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Light Transport in Tissue Integrals over hemispheres The integral of a unit vector $\hat\mathbf{s}$ over a hemisphere with the z-axis as its pole is obtained by summing vectors as in Figure C.2, but omitting the extra factor of $\cos\theta$ $$\int_{2\pi\,\,\mu\ge0} \hat\mathbf{s}\,d\omega=\hat\mathbf{z... ...theta =2\pi\hat\mathbf{z} \int_0^1 \mu\,d\mu=\pi\hat\mathbf{z}$$ (C12) Because the integral over the $\mu\le0$ hemisphere is oriented in the opposite direction from the above integral, care must be taken to ensure that signs remain consistent. In Figure C.3 the vectors in the $\mu\le0$ hemisphere add to a vector in the -z direction. $$\int_{2\pi\,\,\mu\le0} \hat\mathbf{s}\,d\omega=-\hat\mathbf{... ...i(-\hat\mathbf{z}) \int_{-1}^0 (-\mu)\,d\mu=-\pi\hat\mathbf{z}$$ (C13) Clearly, the sum of the integrals over each hemisphere equals zero--the result for the integral over the whole sphere (Equation C.2). The following integrals follow immediately from (C.12) and (C.13) $${1\over4\pi}\int_{2\pi\,\,\mu\ge0} (\hat\mathbf{z}\cdot\hat\mathbf{s})\,d\omega={1\over4}$$ (C14) and $${1\over4\pi}\int_{2\pi\,\,\mu\le0} (-\hat\mathbf{z}\cdot\hat\mathbf{s})\,d\omega={1\over4}$$ (C15) Figure C.3 shows how the vectors may be combined in a similar fashion to evaluate the following integral $$\int_{2\pi\,\,\mu\ge0} \hat\mathbf{s}(\hat\mathbf{z}\cdot\ha... ...\hat\mathbf{z} \int_0^1 \mu^2\,d\mu={2\pi\over3}\hat\mathbf{z}$$ (C16) The integral over the other hemisphere is $$\int_{2\pi\,\,\mu\le0} \hat\mathbf{s}(-\hat\mathbf{z}\cdot\hat\mathbf{s})\,d\omega -\hat\mathbf{z} \int_0^\pi\,d\varphi\int_{\pi/2}^\pi2\cos^2\theta \sin\theta\,d\theta$$ = $$-2\pi\hat\mathbf{z} \int_{-1}^0 \mu^2\,d\mu=-{2\pi\over3}\hat\mathbf{z}$$ = (C17) Figure C.3: Geometry for integral (C.16).

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