How to Integrate Over a Solid Angle

A solid angle measures the size of a region on the unit sphere, just as an ordinary angle measures an arc on the unit circle. Its unit is the steradian, and the full sphere has solid angle $4\pi$.

The usual meaning of “integrate over a solid angle” is to integrate a function of direction over a region of the unit sphere. In spherical coordinates, the essential formula is

$$d\Omega = \sin\theta\,d\theta\,d\phi,$$

where $\theta$ is the polar angle measured down from the positive $z$-axis and $\phi$ is the azimuthal angle in the $xy$-plane.

The solid-angle element

A point on the unit sphere has Cartesian coordinates

$$(x,y,z)=(\sin\theta\cos\phi,\,\sin\theta\sin\phi,\,\cos\theta).$$

A small patch bounded by $d\theta$ and $d\phi$ has side lengths approximately $d\theta$ and $\sin\theta\,d\phi$. The second length contains the factor $\sin\theta$ because circles of constant $\theta$ shrink near the poles. Therefore its area, and hence its solid angle, is

$$d\Omega=\sin\theta\,d\theta\,d\phi.$$

For a function $f(\theta,\phi)$, the solid-angle integral over a region $R$ is

$$\int_R f\,d\Omega = \int_{\phi=\phi_1}^{\phi_2} \int_{\theta=\theta_1(\phi)}^{\theta_2(\phi)} f(\theta,\phi)\sin\theta\,d\theta\,d\phi.$$

The limits may be reversed if the region is more naturally described using $\theta$ as the outer variable.

Basic checks

For the entire sphere, use $0\leq\theta\leq\pi$ and $0\leq\phi\leq 2\pi$:

$$\begin{aligned} \int_{S^2}d\Omega &=\int_0^{2\pi}\int_0^\pi \sin\theta\,d\theta\,d\phi\\ &=\left(\int_0^{2\pi}d\phi\right) \left(\int_0^\pi\sin\theta\,d\theta\right)\\ &=2\pi\cdot 2\\ &=4\pi. \end{aligned}$$

This also shows that integrating the constant function $1$ returns the solid angle of the region:

$$\Omega(R)=\int_Rd\Omega.$$

For example, a hemisphere has solid angle $2\pi$.

Worked example: a circular cone

Consider all directions inside a cone with its vertex at the origin and its axis along the positive $z$-axis. Let the cone have half-angle $\alpha$. Its directions satisfy

$$0\leq\theta\leq\alpha, \qquad 0\leq\phi\leq 2\pi.$$

Its solid angle is

$$\begin{aligned} \Omega &=\int_0^{2\pi}\int_0^\alpha \sin\theta\,d\theta\,d\phi\\ &=2\pi[-\cos\theta]_0^\alpha\\ &=2\pi(1-\cos\alpha). \end{aligned}$$

For a $60^\circ$ cone, $\alpha=\pi/3$, so

$$\Omega=2\pi\left(1-\frac12\right)=\pi.$$

Thus the cone covers a quarter of the full sphere, since $\pi/(4\pi)=1/4$.

As a second example, integrate $\cos\theta$ over the same cone:

$$\begin{aligned} \int_R\cos\theta\,d\Omega &=\int_0^{2\pi}\int_0^\alpha \cos\theta\sin\theta\,d\theta\,d\phi\\ &=2\pi\left[\frac12\sin^2\theta\right]_0^\alpha\\ &=\pi\sin^2\alpha. \end{aligned}$$

For $\alpha=\pi/3$, this is $3\pi/4$.

Integrating a direction-dependent function

If a function is given in Cartesian coordinates as $f(x,y,z)$ but is intended to depend only on direction, restrict it to the unit sphere:

$$x=\sin\theta\cos\phi,\qquad y=\sin\theta\sin\phi,\qquad z=\cos\theta.$$

For example, let $f(x,y,z)=z^2$. Over the full sphere,

$$\begin{aligned} \int_{S^2}z^2\,d\Omega &=\int_0^{2\pi}\int_0^\pi \cos^2\theta\sin\theta\,d\theta\,d\phi\\ &=2\pi\left[-\frac13\cos^3\theta\right]_0^\pi\\ &=\frac{4\pi}{3}. \end{aligned}$$

This agrees with symmetry: on the unit sphere, the integrals of $x^2$, $y^2$, and $z^2$ are equal, and $x^2+y^2+z^2=1$, whose integral is $4\pi$.

Solid angle from an ordinary surface

If a surface patch $S$ is viewed from a point, its oriented (signed) differential solid angle is

$$d\Omega_{\text{signed}}=\frac{\mathbf r\cdot\mathbf n}{\lVert\mathbf r\rVert^3}\,dS,$$

where $\mathbf r$ points from the observation point to the surface element, $\mathbf n$ is a chosen unit normal, and $dS$ is the ordinary surface-area element. The dot product accounts for foreshortening. The ordinary nonnegative solid angle subtended by the patch is obtained by taking the absolute value:

$$d\Omega=\frac{|\mathbf r\cdot\mathbf n|}{\lVert\mathbf r\rVert^3}\,dS.$$

For a flat surface with normal making angle $\gamma$ with $\mathbf r$, this becomes, for the ordinary solid angle,

$$d\Omega=\frac{|\cos\gamma|}{r^2}\,dS.$$

For an oriented solid angle, omit the absolute value and retain the sign. The sign depends on the orientation of the normal. If only the size of the visible solid angle is wanted, take the appropriate absolute value. For a surface that is not star-shaped relative to the observation point, split it into suitable patches and handle visibility and orientation carefully.

Common mistakes

• Omitting $\sin\theta$. The expression $d\theta\,d\phi$ is not the solid-angle element. The factor $\sin\theta$ is required.
• Using the wrong polar range. A full sphere uses $0$ to $\pi$ for $\theta$, not $0$ to $2\pi$.
• Confusing $\theta$ and $\phi$. In the convention used here, $\theta$ is measured from the $z$-axis and $\phi$ rotates around that axis. Some fields swap these names, so always check the convention.
• Adding a radial factor. A solid-angle integral is over directions and has no $r^2$ factor. A volume integral uses $r^2\sin\theta\,dr\,d\theta\,d\phi$ instead.
• Forgetting units or normalization. Solid angle is measured in steradians, but radians and steradians are dimensionless in formulas. A normalized average over a region is $\Omega(R)^{-1}\int_R f\,d\Omega$, not simply $\int_R f\,d\Omega$.

For a region described by a boundary curve on the unit sphere, the practical procedure is therefore: convert the boundary to angular limits, insert $\sin\theta$, and verify the result by testing the constant function $1.