How do we go from integrating over the entire hemisphere H^2 to just integrating over the solid angle Omega? Are those two bounds equivalent?
henryzxu
I believe so. If you notice, we're taking the integral with respect to little omega, which are very small solid angles themselves. You can think about solid angles as sections of area on the surface of the sphere (it's kind of like the 3D generalization of 2D angles). I found the explanations here and here quite helpful.
jeromylui
Is there any intuition the equation on the bottom right, that the derivative of the projected omega is equal to |cos(theta)|dw? I'm having trouble understanding how weighing dw with cos(theta) achieves this.
henryzxu
The cos(theta) can be thought of as our way of computing the component of d-omega that lies on the xy-plane. I found this stackexchange post helpful for visualizing what d-omega and its projection look like.
How do we go from integrating over the entire hemisphere H^2 to just integrating over the solid angle Omega? Are those two bounds equivalent?
I believe so. If you notice, we're taking the integral with respect to little omega, which are very small solid angles themselves. You can think about solid angles as sections of area on the surface of the sphere (it's kind of like the 3D generalization of 2D angles). I found the explanations here and here quite helpful.
Is there any intuition the equation on the bottom right, that the derivative of the projected omega is equal to |cos(theta)|dw? I'm having trouble understanding how weighing dw with cos(theta) achieves this.
The cos(theta) can be thought of as our way of computing the component of d-omega that lies on the xy-plane. I found this stackexchange post helpful for visualizing what d-omega and its projection look like.