Sphere cutout

Sphere sector (blue)

A spherical sector or spherical sector denoted in the mathematics a conical segment from the center of a ball to its surface . The hemisphere is a special case .

Formulas

The following formulas apply to the calculation of volume , lateral area and surface of a spherical section . It denotes the radius of the ball , the radius of the base circle of the ball segment and the height of the ball segment. ${\displaystyle r}$ ${\displaystyle a}$ ${\displaystyle h}$

These three quantities are not independent of one another. The segment of the sphere is determined by any two of these three sizes. The third can be calculated from two of the three quantities. In all formulas - should be taken for ± if the segment of the sphere is less than half the size of the sphere, otherwise + for ±.

${\displaystyle (r-h)^{2}+a^{2}=r^{2}}$

${\displaystyle 2\cdot r\cdot h=a^{2}+h^{2}}$

${\displaystyle h=r\pm {\sqrt {r^{2}-a^{2}}}}$

${\displaystyle h^{2}=2\cdot r\cdot (r\pm {\sqrt {r^{2}-a^{2}}})-a^{2}}$

Instead of and, it is sufficient to specify the angle of the base circle (see illustration). The following applies: ${\displaystyle a}$ ${\displaystyle h}$ ${\displaystyle \theta _{0}}$

${\displaystyle a=r\cdot \sin(\theta _{0})}$

${\displaystyle h=r\cdot (1-\cos(\theta _{0}))}$

There are therefore several formulas, depending on which of the quantities is given.

Sizes of a sphere section with the radius r of the sphere, the radius a of the base circle and the height h
volume	${\displaystyle V={\frac {2\cdot \pi }{3}}\cdot r^{2}\cdot h}$
	${\displaystyle V={\frac {\pi }{6}}\cdot h\cdot (3\cdot a^{2}+h^{2})}$
	${\displaystyle V={\frac {2\cdot \pi }{3}}\cdot r^{2}\cdot (r\pm {\sqrt {r^{2}-a^{2}}})}$
	${\displaystyle V={\frac {2\cdot \pi }{3}}\cdot r^{3}\cdot (1-\cos(\theta _{0}))}$
Area of the lateral surface of the cone	${\displaystyle M_{K}=\pi \cdot r\cdot {\sqrt {(2\cdot r-h)\cdot h}}}$
	${\displaystyle M_{K}=\pi \cdot {\frac {a\cdot (a^{2}+h^{2})}{2\cdot h}}}$
	${\displaystyle M_{K}=\pi \cdot a\cdot r}$
	${\displaystyle M_{K}=\pi \cdot r^{2}\cdot \sin(\theta _{0})}$
Area of the lateral surface of the spherical segment	${\displaystyle M_{S}=2\cdot \pi \cdot r\cdot h}$
	${\displaystyle M_{S}=\pi \cdot (a^{2}+h^{2})}$
	${\displaystyle M_{S}=2\cdot \pi \cdot r\cdot (r\pm {\sqrt {r^{2}-a^{2}}})}$
	${\displaystyle M_{S}=2\cdot \pi \cdot r^{2}\cdot (1-\cos(\theta _{0}))}$
Surface area	${\displaystyle O=\pi \cdot r\cdot (a+2\cdot h)}$
	${\displaystyle O=\pi \cdot {\frac {(a+2\cdot h)\cdot (a^{2}+h^{2})}{2\cdot h}}}$
	${\displaystyle O=\pi \cdot r\cdot (a+2\cdot (r\pm {\sqrt {r^{2}-a^{2}}}))}$
	${\displaystyle O=\pi \cdot r^{2}\cdot (2-2\cdot \cos(\theta _{0})+\sin(\theta _{0}))}$

special cases

For is and the sphere section is a hemisphere : ${\displaystyle h=r}$ ${\displaystyle a=r}$ ${\displaystyle V={\tfrac {2\cdot \pi }{3}}\cdot r^{3},\ M=2\cdot \pi \cdot r^{2},\ O=3\cdot \pi \cdot r^{2}.}$

For is and the sphere section is a whole sphere : ${\displaystyle h=2\cdot r}$ ${\displaystyle a=0}$ ${\displaystyle V={\tfrac {4\cdot \pi }{3}}\cdot r^{3},\ M=O=4\cdot \pi \cdot r^{2}.}$

Derivation

To derive these formulas , one subdivides them into two fields: cone and spherical segment . The cone has the base circle radius and the height . ${\displaystyle a}$ ${\displaystyle r-h}$

The volume of the cone is

${\displaystyle V_{K}={\frac {\pi }{3}}\cdot a^{2}\cdot (r-h)}$

The spherical segment has the volume

${\displaystyle V_{S}={\frac {\pi }{3}}\cdot h^{2}\cdot (3\cdot r-h)}$

So is the volume of the sphere sector

${\displaystyle V=V_{K}+V_{S}={\frac {\pi }{3}}\cdot a^{2}\cdot (r-h)+{\frac {\pi }{3}}\cdot h^{2}\cdot (3\cdot r-h)}$

From the Pythagorean theorem follows . Finally, inserting and releasing the brackets delivers ${\displaystyle a^{2}=2\cdot h\cdot r-h^{2}}$

${\displaystyle V={\frac {2\cdot \pi }{3}}\cdot r^{2}\cdot h}$

Another possibility to calculate the volume is provided by spherical coordinates :

${\displaystyle V=\int _{0}^{\theta _{0}}\int _{0}^{2\cdot \pi }\int _{0}^{r}\rho ^{2}\cdot \sin(\theta )\,\mathrm {d} \rho \,\mathrm {d} \phi \,\mathrm {d} \theta ={\frac {2\cdot \pi }{3}}\cdot r^{3}\cdot \int _{0}^{\theta _{0}}\sin(\theta )\,\mathrm {d} \theta ={\frac {2\cdot \pi }{3}}\cdot r^{3}\cdot (1-\cos(\theta _{0}))}$

where is half the opening angle of the cone part. With follows the above formula for the volume . ${\displaystyle \theta _{0}}$ ${\displaystyle h=r(1-\cos \theta _{0})}$

The outer surface of the cone is

${\displaystyle M_{K}=\pi \cdot a\cdot r}$

and the surface of the spherical segment (without a base circle)

${\displaystyle M_{S}=2\cdot \pi \cdot r\cdot h}$.

That’s the surface

${\displaystyle O=M_{K}+M_{S}=\pi \cdot r\cdot (a+2\cdot h)}$

See also

• Spherical segment
• Spherical layer
• Ball ring
• Ball wedge
• District sector

Web links

• Eric W. Weisstein : Spherical sector . In: MathWorld (English).
• Eric W. Weisstein : Spherical cone . In: MathWorld (English).

literature

• Bronstein-Semendjajew: Paperback of mathematics. Harri-Deutsch-Verlag, 1983, ISBN 3-87144-492-8 , p. 252.
• Small Encyclopedia Mathematics , Harri Deutsch-Verlag, 1977, p. 215.