Cone, spherical cap, hemisphere The solid angle of a
cone with its apex at the apex of the solid angle, and with
apex angle 2, is the area of a
spherical cap on a
unit sphere \Omega = 2\pi \left (1 - \cos\theta \right)\ = 4\pi \sin^2 \frac{\theta}{2}. For small , such that , this reduces to = . The above is found by computing the following
double integral using the unit
surface element in spherical coordinates: \begin{align} \int_0^{2\pi} \int_0^\theta \sin\theta' \, d \theta' \, d \phi &= \int_0^{2\pi} d \phi\int_0^\theta \sin\theta' \, d \theta' \\ &= 2\pi\int_0^\theta \sin\theta' \, d \theta' \\ &= 2\pi\left[ -\cos\theta' \right]_0^{\theta} \\ &= 2\pi\left(1 - \cos\theta \right). \end{align} This formula can also be derived without the use of
calculus. Over 2200 years ago
Archimedes proved that the surface area of a spherical cap is always equal to the area of a circle whose radius equals the distance from the rim of the spherical cap to the point where the cap's axis of symmetry intersects the cap. In the above coloured diagram this radius is given as 2r \sin \frac{\theta}{2}. In the adjacent black & white diagram this radius is given as "t". Hence for a unit sphere the solid angle of the spherical cap is given as \Omega = 4\pi \sin^2 \frac{\theta}{2} = 2\pi \left (1 - \cos\theta \right). When = , the spherical cap becomes a
hemisphere having a solid angle 2. The solid angle of the complement of the cone is 4\pi - \Omega = 2\pi \left(1 + \cos\theta \right) = 4\pi\cos^2 \frac{\theta}{2}. This is also the solid angle of the part of the
celestial sphere that an astronomical observer positioned at latitude can see as the Earth rotates. At the equator all of the celestial sphere is visible; at either pole, only one half. The solid angle subtended by a segment of a spherical cap cut by a plane at angle from the cone's axis and passing through the cone's apex can be calculated by the formula \Omega = 2 \left[ \arccos \left(\frac{\sin\gamma}{\sin\theta}\right) - \cos\theta \arccos\left(\frac{\tan\gamma}{\tan\theta}\right) \right]. For example, if , then the formula reduces to the spherical cap formula above: the first term becomes , and the second .
Tetrahedron Let OABC be the vertices of a
tetrahedron with an origin at O subtended by the triangular face ABC where \vec a\ ,\, \vec b\ ,\, \vec c are the vector positions of the vertices A, B and C. Define the
vertex angle to be the angle BOC and define , correspondingly. Let \phi_{ab} be the
dihedral angle between the planes that contain the tetrahedral faces OAC and OBC and define \phi_{ac}, \phi_{bc} correspondingly. The solid angle subtended by the triangular surface ABC is given by \Omega = \left(\phi_{ab} + \phi_{bc} + \phi_{ac}\right)\ - \pi. This follows from the theory of
spherical excess and it leads to the fact that there is an analogous theorem to the theorem that
"The sum of internal angles of a planar triangle is equal to ", for the sum of the four internal solid angles of a tetrahedron as follows: \sum_{i=1}^4 \Omega_i = 2 \sum_{i=1}^6 \phi_i\ - 4 \pi, where \phi_i ranges over all six of the dihedral angles between any two planes that contain the tetrahedral faces OAB, OAC, OBC and ABC. A useful formula for calculating the solid angle of the tetrahedron at the origin O that is purely a function of the vertex angles , , is given by
L'Huilier's theorem as \tan \left( \frac{1}{4} \Omega \right) = \sqrt{ \tan \left( \frac{\theta_s}{2}\right) \tan \left( \frac{\theta_s - \theta_a}{2}\right) \tan \left( \frac{\theta_s - \theta_b}{2}\right) \tan \left(\frac{\theta_s - \theta_c}{2}\right)}, where \theta_s = \frac {\theta_a + \theta_b + \theta_c}{2}. Another interesting formula involves expressing the vertices as vectors in 3 dimensional space. Let \vec a\ ,\, \vec b\ ,\, \vec c be the vector positions of the vertices A, B and C, and let , , and be the magnitude of each vector (the origin-point distance). The solid angle subtended by the triangular surface ABC is: \tan \left( \frac{1}{2} \Omega \right) = \frac{\left|\vec a\ \vec b\ \vec c\right|}{abc + \left(\vec a \cdot \vec b\right)c + \left(\vec a \cdot \vec c\right)b + \left(\vec b \cdot \vec c\right)a}, where \left|\vec a\ \vec b\ \vec c\right|=\vec a \cdot (\vec b \times \vec c) denotes the
scalar triple product of the three vectors and \vec a \cdot \vec b denotes the
scalar product. Care must be taken here to avoid negative or incorrect solid angles. One source of potential errors is that the scalar triple product can be negative if , , have the wrong
winding. Computing the absolute value is a sufficient solution since no other portion of the equation depends on the winding. The other pitfall arises when the scalar triple product is positive but the divisor is negative. In this case returns a negative value that must be increased by .
Pyramid The solid angle of a four-sided right rectangular
pyramid with
apex angles and (
dihedral angles measured to the opposite side faces of the pyramid) is \Omega = 4 \arcsin \left( \sin \left({a \over 2}\right) \sin \left({b \over 2}\right) \right). If both the side lengths ( and ) of the base of the pyramid and the distance () from the center of the base rectangle to the apex of the pyramid (the center of the sphere) are known, then the above equation can be manipulated to give \Omega = 4 \arctan \frac {\alpha\beta} {2d\sqrt{4d^2 + \alpha^2 + \beta^2}}. The solid angle of a right -gonal pyramid, where the pyramid base is a regular -sided polygon of circumradius , with a pyramid height is \Omega = 2\pi - 2n \arctan\left(\frac {\tan \left({\pi\over n}\right)}{\sqrt{1 + {r^2 \over h^2}}} \right). The solid angle of an arbitrary pyramid with an -sided base defined by the sequence of unit vectors representing edges {{math|{
s1,
s2}, ...
sn}} can be efficiently computed by: Mathematically, this represents an arc of angle swept around a sphere by radians. When longitude spans 2 radians and latitude spans radians, the solid angle is that of a sphere. A latitude-longitude rectangle should not be confused with the solid angle of a rectangular pyramid. All four sides of a rectangular pyramid intersect the sphere's surface in
great circle arcs. With a latitude-longitude rectangle, only lines of longitude are great circle arcs; lines of latitude are not.
Celestial objects By using the definition of
angular diameter, the formula for the solid angle of a celestial object can be defined in terms of the radius of the object, R, and the distance from the observer to the object, d: \Omega = 2 \pi \left (1 - \frac{\sqrt{d^2 - R^2}}{d} \right ) : d \geq R. By inputting the appropriate average values for the
Sun and the
Moon (in relation to Earth), the average solid angle of the Sun is steradians and the average solid angle of the
Moon is steradians. In terms of the total celestial sphere, the
Sun and the
Moon subtend average
fractional areas of % () and % (), respectively. As these solid angles are about the same size, the Moon can cause both total and annular solar
eclipses depending on the distance between the Earth and the Moon during the eclipse. == Solid angles in arbitrary dimensions ==