Expected Value of Area of Triangle

Date: 05/10/2000 at 10:16:17
From: Aled Jones
Subject: Uniform Distribution

AB is the diameter of a circle, with radius r. AP is a chord. X is the 
angle between AB and AP. X is uniformly distributed over the interval 
0 to pi/2. Find the expected value of the area of triangle APB.

Date: 05/10/2000 at 14:59:55
From: Doctor Anthony
Subject: Re: Uniform Distribution

The area of triangle

     APB = (1/2)2r.2r.cos(x).sin(x)

         =  2r^2.cos(x).sin(x)

         =  r^2.sin(2x) 

The expected value of the area is then

       INT(0 to pi/2)[r^2.sin(2x).dx/(pi/2)]

     = (2/pi)r^2.(-1/2)cos(2x)   from  0 to pi/2

     = (-1/pi)r^2[cos(pi) - cos(0)]

     = (-1/pi)r^2[-1 - 1]

     = 2.r^2/pi

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/