Solid angle (and integral of a sine function)

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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+.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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