Volume of a SphereDate: 05/28/99 at 16:36:40 From: Anonymous Subject: Volume of a sphere I know that the volume of a sphere is V = (4 Pi/3)r^3, but I don't know how this was arrived at and what the explanation of the forumla is. I would like to know how the volume of the sphere is arrived and why it is V = (4 Pi/3)r^3. Please respond ASAP. Thank you very much I appreciate your help. Date: 05/28/99 at 20:14:51 From: Doctor Peterson Subject: Re: Volume of a sphere Hi, Usually these volume and area formulas are derived using calculus; but they were first worked out by the Greeks, notably Archimedes, using geometrical methods. The complete proof of the formulas can be a little involved, but I can give you the basic idea. I like to start with the area of the sphere, and get the volume from that - that covers two questions at once. First let's look at the surface area of a sphere. Picture cutting the sphere into many horizontal slices, making a set of sloped rings like this: **************** ***** ***** * + *------- / ***** | ***** \ l | / **************** \ |h /. | .\ | / ..... | ..... |\ | * .................. *----- * | C *| ***** | ***** | ****************** | |<------x------>| Each is almost the frustum of a cone. Let's ignore the curvature and pretend it is. The area of this piece is the product of its height l (along the slant) and its length, that is, its circumference C around the middle. If we cut all the pieces so their actual heights h are the same, then the slant height l of each piece will get larger as it gets closer to the top or bottom and becomes more slanted. In fact, the ratio of the slant height to the actual height is the same as the ratio of the radius of the sphere, r, to the radius of the ring, x: *********** ***** | ***** ****----------+----------****--+ ** | **|h *----------------+----------------* ** | / |** * | r / | * * | / | * * | / | * * | / | * * +----------------+ * * | x * * | * * | * * | * ** | ** * | * ** | ** **** | **** ***** | ***** *********** l r --- = --- h x So the area of the ring is r * h A_ring = C * l = 2 pi x * ----- = 2 pi r h x This means that each ring has the same area as a cylinder with radius r and height h - the cylinder that would be cut from the cylinder that circumscribes the sphere, by the same two horizontal cuts that made the ring. If you put all these cylinders together, you get the whole cylinder: --------------------------- ------- ------- +- *********** -+ | ------- ***** | ***** ------- | | *---------------------------* | | ** | ** | | * | * | | ** | ** | | * | * | |* | *| |* | *| * | * * + * 2r * | * |* | *| |* | *| | * | * | | ** | ** | | * | * | | ** | ** | | **** | **** | | ***** | ***** | +- *****+*****----------------+ ------- r ------- --------------------------- The height of this cylinder is 2r, so its lateral area is A = 2 pi r * 2r = 4 pi r^2 This is the area of the sphere! Now, for the volume, just picture cutting the sphere into lots of little polygons, and connect each of these to the center of the sphere to make a pyramid. *********** ***** ***** **** **** ** ** * * ** ** * * * * * * * * * + * * \ * * \r * * \\ * * \\\ * ** \ \_\ ** * \|_| * ** ** **** **** ***** ***** *********** The area of each pyramid is 1/3 the product of the base area and the height. But the heights are all the same (if they're small enough): the radius of the sphere. And the sum of their base areas is the surface area of the sphere: 4 pi r^2. So the volume is the same as that of a pyramid whose base is the surface area of the sphere and whose height is its radius: V = 1/3 4 pi r^2 * r = 4/3 pi r^3 What I've described makes a lot of assumptions; to turn it into a careful proof I would have to deal with all those curves I had to ignore, and talk more carefully about pieces being "small enough." But this should give you enough reason to believe the formulas are right. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/