integral of the sphere describing lambertian reflectance

Question

A Lambertian surface reflects or emits radiation proportional to the cosine of the angle subtended between the exiting angle and the normal to that surface. The integral of surface of the hemisphere which describes the exiting radiance is supposed to be equal to π. Is there a way I can prove that the surface of the lambertian hemisphere is equal to π? The following is what I have tried. I assume that this hemispherical function can be described as $$ x = \sin(\arccos(y)) = \sqrt{1-y^2} $$ where I would attempt to use integration of a rotational surface to calculate the surface. The normal to the lambertian surface is defined by the y-axis. I want to rotate this surface about the y-axis. I thus use the following equation to calculate the rotational integration $$ A = \int_0^1{2πx\sqrt{1+\left(\frac{dx}{dy}\right)^2}}dy $$ I have used matlab to solve the integral symbolically and used numerical integration to try and get to the value of π, but it doesn't work. I think I am not starting with the right function to describe the surface. The following link may explain this better than above. http://fp.optics.arizona.edu/Palmer/rpfaq/rpfaq.htm#lambertian

Answer 1

Based on the p.571 and 566 in Stroud's 'Engineering Mathematics' I am setting out the answer below. The surface of revolution based on the parametric equation where in our case the rotation is around the y-axis the equation is as follows: $$ A = \int_0^{\pi/2} 2\pi x \sqrt{(\frac{dx}{d\theta})^2 + (\frac{dy}{d\theta})^2}d\theta $$ this is only for a hemisphere with a radius of unity. We require to multiply the integrand by the cosine of $\theta$ as rightly pointed out by Ron above, and this is positive for $\theta$ between 0 and $\pi/2$. As: $ y = cos\theta $ and $x = sin\theta$ we have: $$ A = \int_0^{\pi/2} cos\theta\space2\pi\space sin\theta \sqrt{sin^2\theta + cos^2\theta}\space d\theta $$ This reduces to: $$ A = \space \Biggr|_0^{pi/2} \pi \space sin^2\theta = \pi $$

Answer 2

Note that intensity is proportional to $|\cos{\theta}|$, as this is a positive quantity. The integrated quantity you seek (total intensity divided by radiance) is then

$$\pi \int_0^{\pi} d\theta \, \sin{\theta} |\cos{\theta}| = 2 \pi \int_0^{\pi/2} d\theta \, \sin{\theta} \cos{\theta} = \pi$$

Answer 3

Given your question, I think you need to evaluate

$$ \int_0^{\pi/2} d\theta \int_0^{2\pi} d\phi \sin(\theta) \cos(\theta) $$

$$= \pi \int_0^{\pi/2} \sin(2\theta) d\theta = \pi$$