TITLE: Why does the integral of the surface area of a sphere equal its volume?
QUESTION [5 upvotes]: Why does the integral of the surface area of a sphere equal it's volume? 
$$\int{4\pi r^2 \ \mathrm{d}r =\frac{4}{3}\pi r^3}$$
I don't quite understand why the relationship between surface area and volume is this way. Is this true for any other shapes?

REPLY [3 votes]: The reason is essentially the same as the reason why the volume of an onion is the same as the volume of all its layers. Imagine a spherical ball as being composed of many thin spherical layers, like an onion. The volume of the layer between radius $r$ and radius $r+\delta r$ is (to a close approximation if $\delta r$ is small enough) the area $4\pi r^2$ times its thickness: $4\pi r^2\cdot\delta r$. Summing all these small volumes from the centre of the ball ($r=0$) to its outer radius ($r=a$, say) and taking the limit as the $\delta r$ tend to $0$ gives us the limit
$$\int_0^a 4\pi r^2\mathrm dr.$$As you know, this evaluates to $\frac43\!\pi a^3$, the volume of the sphere.