Question: Is there a general relationship between surface area and volume analogous to the below examples?

Example 1. Consider a ball $B$ centered at the origin of a spherical coordinate system. The ball has volume $V(r) = \frac{4}{3}\pi r^3$, where $r$ is the radial distance from the origin to the boundary $\partial B$ of the ball. Now consider the surface area of $\partial V$. Elementary geometry shows that $A(r) = 4 \pi r^2$. Comparing $A$ and $V$, it follows that

$$ \frac{\partial V}{\partial r} = A. $$

Example 2. Another simple example is an open cylinder of length $l$. Place the cylinder at the origin and then define $s$ as the radial distance from the origin (in cylindrical coordinates). The volume of the cylinder is $V(s) = \pi s^2 l$ and its surface area is $A(s) = 2 \pi s l$. Once again, we have a relationship between volume and surface area with reference to the variable of variation (e.g., $s$ is the variable of variation, since both the volume and cylinder share the common length $l$),

$$ \frac{\partial V}{\partial s} = A. $$

Example 3. Although a cube may seem to stray from this relationship, as $\partial_s (s^3) \neq 6s^2$, this is only due to the lack of symmetry. If we center the cube in a Cartesian coordinate system, and define $a$ to be half the side length $s$, we see that $V(a) = (2a)^3 = 8a^3$ and $A(a) = 6(2a)^2 = 24a^2$. Thus,

$$ \frac{\partial V}{\partial a} = A. $$

Now consider a region $Q \subset \mathbb{R}^3$ with boundary $\partial Q$. The "center of mass" is defined as

$$ \textbf{r}_{cm} = \frac{1}{V(Q)} \int_{Q} \textbf{r} dV, $$

where $\textbf{r} = (x, y, z)$ in Cartesian coordinates. Assume that the object is centered in a coordinate system such that $\textbf{r}_{cm} = \textbf{0}$. Now let a coordinate system be described by a basis $\{\textbf{x}_i\}$, where $i = 1, 2, 3$. In general, when will the following relationship hold between $V = V(Q)$ and $A(\partial Q)$,

$$ \frac{\partial V}{\partial x_i} = A? $$

Also, assuming this relationship holds in some coordinate system, is there only one associated region that satisfies the necessary conditions (e.g., a sphere in spherical coordinates, a cube in Cartesian coordinates)?

  • $\begingroup$ You may find this paper very interesting: math.byu.edu/~mdorff/docs/DorffPaper07.pdf $\endgroup$
    – fixedp
    Aug 21, 2014 at 9:08
  • $\begingroup$ What do you take $\partial V/\partial x_i$ to mean? Without some fixed pattern for forming different shapes $Q$, I find it hard to see how you would find $V$ in terms of the $x_i$. $\endgroup$
    – Jessica B
    Aug 21, 2014 at 9:20
  • 1
    $\begingroup$ There is a non-rigorous intuitive argument. If you take $\partial r$ to be a thickness that you add to each face of the object, then certainly $\partial V = A*\partial r$ because the incremental volume is a thin shell that has area $A$ and thickness $\partial r$, therefore volume $A*\partial r$. This seems clear for convex objects, anyway. $\endgroup$
    – bubba
    Aug 21, 2014 at 12:37
  • $\begingroup$ Partial duplicate: math.stackexchange.com/questions/625/… $\endgroup$ Aug 30, 2022 at 21:00

1 Answer 1


The answer is: YES - see Stokes Theorem and as special case Greens Theorem.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .