Solid angle (and integral of a sine function)

[mnb96]

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

[tex]\int_{\theta_-}^{\theta_+}\sin(\theta) d\theta[/tex]

becomes,

[tex]2 \cos|\theta_{\pm}|[/tex]

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

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

Where is my mistake?

 

[mathman]

I am not sure what they are doing. However the integral of sinθ is -cosθ, so the definite integral is
cosθ_- - cosθ₊.

 

[earnric]

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.

 

[mathman]

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.

 

[earnric]

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!

 

[haruspex]

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

[tex]\int_{\theta_-}^{\theta_+}\sin(\theta) d\theta[/tex]

becomes,

[tex]2 \cos|\theta_{\pm}|[/tex]

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

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

Where is my mistake?

The range of θ is symmetric about θ = π/2, not about θ = 0. So θ_- = π - θ₊.