Okay this seems like a really simple question. Basically I'm adding together 8 spheres (like raindrops coalescing into one bigger drop) and I'm getting two different answers for the new radius. Each individual drop is identical. I start by expressing the new volume in terms of the individual drops' radii, and then the new radius. The individual drops' radii are R. Volume = (4/3)∏(R)^3 * 8 = (4/3)pi(Rf)^3 I work it all out and find that the new radius is 2 times the radius of an individual drop. This is true according to a solution given. But when I try this with surface area.... S.A. = 4∏(R)^2 * 8 = 4∏(Rf)^2 4∏ cancels, then the new radius comes out as R*2*√(2) What am I missing here? Is Surface area not additive, or am I making some calculation error?
Surface area is not additive. Imagine if you took two rectangular prisms and joined them together at a common side: each rectangular prism has lost a side to the interior of the combined object, so the surface area is NOT double the surface area of an individual block. The case with the spheres is similar. The volume should be additive, however. The reason is that, thinking physically, the mass of all the spheres is conserved, so unless the density were to change during the process of joining the droplets together, the volume increases by the same factor as the increase in mass.
Two raindrops, each of radius r, have volume [itex](4/3)\pi r^3[/itex] each. If they "coalesce", because mass is conserved, and the density of water is a constant, the volume will be [itex](8/3)\pi r^3[/itex]. Solve [itex](4/3)\pi R^3= (8/3)\pi r^3[/itex] for the new radius, R.