The volume of an $n$-sphere of radius $R$ is

$$V_n(R) = \frac{\pi^{n/2} R^n}{\Gamma(\frac{n}2+1)}$$

and the surface area is

$$S_n(R) = \frac{2\pi^{n/2}R^{n-1}}{\Gamma(\frac{n}2)} = \frac{n\pi^{n/2}R^{n-1}}{\Gamma(\frac{n}2+1)} = \frac{d V_n(R)}{dR}$$

What is the intuition for this relationship between the volume and surface area of an $n$-sphere? Does it relate to the fact that the $n$-sphere is the most compact shape in $n$ dimensions, or is that merely a coincidence?

Are there other shapes for which it holds, or is this the limiting case of a relationship of inequality? For example, an $n$-cube of side $R$ has volume $V^c_n(R)=R^n$ and surface area $S^c_n(R)=2nR^{n-1}$, so that

$$\frac{dV^c_n(R)}{dR} = n R ^{n-1}$$

and we have $S^c_n = 2 dV^c_n/dR$.

Edit: As pointed out by Rahul Nahrain in the comments below, if we define $R$ to be the half-side length of the unit cube rather than the side length, then we have the relationship $S^c_n(R)=\frac{d}{dR} V^c_n(R)$, exactly as for the sphere. Is there a sense in which relationships like this can be stated for a large class of shapes?

  • 3
    $\begingroup$ See this MO-question and also the isoperimetric problem. $\endgroup$
    – t.b.
    Commented May 9, 2011 at 10:52
  • $\begingroup$ The main takeaway seems to be that this relationship is coincidental and based on using $R$ as the linear measure of distance. Were we to use $D=2R$ instead, we would get $S_n(D)=2\frac{d}{dD}V_n(D)$. However, there is still the interesting (to my mind) question of why the ratio of $S_n(x)$ to $\frac{d}{dx}V_n(x)$ (for some linear measure $x$) is independent of $n$ for the $n$-sphere but seemingly not for other shapes - no? $\endgroup$ Commented May 9, 2011 at 11:04
  • $\begingroup$ Sure! I also voted the question up, by the way. I didn't mean to imply that these links answer the question completely. I just wanted you to see them because they certainly give some food for thought. $\endgroup$
    – t.b.
    Commented May 9, 2011 at 11:06
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    $\begingroup$ Isn't the surface area of an $n$-cube $2nR^{n-1}$, not $6R^{n-1}$? Then $S_n^c = 2dV_n^c/dR$ for all $n$. $\endgroup$
    – user856
    Commented May 9, 2011 at 11:31
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    $\begingroup$ In fact, if you let the parameter be the "radius" of the cube, i.e. half of its side length, then $V_n^c = (2R)^n$ and $S_n^c = 2n(2R)^{n-1}$, and $S_n^c = dV_n^c/dR$ for all $n$, just like the sphere... $\endgroup$
    – user856
    Commented May 9, 2011 at 11:39

3 Answers 3


There is a very simple geometric explanation for the fact that the constant of proportionality is 1 for the sphere's radius and the cube's half-width. In fact, this relationship also lets you define a sensible notion of a "half-width" of an arbitrary $n$-dimensional shape.

Pick an arbitrary shape and a point $O$ inside it. Suppose you enlarge the shape by a factor $\alpha \ll 1$ keeping $O$ fixed. Each surface element with area $dA$ at a position $\vec r$ relative to $O$ gets extruded into an approximate prism shape with base area $dA$ and offset $\alpha \vec r$. The corresponding additional volume is $\alpha \vec r\cdot \vec n dA$, where $\vec n$ is the normal vector at the surface element.

Now the quantity $\vec r \cdot \vec n$, call it the projected distance, has a natural geometric interpretation. It is simply the distance between $O$ and the tangent plane at the surface element. (Observe that for a sphere with $O$ at the center, it is always equal to the radius, and for a cube with $O$ at the center, it is always equal to the distance from the center to any face.)

