e-MATH
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 the surface 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 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

\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 $\cos^2\theta - \sin^2\theta = \cos(2\theta)$ 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.

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 calculated as before but using .715 instead of 1. It comes out to be $\frac{\pi}{6}((1+4\cdot .715^2)^{\frac{3}{2}}-1)=2.258...$.

Back to Sphericon page 2.

© copyright 1999, American Mathematical Society.