Spherical Segment

A spherical segment is the solid defined by cutting a Sphere with a pair of Parallel Planes. It can be thought of as a Spherical Cap with the top truncated, and so it corresponds to a Spherical Frustum. The surface of the spherical segment (excluding the bases) is called a Zone.

Call the Radius of the Sphere $R$ and the height of the segment (the distance from the plane to the top of Sphere) $h$. Let the Radii of the lower and upper bases be denoted $a$ and $b$, respectively. Call the distance from the center to the start of the segment $d$, and the height from the bottom to the top of the segment $h$. Call the Radius parallel to the segment $r$, and the height above the center $y$. Then $r^2=R^2-y^2$,

$$V = \int_d^{d+h} \pi r^2\,dy = \pi \int_d^{d+h} (R^2-y^2)\,dy$$

$$= \pi\left[{R^2y-{\textstyle{1\over 3}} y^3}\right]_{d}^{d+h}= \pi \{R^2h-{\textstyle{1\over 3}}[(d+h)^3-d^3]\}$$

$$= \pi [R^2h-{\textstyle{1\over 3}}(d^3+3d^2h+3h^2d+h^3-d^3)]$$

$$= \pi(R^2h-d^2h-h^2d-{\textstyle{1\over 3}}h^3)$$

$$= \pi h(R^2-d^2-hd-{\textstyle{1\over 3}} h^2).$$ (1)

Using

$$a^2 = R^2-d^2$$ (2)

$$b^2 = R^2-(d+h)^2 = R^2-d^2-2dh-h^2,$$ (3)

gives

$$a^2+b^2 = 2R^2-2d^2-2dh-h^2$$ (4)

$$R^2-d^2-dh = {\textstyle{1\over 2}}(a^2+b^2+h^2),$$ (5)

so

$$V = \pi h[{\textstyle{1\over 2}}(a^2+b^2+h^2)-{\textstyle{1\over 3}} ... ...{\textstyle{1\over 2}}a^2+{\textstyle{1\over 2}}b^2+{\textstyle{1\over 6}} h^2)$$

$$= {\textstyle{1\over 6}}\pi h(3a^2+3b^2+h^2).$$ (6)

The surface area of the Zone (which excludes the top and bottom bases) is given by

$$S=2\pi Rh.$$ (7)

See also Archimedes' Problem, Frustum, Hemisphere, Sphere, Spherical Cap, Spherical Sector, Surface of Revolution, Zone

References

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