Let r be radius. Let us assume we already know the results of basic circles (I can't derive everything, you know!) such as Circumference =2π*r and Area=π*r^2 as well as squares such as Area=(side)^2. We can find the surface area of a sphere first. Since the surface area of a square box is 6(side)^2 (we are just adding 6 square areas up), we should expect our answer for surface area of a sphere to have the form of constant*(side)^2. Our first guess, a dumb guess if you will, would be (Circumference)^2=4π^2*r^2 since like the case of the square we have our length squared. Geometrically we can think of the circumference "sweeping" out area as it rotates. A link to a Mathematica file I made of this phenomena is right here. However, we can see that during the first animation that there is an overlap where area is counted more than one time. Upon measurement we would see that experiment matches this observation. Adjusting our thinking slightly, we can see that we don't need one of the circumferences to sweep out entirely. In fact one of these terms can simply be the diameter (2*r) while we let the other sweep. Physically this means that we can let the sweeping occur around a diameter. The second animation shows this accurately. So now we have
Upon measurement, we can agree that this is the correct expression.
Now onto volume. I can postulate on the volume of sphere by the fact that I know the volume of a cylinder with the length 2r (Volume=Area*2r=2pi*r^3). We know that the sphere could not have a greater volume since it would sit inside the cylinder. A VERY VERY keen observer would notice while looking at the cross section of a cylinder, a sphere, and cone all with the same radius, the cross sectional areas of each three objects would have the relation.
So a dumb guess would be that same relation holds for the volume (not unfounded since the volumes and cross areas differ only from a "sweeping" mechanism), so we would have
And by golly we have it.