
 Originally Posted by 
odinn7
					 
				 
				We're not allowed to give you the answer. You have to try to figure it out first, post your answer and how you got it, then someone will come along and help you out with it.
			
		 
	 
 I never knew. All right. I tried. I think we should use the equation for calculating the volume of frustum: V= h*pi/3*{R^2 + R*r + r^2} 
Let r1 be the radius of the area. The volume of the upper section is: 
V1 = h1*pi/3*{r^2 + r1*r + r1^2} ,
and the volume of the lower section is: 
V2 = h2*pi/3*{r1^2 + r1*R + R^2} 
We know V1 + V2 = V and h1 +h2 = h. 
So, h1*pi/3*{r^2 + r1*r + r1^2} + h2*pi/3*{r1^2 + r1*R + R^2} = V 
After distributing and simplifying we get: 
h*r1^2 + r1*{r*h1 + R*h2} +h1*r^2 + h2*R^2 - 3*V/pi = 0
This is a quadratic equation of the form ax2 + bx + c = 0 where, 
a = h
b = {r*h1 + R*h2}
c= h1*r^2 + h2*R^2 - 3*V/pi 
x = (-b + sqrt{b^2 - 4ac})/2a , x = (-b - sqrt{b^2 - 4ac})/2a 
By substituting a, b, c in these solutions, we should get a parametric answer for 
radius r1. I haven't done that yet. 
So is this correct?