Multiple Dimensions - Moving On To 3

Here is a more complete solution. Of course, no solution will be complete unless it starts by explaining why the integral of the surface area = volume.

Suppose surface area of the sphere = $f(r) = 4\pi r^2$

Now consider the volume of a very thin spherical shell or coating around this surface area, with thickness dr. This is like a thin sheet with a surface area and thickness.

The volume of this thin spherical shell $\approx$ surface area x thickness

$dV = 4\pi r^2 dr$

To find the volume of the whole sphere, V from the above, we must sum the volumes of all the thin shells or layers making up the whole sphere, starting from 0 up to r, and we do this by integrating both sides:

$V = \int dV = \int 4\pi r^2 dr = {4 \over 3}\pi r^3$

integrating 4pir^2 with respect to r we get 4/3pir^2

It is known that the surface area of the sphere radius r is 4pir^2 and its volume, 4/3pir^3.

integrate the surface area wrt r=4/3pir^2