Find an ellipse whose length is the same as the outer rim of the monkey saddle 
Given the monkey saddle $z=x^3-3xy^2$ over the unit circle $x^2+y^2
\leq 1$, find an ellipse whose length is the same as the length of the
  outer edge of the monkey saddle.

I've already found a parameterization for the monkey saddle in cylindrical coordinates:
$x=r\cos \theta$
$y = r\sin\theta$
$z = r^3\cos3\theta$
And I've found the area of the monkey saddle in the region: $\frac{\pi}{6}[3\sqrt{10}+log(3+\sqrt{10})]$
And I know that the arc length of the ellipse $x^2/a^2+y^2/b^2=1$ is $\int_0^{2\pi}{\sqrt{a^2\cos^2\theta+b^2\sin^2\theta} d\theta}$

The outer edge of the monkey saddle is at $r=1$, and thus $(x,y,z)=(\cos\theta,\sin\theta,\cos3\theta)$. The total arc length is
$$
\begin{align}
\int\mathrm ds
&=
\int\sqrt{\mathrm dx^2+\mathrm dy^2+\mathrm dz^2}
\\
&=
\int_0^{2\pi}\sqrt{\left(\frac{\mathrm dx}{\mathrm d\theta}\right)^2+\left(\frac{\mathrm dy}{\mathrm d\theta}\right)^2+\left(\frac{\mathrm dz}{\mathrm d\theta}\right)^2}\mathrm d\theta
\\
&=
\int_0^{2\pi}\sqrt{\sin^2\theta+\cos^2\theta+9\sin^2(3\theta)}\,\mathrm d\theta
\\
&=
\int_0^{2\pi}\sqrt{1+9\sin^2(3\theta)}\,\mathrm d\theta
\\
&=
\int_0^{2\pi}\sqrt{1+9\sin^2\theta}\,\mathrm d\theta\;.
\\
&=
\int_0^{2\pi}\sqrt{\cos^2\theta+10\sin^2\theta}\,\mathrm d\theta\;.
\end{align}
$$
What you wrote in the last line of the question is actually the arc length of an ellipse, not its area (as you can tell just from the units of length). Thus an ellipse with semi-axes $1$ and $\sqrt{10}$ has the same length as the outer edge of the monkey saddle.
