How to calculate the solid angle of a cone with cone axis arbitrary?

soljiang
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I read that if the cone with apex angle 2α whose central axis is vertical, apex at the origin, then one can use spherical coordinate to calculate the solid angle of the cone

02∏0αsin\varphid\thetad\varphi

However, what if the central axis is align to y-axis horizontally, instead of z-axis.
My question is with angle 2α, if the central axis goes through (θ0, \varphi0), then how to set the intragation boundaries? It should give the same answer but I need to now how to set up the equation. Can one still use spherical coordinate?

Thank you so much if I get some help.


Charles
 
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I would simply choose a different coordinate system, to have the same orientation (but in the new system) again. No need to make it more complicated than necessary.
 
So does it mean spherical coordinate is only useful for calculating shapes/surface that's axially symmetric about z axis but not about x or y axis?
 
If something is symmetric around a different axis, I would choose a different coordinate system - a symmetry around the z-axis (which you can define as you like) is easier to handle in spherical coordinates.
 
Hi, mfb, so it's true that if a shape is only symmetric about x or y axis, it's not easy to integrate the surface area using spherical coordinate, unless one redefine the orientation of coordinate. Is that correct? Thank you.
 
That depends on your definition of easy, but it is certainly easier with a different coordinate system.
 
Hello! There is a simple line in the textbook. If ##S## is a manifold, an injectively immersed submanifold ##M## of ##S## is embedded if and only if ##M## is locally closed in ##S##. Recall the definition. M is locally closed if for each point ##x\in M## there open ##U\subset S## such that ##M\cap U## is closed in ##U##. Embedding to injective immesion is simple. The opposite direction is hard. Suppose I have ##N## as source manifold and ##f:N\rightarrow S## is the injective...

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