# Double Integration Problem for Buffon's Needle experiment

Numberphile has a video about the Buffon's needle experiment (Video). I am writing an essay on determining $\pi$ using probability and I need to show my understanding of the topic. I kind of already understand how we get $\pi$ from the randomness however I need some clarification on how the following was evaluated:

$$\int_{\theta=0}^{{\pi\over2}}\int_{x=0}^{{l\over2}sin\theta}P_x P_{\theta} dx d\theta$$

$P_x$ being ${1\over l}$ where $l$ is the length of the match (this should be irrelevant since it will cross out)

$P_{\theta}$ being ${2 \over \pi}$

The result should be ${1\over\pi}$

I would recommend watching the video, start at 3 min if you only want the math behind it.

I need someone to explain a step by step of how we got to the result, specifically how double integration works in this case. I do understand some calculus so no need to go to extreme details.

If indeed $P_\theta = \pi/2$ here is what I get. $$\begin{split} \int_0^{\pi/2} \int_0^{\frac{L}{2} \sin \theta} \frac{1}{L} \frac{\pi}{2} dx d\theta &= \frac{\pi}{2L} \int_0^{\pi/2} \left[ \int_0^{\frac{L}{2} \sin \theta} dx \right] d\theta \\ &= \frac{\pi}{2L} \int_0^{\pi/2} \frac{L}{2} \sin \theta d\theta \\ &= \frac{\pi}{4} \int_0^{\pi/2} \sin \theta d\theta \\ &= \frac{\pi}{4} \left[ -\cos \theta \right]_0^{\pi/2} \\ &= \pi/4. \end{split}$$ But if $P_\theta = 2/\pi$ (which I think is right), the same technique produces $$\begin{split} \int_0^{\pi/2} \int_0^{\frac{L}{2} \sin \theta} \frac{1}{L} \frac{2}{\pi} dx d\theta &= \frac{2}{L\pi} \int_0^{\pi/2} \left[ \int_0^{\frac{L}{2} \sin \theta} dx \right] d\theta \\ &= \frac{2}{L\pi} \int_0^{\pi/2} \frac{L}{2} \sin \theta d\theta \\ &= \frac{1}{\pi} \int_0^{\pi/2} \sin \theta d\theta \\ &= \frac{1}{\pi} \left[ -\cos \theta \right]_0^{\pi/2} \\ &= 1/\pi. \end{split}$$
• Yes I just noticed that mistake it is definitely $2/\pi$. Give me a couple of minutes to look through the steps and see if I can understand everything. – bukka Dec 17 '13 at 14:13
• Alright I accepted this as an answer, thanks a lot. Just one more question. At the end when you derive $\sin$ into $\left[ -\cos \theta \right]_0^{\pi/2}$ What is that $\cos$ ? I mean why is the $\pi / 2$ and $0$ shifted there? – bukka Dec 17 '13 at 14:24
• Ok I went back into it once again and one thing still bugs me. Why do we get ${{L \over 2}\sin \theta}$ after we integrate the empty integral? – bukka Dec 17 '13 at 22:10
• @bukka $$\int_0^a dx = x|_0^a = (a-0) = a.$$ – gt6989b Dec 17 '13 at 22:54