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On a geometry test I forgot the formula for the surface area of a sphere so I derived it and ended up being right. But it seems like my derivation is wrong. I got the surface area formula by taking the derivative of the volume formula, $\frac{4}{3}\pi r^3$. My reasoning was that taking the derivative is equivalent taking the volume of a sphere with radius $r+h$ minus the volume of a sphere with radius $r$ and then dividing by $h$ as $h$ goes to $0$. I figure that if you took the part of the larger sphere that is not in the smaller sphere and laid it flat it would be a prism with height $h$, so its base area (i.e. surface area of the sphere) would be the volume divided by $h$. So my questions are: First, why must $h$ go to $0$? First I tried it with $h = 1$ thinking, since the height is $1$, the base area (i.e. surface area of sphere) would equal the volume. But when $h=1$ you get the wrong formula. Secondly, is the rest of the reasoning correct? Thanks, Elliot

Elliot Gorokhovsky
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The size of the boundary times the rate of motion of the boundary equals the rate of change of size of the bounded region.

This gets you

  • the surface area of a sphere;
  • the product rule (apply it to a rectangle with two moving sides);
  • the fundamental theorem of calculus.

I call this the "boundary rule". I'd give it a prominent position early in a calculus course (in particular, in a textbook).

So let's address your question: why must $h$ go to $0$? The reason is that as the radius grows from $r$ to $r+h$, the surface area changes. On a spherical earth with an atmosphere of depth $h$, the volume of the atmosphere divided by $h$ is in units of area, and would be the surface area if the surface area were the same at all altitudes. If you want the surface area only at altitude $0$, then you can approximate it as closely as desired by making $h$ close enough to $0$. In other words, you take the limit as $h\to0$.

Michael Hardy
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