Surface Area and Volume of a SphereDate: 08/30/99 at 03:21:08 From: Tuan Vu Subject: Surface and volume of spheres Besides using integration, is there an intuitive way of seeing why the surface area of a sphere = 4(pi)r^2 and the volume = (4/3)(pi)r^3? Date: 08/30/99 at 08:41:17 From: Doctor Peterson Subject: Re: Surface and volume of spheres Welcome to Doctor Math. Usually these 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. This will not be exactly intuitive, but certainly more so than calculus methods. 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. If you put all these cylinders together, you get the cylinder that circumscribes the sphere: --------------------------- ------- ------- +- *********** -+ | ------- ***** | ***** ------- | | *---------------------------* | | ** | ** | | * | * | | ** | ** | | * | * | |* | *| |* | *| * | * * + * 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/ |
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