Let $\hat r = A^{-1} \int \vec r \cdot \vec n dA$ be the mean projected distance over the surface of the shape. Then the change in volume by a scaling of $\alpha$ is simply $\delta V = \alpha \int \vec r\cdot \vec n dA = \alpha \hat r A$. In other words, a change of $\alpha \hat r$ in $\hat r$ corresponds to a changed of $\alpha \hat r A$ in $V$. So if you use $\hat r$ as the measure of the size of a shape, you find that $dV/d\hat r = A$. And since $\hat r$ equals the radius of a sphere and the half-width of a cube, the observation in question follows. This also implies the distance-to-face measure for regular polytopes that user9325 mentioned, but generalizes to other polytopes and curved shapes. (I'm not completely happy with the definition of $\hat r$ because it's not obvious that it is independent of the choice of $O$. If someone can see a more natural definition, please let me know.)


Some remarks:

  1. It is not necessary to regard solids that generalize to $n$ dimensions, so one can start with shapes in 2 dimensions.

  2. It is very natural to always use a radius-type parameter to scale the figure because the intuition is that the figure gets growth rings that have the size of the surface.

  3. Unfortunately, the property is no longer true if you replace a square with a rectangle or a circle with an ellipse.

  4. For a regular polygon, you can always take as parameter the distance to an edge. This will work similarly for Platonic solids or any polygons/polyhedra that contain a point that is equidistant to all faces.

  5. This argument does not look good for general curves, because intuitively the edges of a smooth curve are infinitesimally small, so the center should be equidistant to all points, but of course, we could identify cases where the changing thickness of the growth ring averages out.

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    $\begingroup$ +1. Point 2 is the intuitive explanation, while points 3 and 5 are the essential caveats. $\endgroup$
    – Henry
    Commented May 9, 2011 at 12:32

What you're seeing with spheres is an instance of something really important called the coarea formula. You're looking at a function like $f(x,y,z) = r = \sqrt{x^2 + y^2 + z^2}$ and you're comparing the volumes of the lower contour sets $f(x,y,z) \leq R$ (in this case they are balls) to the areas of the level sets $f(x,y,z) = R$.

In the instance with cubes, you're considering a different function whose level sets are cubes.

The coarea formula states that (subject to some assumptions that cause both sides to make sense),

$Vol(f(x, y, z) \leq R) = \int_{-\infty}^R \left[ \int_{f(x,y,z) = t} |\nabla f|^{-1} d\sigma \right] dt $

You can also differentiate this formula with respect to $R$ to get the differential form you were observing.

Just as a dummy check, note that $t$ has the same units as $f$, so if you think of $f$ and the coordinates as having units, the dimensional analysis checks out and both sides have units of "length^3". (The formula is also valid in higher dimensions.)

In the examples you gave, the gradient of $f$ has length $1$, so when you integrate $1$ over the level sets $\{f(x, y, z) = t \}$ you're just getting the surface area.

The proof of this formula boils down to asking "what's the difference between $Vol( f(x,y,z) \leq R )$ and $Vol(f(x,y,z) \leq R + h$ if $h$ is small?". The two regions are very similar, differing only around the boundary $f(x,y,z) = R$. This is particularly clear in the case of a sphere. If you think about a small portion of that boundary of "width" $d\sigma$ in the directions of constant $f$, then the difference of the two regions has height $~ \frac{h}{|\nabla f|}$ in the direction of increasing $f$.

Another dummy check: The inverse makes sense because if $f$ is increasing extremely quickly, then increasing the value of $f$ will hardly increase the region $\{ f(x,y,z) \leq R \}$.

Note, the statement of the coarea is really a statement about the function and not just about the shapes of the level sets -- you are asking about the parameter $R$ by which these level sets are labeled, but you could pick a different function (like $\tilde{f} = e^f$ which has the same level sets but labels them differently.


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