Skip to main content
8 events
when toggle format what by license comment
Apr 13, 2020 at 15:15 comment added boyfarrell Yes! I that’s correct. A solid angle defines a “volume” of directions from a starting point. The first integral in actually over $\theta$, it sweeps out a “standard” angle (it looks like a “plane” of directions from $\theta_c$ to $\pi/2$). The second integral, over $\phi$, rotates the “plane” to define a “volume” of angles: a solid angle. Yes a solid angle is defined by a two angular ranges: one over $\theta$ the other over $\phi$.
Apr 13, 2020 at 15:03 comment added Mathrix Thank you! I don't know if I was clear so I am asking again: By making the second integral's boundaries $\frac{\pi}{2}$ to $\theta_c$ we kind of calculated the area in two dimensions. And by making the first integral's boundaries $2\pi$ to $0$, we integrated the value (I don't really know what it is :/) over the second angle of the solid angle. I am sorry this is not my first language but what I want to say at short is that solid angle has two angles and that's why we integrate it twice. Is that it?
Apr 13, 2020 at 14:54 comment added boyfarrell These are not length integrals they are direction (solid angle) integrals. Imagine the integral is sweeping out a “volume”, if you use the limits above it sweeps out the yellow section in the diagram. This is what I was trying to show with that.
Apr 13, 2020 at 14:50 comment added Mathrix I think I understand why you put these limits but could you please explain why did you choose these ones? Especially for the first integral. Is it because it's a hemisphere with a radius of $\frac{thickness\space of\space the\space liquid\space layer}{2}\space \space \space$?
Apr 10, 2020 at 8:16 history edited boyfarrell CC BY-SA 4.0
deleted 4 characters in body
Apr 9, 2020 at 22:52 vote accept Mathrix
Apr 9, 2020 at 17:17 history edited boyfarrell CC BY-SA 4.0
added 1 character in body; edited body
Apr 9, 2020 at 12:25 history answered boyfarrell CC BY-SA 4.0