Q: Is it a coincidence that a circles circumference is the derivative of its area, as well as the volume of a sphere being the antiderivative of its surface area? What is the explanation for this?

Physicist: For those of you not hip to the calculus groove, here’s what’s going down:  The derivative of Y with respect to X, written $\frac{dY}{dX}$, is just a description of how fast Y changes when X changes.  It so happens that if $Y=X^N$, then $\frac{dY}{dX}=NX^{N-1}$.  So, for example, if $Y=5X^3$, then $\frac{dY}{dX} = 15X^2$.  The area of a circle is $\pi R^2$, and the circumference is $2\pi R$, which is the derivative.  The volume of a sphere is $V = \frac{4}{3}\pi R^3$, and the surface area is $S = 4\pi R^2$, which is again the derivative.  This, it turns out, is no coincidence!

If you describe volume, V, in terms of the radius, R, then increasing R will result in an increase in V that’s proportional to the surface area. If the surface area is given by S(R), then you’ll find that for a tiny change in the radius, dR, $dV = S(R)dR$, or $\frac{dV}{dR} = S(R)$.

You can think of a sphere as a series of very thin surfaces added together.  This is another, equivalent, way of describing the situation.  Each layer adds (surface area of layer)x(thickness of layer) to the volume.

You can think of this like painting a spherical tank. The increase in volume, dV, is the amount of paint you use, and the amount of paint is just the surface area, S(R), times the thickness of the paint, dR.  This same argument can be used to show that the volume is the integral of the surface area (just keep painting layer after layer).

It’s a little easier to keep track of what’s going on with circles.

If you increase the radius of a circle by a tiny amount, dR, then the area increases by (2πR)(dR).

The same “derivative thing” holds up for the circumference vs. the area of a circle.  The change in area, dA, is dA = (2πR)dR.  So, $\frac{dA}{dR}=2\pi R$.  That is, the derivative of the area is just the circumference.

This, by the way, is one of the arguments for using “τ” instead of “π”. τ = 2π, so the area of a circle is $\pi R^2 = \frac{1}{2}\tau R^2$. This makes the “differential nature” of the circumference a little more obvious.

Answer gravy: By the way, this argument is only exact when the thickness of the new layer “goes to zero“.  Basically, the top of the new layer is a little longer/bigger than the bottom of the new layer.  So if the area is $A=\pi R^2$, and the change in radius is $\Delta R$, then $\Delta A = \pi(R+\Delta R)^2 - \pi R^2 = \pi R^2+2\pi R\Delta R + \pi (\Delta R)^2-\pi R^2 = 2\pi R\Delta R + \pi (\Delta R)^2$

This extra little $\pi (\Delta R)^2$ is a result of the top and bottom of the new layer being very slightly different lengths.  But when ΔR is small, (ΔR)2 is really small.  For example, if ΔR is one thousandth, then (ΔR)2 is one millionth.  The whole idea behind calculus is that when the scales get very small you can just ignore these “extra tiny” terms.  In fact, this is the essential difference between dA and ΔA, and how the derivative is defined:

$\frac{dA}{dR}=\lim_{\Delta R\to 0}\frac{\Delta A}{\Delta R} = \lim_{\Delta R\to 0}\frac{1}{\Delta R}\left(2\pi R\Delta R + \pi (\Delta R)^2\right)$

$= \lim_{\Delta R\to 0} 2\pi R + \pi\Delta R = 2\pi R$

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5 Responses to Q: Is it a coincidence that a circles circumference is the derivative of its area, as well as the volume of a sphere being the antiderivative of its surface area? What is the explanation for this?

1. Joe says:

Cool. Wonderful answer to my question.

2. juges debnath says:

It happens only in calculus. I only thing I would like to say is thanks to Newton for his ingenious discovery of Calculus !

3. Patrick says:

Very cool. I never thought about this before. But after this explanations it fits together so easily. Thank you

4. asdfasd says:

“Volume is the integral of surface area.”

To me, that makes so much more sense that “surface area is the derivative of volume.” Either way, this answers something I’ve been wondering about for a while. thanks man.

5. K.N. says:

It’s easy to visualize in 1D, too. The sphere of radius r and center x is an interval (x-r,x+r), which has ‘volume’ 2r and ‘surface area’ 2: the two endpoints is the boundary. And you can get the former by integrating the latter in the obvious way.

Hyperspheres somewhat less easy, though.