Light Transport in Tissue

Integrals over entire spheres

There are several common integrals over solid angles. The simplest is the integral of a constant over all $4\pi$ steradians

$\begin{displaymath} \int_{4\pi}\,d\omega=\int_{-\pi}^\pi\,d\varphi\int_0^\pi\sin... ...theta =2\pi\int_{-1}^1d(\cos\theta)=2\pi\int_{-1}^1\,d\mu=4\pi \end{displaymath}$

(C1)

Here the $4\pi$ beneath the integral is used to indicate that the integral is done over all $4\pi$ steradians. The angles $\varphi$ and $\theta$ refer to the usual azimuthal and longitudinal angles in a spherical geometry. The differential $d\omega=\sin\theta\,d\theta$ is a differential solid angle with $\hat\mathbf{s}$ as an outward normal unit vector. The substitution $\mu=\cos\theta$ has been made.

The integral of $\hat\mathbf{s}$ over all angles is zero, by symmetry

$\begin{displaymath} \int_{4\pi} \hat\mathbf{s}\,d\omega=0 \end{displaymath}$

(C2)

The integral is zero because for each $\hat\mathbf{s}$ in the upper hemisphere there is a unit vector $\hat\mathbf{s}'$ in the lower hemisphere pointed in the opposite direction (Figure C.1a). When $\hat\mathbf{s}$ is integrated over all angles then the contribution from the upper hemisphere is exactly cancelled by that from the lower hemisphere.

The next integral is slightly more complicated

$\begin{displaymath} \int_{4\pi} \hat\mathbf{s}(\hat\mathbf{s}\cdot\mathbf{A})\,d\omega={4\pi\over3}\mathbf{A} \end{displaymath}$

(C3)

The vertical axis is chosen parallel to the arbitrary vector A (Figure C.1b). The vector $\hat\mathbf{s}(\hat\mathbf{s}\cdot\mathbf{A})$ has a magnitude of $\vert\mathbf{A}\vert\cos\theta$ . Now choose a vector $\hat\mathbf{s}'$ , such that $\hat\mathbf{s}'\cdot\mathbf{A}=\hat\mathbf{s}\cdot\mathbf{A}$ and such that $\hat\mathbf{s}'$ , $\hat\mathbf{s}$ and A are co-planar. From Figure C.1b it is evident that adding $\hat\mathbf{s}'(\hat\mathbf{s}'\cdot\mathbf{A})$ to $\hat\mathbf{s}(\hat\mathbf{s}\cdot\mathbf{A})$ results in a vector in the A direction. The magnitude of the vector sum is the projection of each of these vectors onto A. Since the magnitude of $\hat\mathbf{s}'(\hat\mathbf{s}'\cdot\mathbf{A})$ is equal to that of $\hat\mathbf{s}(\hat\mathbf{s}\cdot\mathbf{A})$ , we have

$\begin{displaymath} \hat\mathbf{s}(\hat\mathbf{s}\cdot\mathbf{A})+\hat\mathbf{s}... ...{\mathbf{A}\over\vert\mathbf{A}\vert}=2\cos^2\theta \mathbf{A} \end{displaymath}$

(C4)

The integral (C.3) is then

$\begin{displaymath} \int_{4\pi} \hat\mathbf{s}(\hat\mathbf{s}\cdot\mathbf{A})\,d... ...=2\pi\mathbf{A}\int_{-1}^1 \mu^2\,d\mu ={4\pi\over3}\mathbf{A} \end{displaymath}$

(C5)

The azimuthal integral in $\varphi$ is done only from 0 to $\pi$ to account for adding $\hat\mathbf{s}'$ and $\hat\mathbf{s}$ in Equation (C.4).

**Figure C.1:** Geometry used to evaluate integrals (C.2) and (C.3).
$\includegraphics [scale=0.902]{figA31.eps}$

The next integral is

$\begin{displaymath} \int_{4\pi} \hat\mathbf{s}(\hat\mathbf{s}\cdot\mathbf{A})(\hat\mathbf{s}\cdot\mathbf{B})\,d\omega=0 \end{displaymath}$

(C6)

This integral is evaluated by decomposing B into components parallel and perpendicular to A

$\begin{displaymath} \mathbf{B}=\mathbf{B}_\bot +\mathbf{B}_\Vert \end{displaymath}$

(C7)

The integral in Equation (C.6) becomes

$\begin{displaymath} \int_{4\pi} \hat\mathbf{s}(\hat\mathbf{s}\cdot\mathbf{A})(\h... ...ot\mathbf{A})(\hat\mathbf{s}\cdot\mathbf{B}_\bot)\,d\omega = 0 \end{displaymath}$

(C8)

The first integral is very similar to Equation (C.3). The difference is an additional factor of $\mathbf{B}\cos\theta$ , i.e.,

$\displaystyle \int_{4\pi} \hat\mathbf{s}(\hat\mathbf{s}\cdot\mathbf{A})(\hat\mathbf{s}\cdot\mathbf{B}_\Vert)\,d\omega$	=	$\displaystyle \int_0^\pi\,d\varphi\int_0^\pi 2\mathbf{A}\cos^2\theta(\vert\mathbf{B}_\Vert \vert\cos\theta) \sin\theta\,d\theta$
	=	$\displaystyle 2\pi\mathbf{A}\vert\mathbf{B}_\Vert \vert \int_{-1}^1 \mu^3\,d\mu=0$	(C9)

The second integral is evaluated by referring to Figure C.2a. If $\hat\mathbf{s}'$ is chosen as in Figure C.2a, then

$\displaystyle \hat\mathbf{s}\cdot\mathbf{A} =\vert\mathbf{A} \vert\cos\theta \qquad\qquad \hat\mathbf{s}\cdot\mathbf{B}_\bot=\vert\mathbf{B}_\bot \vert\sin\theta$
$\displaystyle \hat\mathbf{s}'\cdot\mathbf{A}=-\vert\mathbf{A} \vert\cos\theta \... ...qquad \hat\mathbf{s}'\cdot\mathbf{B}_\bot=-\vert\mathbf{B}_\bot \vert\sin\theta$

It is clear that $\hat\mathbf{s}(\hat\mathbf{s}\cdot\mathbf{A})(\hat\mathbf{s}\cdot\mathbf{B}_\bo... ...thbf{s}'(\hat\mathbf{s}'\cdot\mathbf{A})(\hat\mathbf{s}'\cdot\mathbf{B}_\bot)=0$ since both vectors have the same magnitude and are pointed in opposite directions.

**Figure C.2:** Figures for integrals (C.6) and (C.12).
$\includegraphics [scale=0.902]{figA32.eps}$

Using the previously computed vector integrals two common scalar integrals may be found. For example, using Equation (C.2)

$\begin{displaymath} \int_{4\pi} (\hat\mathbf{s}\cdot\mathbf{A})\,d\omega=\mathbf{A}\cdot\int_{4\pi} \hat\mathbf{s}\,d\omega \end{displaymath}$

(C10)

and using Equation (C.3)

$\begin{displaymath} \int_{4\pi} (\hat\mathbf{s}\cdot\mathbf{A})(\hat\mathbf{s}\c... ...ot\mathbf{A})\,d\omega={4\pi\over3}(\mathbf{A}\cdot\mathbf{B}) \end{displaymath}$

(C11)

S. A. Prahl."Light Transport in Tissue," PhD thesis, University of Texas at Austin, 1988.