# Probability density function of $X$ with a uniform distribution on a unit sphere

Given that $$f_\Phi(\phi)=\dfrac{\sin(\phi)}{2}$$ for $$\Phi\in[0,\pi]$$, and $$f_\Theta(\theta)=\dfrac{1}{2\pi}$$ for $$\Theta\in[0,2\pi]$$, where $$\Phi$$ and $$\Theta$$ are independent. What is the PDF of $$X=\sin(\Phi)\cos(\Theta)$$?

So far, I have managed to obtain the PDFs of $$U=\sin(\Phi)$$ and $$W=\cos(\Theta)$$ as $$f_U(u)=\frac{u}{\sqrt{1-u^2}}, \quad\text{and}\quad f_W(w)=\frac{1}{\pi\sqrt{1-w^2}}.$$

I have attempted to follow the product distribution as described here. I know the answer should be $$f_X(x)=\frac{1}{2}$$, but I have been unable to progress further. How should I proceed?

• Is there an independence assumption? – Kavi Rama Murthy Dec 13 '18 at 7:15
• Yes, sorry, phi and theta are independent. – Dracolich56 Dec 13 '18 at 7:19

The easiest solution returns to the sphere's geometry. The part of the surface ranging from $$x$$ to $$x+dx$$ is a ribbon of radius $$\sqrt{1-x^2}$$ and thickness $$d\arcsin x=\frac{dx}{\sqrt{1-x^2}}$$, so its area is $$2\pi dx$$. This can be used to prove the classical fact that a sphere's area matches that of a cylinder just long and wide enough to hold it, because each ribbon on the sphere's surface has the same area as its shadow on the cylinder. 3Blue1Brown has a great video on it here.