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Area and arclength calculations for:

 The Differential Geometry of the Sphericon.
\vs


The area of the graph of a differentiable function $f(x,y)$
is given by
$$\int\int \sqrt{1+f_x^2 + f_y^2}~dx dy$$
where $f_x$ and $f_y$ are the partial derivatives of $f$ with respect to $x$
and to $y$.\vs

For the positive curvature example ($f(x,y)=-x^2-y^2$),
the circle $x^2+y^2=1$ fits in the
surface exactly at height $-1$. The area enclosed is
$$\int\int_D \sqrt{1+4x^2+ 4y^2}~dx dy$$ where $D$ is the disc
of radius 1. Using polar coordinates $x=r \cos\theta, y=r\sin\theta$
the integral becomes $$\int_0^{2\pi}\int_0^1 \sqrt{1+4r^2}~ r~ dr d\theta$$
which can be evaluated (use the substitution $u=1+4r^2$) as
$\frac{\pi}{6}(5^{\frac{3}{2}}-1)=5.335...$.\vs

The negative example is more complicated because first one must
find the circle in the $(x,y)$-plane which gives a circle in the
graph of circumference $2\pi$. I had to do it by trial and
error. The circle of radius $r$ in the $(x,y)$-plane  can be
parametrized as $x = r\cos\theta, y =r\sin\theta$, $0\leq\theta\leq 2\pi$.
In the graph this circle becomes $( r\cos\theta, r\sin\theta,
r^2\cos^2\theta -r^2\sin^2\theta)$, $0\leq\theta\leq 2\pi$. The
length of a parametrized curve $(x(\theta),y(\theta),z(\theta))$
is given by the integral
$$ \int \sqrt{\frac{dx}{d\theta}^2 +\frac{dy}{d\theta}^2 + \frac{dz}{d\theta}^2}
~d\theta.$$
In this case the length is
$$\int_0^{2\pi}\sqrt{r^2 +2r^4\sin^2(2\theta)}~
d\theta,$$
using the identity $\cos^2\theta - \sin^2\theta = \cos(2\theta)$ before
differentiating, and the identity $\cos^2\theta + \sin^2\theta = 1$
after. For $r=.715$ one gets length 6.28 by numerical integration
which is close enough. \vs

The area computation goes the same way as for positive curvature,
since the quantity $1+f_x^2 + f_y^2$ is the same for both functions.
The area enclosed by the circle of circumference 6.28 is calculated
as before but using .715 instead of 1. It comes out to be
$\frac{\pi}{6}((1+4\cdot .715^2)^{\frac{3}{2}}-1)=2.258...$.







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