Sweep of solid angle along a line I have a planar surface(triangle) $S$ that lies on the $XY$ plane at $z=z_{0}$ and subtends a solid angle $\Omega$ at the origin. Assuming the surface to move in a straight line say $z$ axis, I have to compute the integral of solid angle as the surface moves from $z_{0}$ to $z_{1}$. In essence I would like to compute
$\int_{z_{0}}^{z_{1}}{\int_{S}}$ $\frac{cos(\theta)}{r^{2}}dSdz$
Is there anyway to compute the total solid angle subtended as the surface is swept?
 A: I will not do the detailed calculation here, but give its principles.
A solid angle, by definition, is the area on the unit sphere being cut by the cone of vision issued from the eye of the observer to the object. Here triangle $A'B'C'$ will generate, on the unit sphere centered on the eye, a spherical triangle $ABC$ with angles $\alpha, \beta, \gamma$. 
The area of spherical triangle $ABC$ can be shown to be equal to the "spherical excess" $\pi - (\alpha+\beta+\gamma)$. 
See e.g. http://mathworld.wolfram.com/SphericalTriangle.html
So we are back at integrating 3 apparent angles. 
The best is probably to consider that the observer and his accompanying sphere are fixed, and that it's the triangle that moves. 
Let us indicate how one would proceed with the integration at angle $\alpha$ (at vertex $A$).
Let $\vec{u}=\vec{AB}$ and $\vec{v}=\vec{AC}$. One has first to compute the apparent angle which is obtained by orthogonaly projecting $\vec{u}$ and $\vec{v}$ onto the tangent plane to the sphere at point $\vec{w}=\frac{\vec{OA}}{\|\vec{OA}\|}$:
Let us "technicaly" consider $\vec{w}$ as a column vector $w$. The orthogonal projection matrix is known to be $P=I_3-ww^T$. Then we project $\vec{u}$
and $\vec{v}$ using $P$ into $\vec{u_1}$
and $\vec{v_1}$. All we have to do now is to integrate between $z_0$ and $z_1$ the expression: $\arccos \dfrac{\vec{u_1}.\vec{v_1}}{\|\vec{u_1}\|\|\vec{v_1}\|}$ 
(using the dot product formula $\vec{u_1}.\vec{v_1}=\|\vec{u_1}\|\|\vec{v_1}\| \cos\alpha$).
