# 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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### 10 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.

6. Derpo says:

not to mention the derivative of the circumference is 2 pi which is the amount of radians equal to 360 degrees

7. Tsi says:

Thank you for this wonderful explanation. However, I don’t know calculus. So can you explain some of these relationships between radius, circumference, area, circle, sphere surface area, and sphere volume in other terms, possible algebraic or geometric?

To me, pi is the factor that accounts for the curvature of a circle as opposed to the straight lines of a square. That is, instead of four times the side of a square, which gives you the perimeter of that square, you use pi times the side(side being equal to diameter of inscribed circle) to give you the circumference of the inscribed circle.
ie: circumference = 2πr = πD perimeter = 4D Circle/Square = πD/4D =π/4

Should not this relationship, somehow, carry through to a sphere vs a cube??

8. Angel says:

@Tsi:

There is no algebraic explanation for it given that it’s a circle, or namely, a curve.

Say we have a cube C with an inscribed sphere P in it. We know V(P)=4pi*r^3/3 and V(C)=s^3, with s=2r=d, so that V(C)=(2r)^3=8r^3. If you want the ratio of the volume of P to the volume of C, then:

V(P)/V(C)=(4pi*r^3/3)/(8r^3)=4pi/(3*8)=pi/(3*2)=pi/6

If you want to do the same for the area as you did with the circles and squares, then:

S(C)=6s^2=6(2r)^2=24r^2; S(P)=4pi*r^2. You want the ratio, so:

S(P)/S(C)=4pi*r^2/24r^2=pi/6, which equals the ratio in volumes as well.

Circumference and perimeter are not directly related to surface area here nor to volumes, so that’s why pi/4 cannot be brought over. However, because the surface areas are directly related to each other via intergration/differentiation, we can still see a common ratio. The are of a circle differs from the area of a sphere, and so does the area of square from that of a cube. If you were to obtain circumference from a sphere based on surface area, for instance, the formula would be (after much simplification) C=8pi*r, which is 4 times bigger than 2pi*r . Taking the perimeter of a cube based on its surface area, would give us P=48r, which is 6 times bigger than the usual 8r.

Although, you could make the case that because the circumference is now 4 times as big, and the perimeter six times as big, you could use proportions to relate the ratios for sphere:cube and circle:sqaure by 4/6=2/3. Since you obtained pi/4, then the ratio of sphere:cube must be 2/3*pi/4=2pi/12=pi/6, which is what I obtained earlier. So, with the right calculations an proportions, you could relate the ratio of circumference:perimeter to the ratio of sphere’s volume: cube’s volume. Very interesting question.

9. Ruvian says:

For everyone, to understand better