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Volume and surface area of a sphere
Posted:
May 5, 1998 1:11 PM


Murray Siegel accurately points out:
One gets half the surface area 3 X^2.
Similarly, the derivative of the area of a circle with respect to its radius is the circumference while the derivative of the area of a square with respect to its side is only half of the perimeter.
But then inaccurately concludes: What this demonstrates is that the circle/sphere are more efficient than the square/cube.
What this demonstrates is that you need to choose the right coordinates. This fact (derivative of volume being surface area, and derivative of area being circumference or perimeter) results from the fundamental theorem of calculus. It just needs the area or volume to be made from an appropriate integral.
So the radius (or side length, or whatever) must be perpendicular to the expanding surface of the sphere or cube.
Also it must increase in a direction such that all of the side lengths are increased.
So for a square or cube, you need to use as the "radius" the segment from the center perpendicular to each face, which is half as long as the side of the square or cube.
Area of square = 4x^2 perimeter of square = 8x
Volume of cube = 8x^3 Surface area = 24x^2
works fine.
Joshua Zucker



