The Renewal Equation

A math odyssey

Sphere versus Cube

Today I was able to spend more than an hour on calculus exercises. (Yay!) Now I am pondering the following:

The volume of a sphere of radius $r$ is given by $V=\frac{4}{3}\pi r^3$. The surface area of the same sphere is given by $A=4\pi r^2$, the derivative of the formula for the volume. This makes sense, as a small change in volume involves adding a thin layer to the surface of the sphere. In the limit, that layer will be identical to the sphere’s surface.

On the other hand, the surface area of a cube of side length $x$ is given by $A=6x^2$. This is not the derivative of $V=x^3$, the formula for the cube’s volume. Why?1

  1. To do. ↩︎
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Responses

  1. Lucian S

    Hmm! Conceptually, I’m going to wave my hands and say it works for the sphere because the center is equidistant from the center, so that the volume increases exactly proportionally to that radius. ‘x’ for a square, however, is not related to the radius of the object, and increasing it slightly doesn’t match a proportional change in its distance from its center.

    I’m going to guess that additionally, if you reformulate the sphere problem to use diameter instead of radius, it similarly won’t work.

    However, these are mostly post-hoc guesses based on the fact that I know it’s true, and am trying to guess why, not something I could come up with from first principles.

    Reply
    1. Olly

      Good thoughts. I think it is probably something like this, especially given yerpa’s discovery below.

      Reply
  2. yerpa

    let’s talk about the cube that just encloses the sphere with radius r.
    Its volume V = (2r)^3 = 8•r^3.
    Its area A = 6•(2r)^2 = 24•r^2.
    I think that is the derivative of V

    Reply
    1. Olly

      Even more to think about!

      Reply

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