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

Let us first rewrite the curve in the hope to perhaps parametrize it.
\begin{align*}
x^4+x^2+y^4+y^2=2\iff{}&\left(x^2+\frac12\right)^2+\left(y^2+\frac12\right)^2-\frac12=2\iff{} \\
{}\iff{}&\frac25\left(x^2+\frac12\right)^2+\frac25\left(y^2+\frac12\right)^2=1.
\end{align*}
So we must have $t$ such that:
$$\left\{\begin{array}{c}
\sqrt{\frac25}\left(x^2+\frac12\right)=\cos t \\
\sqrt{\frac25}\left(y^2+\frac12\right)=\sin t \\
\end{array}\right.
\iff
\left\{\begin{array}{c}
x^2=-\frac12+\sqrt{\frac52}\cos t \\
y^2=-\frac12+\sqrt{\frac52}\sin t \\
\end{array}\right.$$
Split a couple of cases depending on sign of $x,y$ (4 cases, I should think), then calculate the integrals of:
$$\gamma(t)=\left(\pm\sqrt{-\frac12+\sqrt{\frac52}\cos t},\pm\sqrt{-\frac12+\sqrt{\frac52}\sin t}\right)$$
with signs depending on the cases, sum them, and you're done.
