Deriving+the+volume+of+a+Cone+andSphere



There are a number of ways to derive the volume (**V**) of a cone or pyramid with a height **h**, and a base with area **B**. Here is how I did it, in high school. I divide the object (cone or pyramid) up into many layers (**n** of them, actually). And I approximate each layer with prism-like object, the same shape as the base, with the sides vertical, not slanting. By "prism-like," I mean that if the original object is a square pyramid, then each layer is now a rectangular solid. If the original object is a circular cone, then each layer is now a circular cylinder. If the original object is some pyramid, with an odd-shaped base, then each layer is now a prism, with a base of the same odd shape. Let's say that each little layer has a height of **x**. Then, **h** (the height) is equal to **nx**. In other words, **x=h/n**. Each part (side, diameter, whatever) of the base of each layer follows an arithmetic progression, from the smallest at the top, to the largest at the bottom. As **i** (a variable that we are using to count the layers) varies from 1 to **n**, the area of the base of each layer is **Bi²/n²**. And the volume of layer **i** is **Bhi²/n³**. The sum of all of the layers is the sum (as **i** goes from 1 to **n**) of this **Bhi²/n³**. Well, for any given cone or pyramid, all of those letters are constants, except for **i**. So, that sum is **Bh/n³** times the sum of all of **i²** (where **i** varies from 1 to **n**). It turns out that it is fairly easy to show that this sum of **i²**, where **i** varies from 1 to **n**, is **n(n+1)(2n+1)/6**. I will not prove that here. But, it can be proven by mathematical induction, as well as other ways. So, the volume of our approximation of a cone, or pyramid, is **Bhn(n+1)(2n+1)/6n³**. Dividing the numerator by **n**, three times, we can rewrite that as **Bh(1+1/n)(2+1/n)/6**. We want our approximation of a cone, or a pyramid, to no longer be an approximation, but to be exact. To do that, we need **n** to become infinite. In that case **1/n** becomes zero. And the above expression simplifies to **V=Bh(1)(2)/6=Bh/3**. The volume of a cone or pyramid is 1/3 the area of the base times the height.

http://www.jimloy.com/geometry/cone1.htm

 Start with a hemisphere of radius r and construct a cylinder of radius r and height r. The diagram below contains a cutaway view of the cylinder. Figure 1 Now place an inverted cone with height r and base radius r inside the cylinder.
 * [[image:http://mathcentral.uregina.ca/QQ/database/QQ.09.99/partridge1.1.gif width="204" height="186"]] || [[image:http://mathcentral.uregina.ca/QQ/database/QQ.09.99/partridge1.4.gif width="204" height="186"]] ||

I am interested in the volume of the region between the cone and the cylinder.

Slice through both figures with a plane parallel to the base and at a height h units above the base. Peal back what you have cut off and look down from above. What you see on the left is a circular disk and on the right is a region between two concentric circles. I want to compare the areas of these two regions. Figure 2 If the radius of the circle on the left is p then its area is pi times p^2. The radius p can be found from a cutaway view of the hemisphere. Figure 3 By the theorem of Pythagoras p^2 = r^2 - h^2 and hence the area of the circular disk is pi(r^2 - h^2). For the region between the concentric circles the radius of the outside circle is the radius of the cylinder,r. Since the radius of the base and the height of the cone are both r, a cutaway view of the cone inside the cylinder shows similar triangles with the height equal to the base. Hence the radius of the inside circle is h. Thus the area of the region between the concentric circles is pi*r^2 -pi*h^2 = pi(r^2 - h^2).

Figure 4 If you think of the hemisphere and the region between the cone and the cylinder in Figure 1 as being constructed of layers as in figure 2 then the volume of the hemisphere is equal to the volume of the region between the cone and the cylinder. The volume of the cylinder is pi*radius2*height = pi*r^3. The volume of the cone is (1/3)*pi*radius^2*height = (1/3)*pi*r^3. Hence the volume of the hemisphere is pi*radius^2*height = pi*r3^ - (1/3)*pi*r^3 = (2/3)*pi*r^3 Thus the volume of the sphere of radius r is (4/3)*pi*r^3
 * Therefore the areas of the two regions in figure 2 are equal.**

From: http://mathcentral.uregina.ca/QQ/database/QQ.09.99/partridge1.html