 + 1/2 &= \sqrt{9/4 - x^2 - x^4} \iff \\
y &= \sqrt{\sqrt{9/4 - x^2 -x^4} - 1/2} \quad (**) \\
\end{align}
This leads to the derivative
\begin{align}
y' &= \frac{1}{2\sqrt{\sqrt{9/4 - x^2 -x^4} - 1/2}} 
\frac{-2x-4x^3}{2\sqrt{9/4 - x^2 -x^4}} \\
&= - \frac{\sqrt{2}\,x\,(1 + 2x^2)}{\sqrt{\left(9 - 4x^2(1+x^2)\right)
\left(\sqrt{9 - 4x^2(1+x^2)} - 1\right)}}
\end{align}
The arc length is
\begin{align}
s &= 4 \int\limits_0^1\sqrt{1 + (y')^2} \, dx \\
  &= 4 \int\limits_0^1\sqrt{1 + \frac{2\,x^2\,(1 + 2x^2)^2}{\left(9 - 4x^2(1+x^2)\right)
\left(\sqrt{9 - 4x^2(1+x^2)} - 1\right)}} \, dx \\
&= 6.68188
\end{align}
where the rounded numerical value was obtained from WolframAlpha.
An alternative query using the simpler equation $(**)$ is this. The result has to be multiplied by $4$ however.
Addendum
Alex Jordan pointed out that $y' \to -\infty$ for $x \to 1$, see WA plot. So we go only for the first 1/8 of the arc, were there is a point $(\xi,\xi)$:
$$
2 = \xi^4 + \xi^2 + \xi^4 + \xi^2 = 2\xi^2(\xi^2+1) \iff \\
1 = \xi^4 + \xi^2 = (\xi^2 + 1/2) - 1/4 \Rightarrow \\
\xi = \sqrt{(\sqrt{5} - 1)/2} = \sqrt{\varphi - 1} = 0.78615\ldots
$$
where $\varphi = 1.618\ldots$ is the golden ratio.
This leads to
\begin{align}
s &= 8 \int\limits_0^\xi\sqrt{1 + (y')^2} \, dx \\
  &= 8 \int\limits_0^\xi\sqrt{1 + \frac{2\,x^2\,(1 + 2x^2)^2}{\left(9 - 4x^2(1+x^2)\right)
\left(\sqrt{9 - 4x^2(1+x^2)} - 1\right)}} \, dx \\
\end{align}
New query: link
