Solid angle (and integral of a sine function)

AI Thread Summary
The discussion revolves around the derivation of the solid angle for a right rectangular pyramid, specifically focusing on the integral of the sine function. The user is confused about how the integral from θ- to θ+ transforms into 2 cos|θ±|, questioning their own calculation that suggests a different result. A key point is that the range of θ is symmetric about θ = π/2, not zero, which clarifies the transformation of the integral. The conversation also highlights the distinction between odd and even functions, correcting a misunderstanding about the symmetry of the sine function. This clarification is crucial for understanding the solid angle derivation accurately.
mnb96
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Hello,

I was following the derivation of the solid angle of right rectangular pyramid that I found at http://www.slac.stanford.edu/~bgerke/notes/solid_angle.pdf" .

I don't quite understand the step between the 3rd to the 4th equation. In particular how the integral

\int_{\theta_-}^{\theta_+}\sin(\theta) d\theta

becomes,

2 \cos|\theta_{\pm}|

Where \theta_{\pm} = \cot^{-1} (\tan\left( \pm \alpha/2)cos\phi \right)

According to my calculation it should be:
2 \left( 1 - \cos(\theta_+) \right)

Where is my mistake?
 
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I am not sure what they are doing. However the integral of sinθ is -cosθ, so the definite integral is
cosθ- - cosθ+.
 
Since Cos[t] = Cos[-t], the integral should be identically 0. If you think about it, Sin is symmetric about the origin: hence any definite integral centered on 0 (i.e. - {-t -> t}) MUST 'sum' to 0.
 
earnric said:
Since Cos[t] = Cos[-t], the integral should be identically 0. If you think about it, Sin is symmetric about the origin: hence any definite integral centered on 0 (i.e. - {-t -> t}) MUST 'sum' to 0.

Your point is valid for the sine, except that it is an odd function, not an even (symmetric) function.
 
mathman said:
Your point is valid for the sine, except that it is an odd function, not an even (symmetric) function.

Oops! My bad: symmetric is the wrong word. As you realized, Sin reflects THRU the origin -- about both x and y... not just the y-axis as I implied.

Sorry!
 
mnb96 said:
Hello,

I was following the derivation of the solid angle of right rectangular pyramid that I found at http://www.slac.stanford.edu/~bgerke/notes/solid_angle.pdf.

I don't quite understand the step between the 3rd to the 4th equation. In particular how the integral

\int_{\theta_-}^{\theta_+}\sin(\theta) d\theta

becomes,

2 \cos|\theta_{\pm}|

Where \theta_{\pm} = \cot^{-1} (\tan\left( \pm \alpha/2)cos\phi \right)

According to my calculation it should be:
2 \left( 1 - \cos(\theta_+) \right)

Where is my mistake?
The range of θ is symmetric about θ = π/2, not about θ = 0. So θ- = π - θ+.
 
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