Possible Duplicate:
Why is the derivative of a circle's area its perimeter (and similarly for spheres)?
We all know that the volume of a sphere is:
$V = \frac{4}{3}\pi r^{3}$
and its surface area is
$S = 4 \pi r^2$
Now we see that
$\frac{dV}{dr} = S$
As well, the area of a circle is
$A = \pi r^2$
The circumference is
$C = 2 \pi r$
Now we see again that
$\frac{dA}{dr}=C$
There may be more that I have not noticed. Why does this relationship occur?
Let $K_1$ be a pleasant region, fixed from now on, like a sphere of radius $1$, or a half-sphere of radius $17$, or a regular tetrahedron of side $9.5 \pi$.
Now let $K_r$ be the region obtained by scaling all linear dimensions of $K_1$ by the scaling factor $r$.
Let $V(r)$ be the volume of $K_r$, and let $S(r)$ be the surface area of $K_r$. Note that because of the way volumes and surface areas scale, $$V(r)=r^3 V(1)\quad\text{and}\quad S(r)=r^2S(1).$$ Thus $$V'(r)=3r^2 V(1)=r^2 S(1)\tfrac{3V(1)}{S(1)}=\tfrac{3V(1)}{S(1)}S(r).$$ Thus for any starting body $K_1$, the derivative $V'(r)$ is a constant times the surface area $S(r)$.
In the case where $K_1$ is the sphere of unit radius, the constant happens to be $1$. For other starting shapes $K_1$, or other measures of linear dimension, the constant will be different. But the important point is that there always is a constant $C=C_{K_1}$ such that $V(r)=C S(r)$.
Similar considerations apply in two dimensions to $S'(r)$ and a linear feature such as the perimeter $P(r)$ of $K_r$. The idea can be extended to higher dimensions.
