A conical surface may contain a certain portion $\Omega$ of the space $\mathbb{R}^3$ . This portion is called solid angle or space angle. If the conical surface contains a portion $A$ of a spherical surface with radius $R$ and with centre $P$ in the vertex of the solid angle, then the magnitude of the solid angle is given by $$\Omega = \frac{A}{R^2}$$ which is independent on the radius $R$ .The spherical surface can be replaced by any surface $a$ , through which all the half-lines originating from $P$ and being contained in the solid angle go. Then the solid angle may be computed from the surface integral

(1)

The SI unit of solid angle, analogous to the angle unit radian, is the steradian ($= 1\;\mbox{sr}$ ). The steradian takes a proportion $\frac{1}{4\pi}$ , or approximately 7.957747 %, of the surface area of a sphere.

If the solid angle is bounded by three planes having exactly one common point, it may be called a trihedral angle; cf. the example 2!

Example 1. The solid angle determined by a right circular cone with the angle $\alpha$ between its axis and side line is equal to $2\pi(1\!-\cos\alpha)$ , i.e. $\displaystyle 4\pi\sin^2{\frac{\alpha}{2}}$ .

Example 2. Let $\vec{r_1},\,\vec{r_2},\,\vec{r_3}$ be the position vectors of three points in $\mathbb{R}^3$ and $r_1,\,r_2,\,r_3$ their lengths. Then the solid angle $\Omega$ of the tetrahedron spanned by the vectors $\vec{r_i}$ is obtained from the equation

(2)

Example 3. Using (2), one can easily get the apical solid angle of a right pyramid with square base: $$\Omega = 4\arctan{\frac{a^2}{2h\sqrt{2a^2+4h^2}}} = 4\arcsin\frac{a^2}{a^2+4h^2}$$ Here $a$ is the side of the base square and $h$ is the height of the pyramid.