Skip to Main Content

sphericon 8

 The Differential Geometry of the Sphericon

Area and arclength calculations for:

The Differential Geometry of the Sphericon.



The area of the graph of a differentiable function f(x,y)is given by

\begin{displaymath}\int\int \sqrt{1+f_x^2 + f_y^2}~dx dy\end{displaymath}

where fx and fy are the partial derivatives of f with respect to xand to y.

For the positive curvature example f(x,y)=-(x2+y2),the circle x2+y2=1 fits in thesurface exactly at height -1. The area enclosed is

\begin{displaymath}\int\int_D \sqrt{1+4x^2+ 4y^2}~dx dy\end{displaymath}

where D is the disc of radius 1. Using polar coordinates $x=r \cos\theta, y=r\sin\theta$ the integral becomes

\begin{displaymath}\int_0^{2\pi}\int_0^1 \sqrt{1+4r^2}~ r~ dr d\theta\end{displaymath}

which can be evaluated (use the substitution u=1+4r2) as $\frac{\pi}{6}(5^{\frac{3}{2}}-1)=5.335...$.

The negative example is more complicated because first one mustfind the circle in the (x,y)-plane which gives a circle in thegraph of circumference $2\pi$.I had to do it by trial anderror. The circle of radius r in the (x,y)-plane can beparametrized as $x=r \cos\theta, y=r\sin\theta$,$0\leq\theta\leq 2\pi$.In the graph this circle becomes $( r\cos\theta, r\sin\theta,r^2\cos^2\theta -r^2\sin^2\theta)$,$0\leq\theta\leq 2\pi$.Thelength of a parametrized curve $(x(\theta),y(\theta),z(\theta))$is given by the integral

\begin{displaymath}\int \sqrt{\frac{dx}{d\theta}^2 +\frac{dy}{d\theta}^2 + \frac{dz}{d\theta}^2}~d\theta.\end{displaymath}

In this case the length is

\begin{displaymath}\int_0^{2\pi}\sqrt{r^2 +2r^4\sin^2(2\theta)}~d\theta,\end{displaymath}

using the identity $\cos^2\theta - \sin^2\theta = \cos(2\theta)$beforedifferentiating, and the identity $\cos^2\theta + \sin^2\theta = 1$after. For r=.715 one gets length 6.28 by numerical integrationwhich is close enough.

The area computation goes the same way as for positive curvature,since the quantity 1+fx2 + fy2 is the same for both functions.The area enclosed by the circle of circumference 6.28 is calculatedas before but using .715 instead of 1. It comes out to be$\frac{\pi}{6}((1+4\cdot .715^2)^{\frac{3}{2}}-1)=2.258...$.