The Horn Torus

The Horn Torus is a solid that if formed by the revolution of a circle around a point on its circumference. Suppose the starting circle has a diameter of 1 unit and thus a circumference of \(\pi\) units. Consider a small rotation of length \( \delta x\) along the outer circumference that produces a wedge-shaped solid that is equivalent to a cylinder of curved surface area \(\pi  \delta x/2\). The total surface area of the resultant torus is given by:$$\begin{align} \text{Surface Area }&=\lim_{\delta x \rightarrow 0} \sum_0^{2\pi} \pi \, \delta x/2 \\ &=\int_0^{2\pi} \! \! \pi/2 \, dx\\&=\bigg [ \pi \, x/2 \bigg ]_0^{ 2\pi}\\&=\pi^2 \end{align}$$This is beautifully simple. If we envisage \(\pi\) as the area of a circle with unit radius then \( \pi^2 \) can be envisaged as the surface area of a horn torus with unit tube diameter. 

While we're here, we may as well calculate the volume of a horn torus with a unit tube diameter. The calculation of volume is very similar to that of the surface area. Consider a small rotation of length \( \delta x\) along the outer circumference that produces a wedge-shaped solid that is equivalent to a cylinder of cross-sectional area \( \pi/4\) and thickness \( \delta x/2\). The volume of this wedge shape is \(\pi \, \delta x/8\). The total volume of the resultant torus is given by:$$\begin{align} \text{Volume }&=\lim_{\delta x \rightarrow 0} \sum_0^{2\pi} \pi \, \delta x/8 \\ &=\int_0^{2\pi} \!  \! \pi/8 \, dx\\&=\bigg [ \pi \, x /8 \bigg ]_0^{ 2\pi}\\&=\pi^2/4\end{align}$$