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The Surface Area of a Spherical Segment. Call the Radius of the Sphere $R$, the upper and lower Radii $b$ and $a$, respectively, and the height of the Spherical Segment $h$. The zone is a Surface of Revolution about the z-Axis, so the Surface Area is given by

S=2\pi\int x\sqrt{1+x'^2}\,dz.
\end{displaymath} (1)

In the $xz$-plane, the equation of the zone is simply that of a Circle,
\end{displaymath} (2)

$\displaystyle x'$ $\textstyle =$ $\displaystyle -z(R^2-z^2)^{-1/2}$ (3)
$\displaystyle x'^2$ $\textstyle =$ $\displaystyle {z^2\over R^2-z^2},$ (4)

$\displaystyle S$ $\textstyle =$ $\displaystyle 2\pi\int^{\sqrt{R^2-b^2}}_{\sqrt{R^2-a^2}} \sqrt{R^2-z^2}\sqrt{1+{z^2\over R^2-z^2}}\,dz$  
  $\textstyle =$ $\displaystyle 2\pi R\int^{\sqrt{R^2-b^2}}_{\sqrt{R^2-a^2}} dz = 2\pi R(\sqrt{R^2-b^2}-\sqrt{R^2-a^2}\,)$  
  $\textstyle =$ $\displaystyle 2\pi Rh.$ (5)

This result is somewhat surprising since it depends only on the height of the zone, not its vertical position with respect to the Sphere.

See also Sphere, Spherical Cap, Spherical Segment, Zonohedron


Beyer, W. H. (Ed.). CRC Standard Mathematical Tables, 28th ed. Boca Raton, FL: CRC Press, p. 130, 1987.

© 1996-9 Eric W. Weisstein