Question 1123579
<font face="Times New Roman" size="+2">
 *[illustration ConeWithWater.jpg].


The illustration is a cross-section of your cone through the axis.  Note that the altitude of the cone and a radius of the base form a triangle, and the depth of the water and the radius of the surface of the water form a similar triangle, similar being used in the formal sense so that the corresponding sides of the two triangles are proportional.


Hence,  *[tex \LARGE \frac{r}{h}\ =\ \frac{2}{4}\ \ ] from which we can easily derive a relationship between *[tex \Large r] and *[tex \Large h]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  r\ =\ \frac{h}{2}]


Considering the formula for the volume of a cone of radius *[tex \Large r] and height *[tex \Large h]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V(r,h)\ =\ \frac{1}{3}\pi{r^2}h]


and the relationship between *[tex \Large r] and *[tex \Large h] just established, we can make the substitution of *[tex \Large \frac{h}{2}] for *[tex \Large r].


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V(h)\ =\ \frac{1}{3}\pi{\(\frac{h}{2}\)^2}h]


Which simplifies to:


I will leave the rest as an exercise for the student.
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
{{n}\choose{r}}

</font>