Why is there an angle of $\frac{\pi}{6}$, and not $\frac{\pi}{2}$, in this problem?

Question

I struggle with the following explanation, and would be grateful if you could help. Thank You.

"Consider a small hole with area $a$ on the surface of the metal cube through which radiation can escape. We want to calculate the radiated power though this hole. For radiation at an angle $θ$ to the normal of the black body surface, the effective area of that hole will be $a \cdot \cos(θ)$. Integrating over all angles in a hemisphere, we find the integrated effective area $a_{eff} = \frac{a}{2}$"

I don't understand why we get that result. If we integrate $a \cdot \cos(θ)$, we get $a \cdot \sin(θ)$, and so to get $a_{eff} = \frac{a}{2}$, we need to integrate from $0$ to $\frac{\pi}{6}$. And my question is: if we integrate over all angles in a hemisphere, why do we integrate to $\frac{\pi}{6}$ and not $\frac{\pi}{2}$ ?

Answer 1

The problem is that you are thinking in a 2D world, when the problem is 3D. When you integrate a function $f$ over all angles on a hemisphere you don't do $$\int_0^{\pi/2} f(\theta)d\theta$$ You instead need to do $$\int_0^{2\pi} d\varphi\int_0^{\pi/2}f(\theta,\varphi)\sin\theta d\theta$$ In your case $f(\theta,\varphi)=a\cos\theta$, which does not depend on $\varphi$, so you have the integral over $\varphi$ equal to $2\pi$. Then $$\int_0^{\pi/2}\sin\theta\cos\theta d\theta=\frac 12$$ So the effective area is $$a_{eff}=\frac{a2\pi\frac 12}{(4\pi)/2}=\frac a2$$ The denominator is the solid angle of a hemisphere (half of $4\pi$)

Answer 2

The angle isn't the issue; you need to multiply by $\sin\theta$ beore you integrate, viz. $d^3x=r^2dr\sin\theta d\theta d\phi$. Note that$$\frac{\int_0^{\pi/2}\sin\theta\cos\theta d\theta}{\int_0^{\pi/2}\sin\theta d\theta}=\frac{\int_0^{\pi/2}\frac12\sin2\theta d\theta}{\int_0^{\pi/2}\sin\theta d\theta}=\frac{\left[-\frac14\cos 2\theta\right]_0^{\pi/2}}{[-\cos\theta]_0^{\pi/2}}=\frac{\frac14(1-(-1))}{1-0}=\frac12.$$