- #1
a_Vatar
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Hi,
I'm trying to evaluate this integral over a hemisphere:
[tex]\int cos(\theta)^{a*cos^2(\phi) + b*sin^2(\phi)} dw[/tex]
where dw - solid angle measure,[tex]\phi[/tex] is azimuthal angle and [tex]\theta[/tex]
is polar angle.
Thus we have:
[tex]\int cos(\theta)^{a*cos^2(\phi) + b*sin^2(\phi)} dw[/tex] = [tex]\int \int cos(\theta)^{a*cos^2(\phi) + b*sin^2(\phi)} * sin(\theta) d\theta d\phi[/tex].
over hemishere.
Integral = [tex]\int^{2 * pi}_{0} \int^{pi/2}_{0}cos(\theta)^{a*cos^2(\phi) + b*sin^2(\phi)} * sin(\theta) d\theta d\phi[/tex]
evaluate inner integral:
making substitution u = cos([tex]\theta[/tex]),
[tex]\int^{pi/2}_{0}cos(\theta)^{a*cos^2(\phi) + b*sin^2(\phi)} * sin(\theta) d\theta =
- \int^{pi/2}_{0}u^{a*cos^2(\phi) + b*sin^2(\phi)} du = \frac{1}{a*cos^2(\phi) + b*sin^2(\phi) + 1}[/tex]
That's where I get stuck with integration wrt [tex]\phi[/tex]
Apparently this integral has to evaluate to [tex]\frac{1}{\sqrt{(a+1)(b+1)}}[/tex]
I'm trying to evaluate this integral over a hemisphere:
[tex]\int cos(\theta)^{a*cos^2(\phi) + b*sin^2(\phi)} dw[/tex]
where dw - solid angle measure,[tex]\phi[/tex] is azimuthal angle and [tex]\theta[/tex]
is polar angle.
Thus we have:
[tex]\int cos(\theta)^{a*cos^2(\phi) + b*sin^2(\phi)} dw[/tex] = [tex]\int \int cos(\theta)^{a*cos^2(\phi) + b*sin^2(\phi)} * sin(\theta) d\theta d\phi[/tex].
over hemishere.
Integral = [tex]\int^{2 * pi}_{0} \int^{pi/2}_{0}cos(\theta)^{a*cos^2(\phi) + b*sin^2(\phi)} * sin(\theta) d\theta d\phi[/tex]
evaluate inner integral:
making substitution u = cos([tex]\theta[/tex]),
[tex]\int^{pi/2}_{0}cos(\theta)^{a*cos^2(\phi) + b*sin^2(\phi)} * sin(\theta) d\theta =
- \int^{pi/2}_{0}u^{a*cos^2(\phi) + b*sin^2(\phi)} du = \frac{1}{a*cos^2(\phi) + b*sin^2(\phi) + 1}[/tex]
That's where I get stuck with integration wrt [tex]\phi[/tex]
Apparently this integral has to evaluate to [tex]\frac{1}{\sqrt{(a+1)(b+1)}}[/tex]