# The Derivative of Volume and Surface Area

The derivative of the volume $V=\frac{4}{3}\pi r^3$ of a sphere of radius $r$ with respect to $r$ is the surface area $S=4\pi r^2$. Is this a coincidence? It is not. There is a good reason why it happened that way. To understand it more easily, let us take a look at it’s lower dimensional analogue, namely the derivative of the area $\pi r^2$ of a circle of radius $r$ is the circumference $2\pi r$. Let me first explain why we get that. Let $A(r)=\pi r^2$. Then
$$\frac{dA}{dr}=\lim_{\Delta r\to 0}\frac{\pi(r+\Delta r)^2-\pi r^2}{\Delta r}.$$
$\pi(r+\Delta r)^2-\pi r^2$ is the area (shape of washer) between two circles both centered at the origin with radii $r+\Delta r$ and $r$, respectively as seen in the following figure:

As $\Delta r$ gets smaller, you see that the washer gets thiner. So if $\Delta r\to 0$, the washer becomes circle of radius $r$ resulting $\frac{dA}{dr}$ its circumference. By doing the same analysis, you can see why the derivative of the volume of sphere is its surface area.

Do any other objects share the relationship? Sure! For instance, the derivative of the area $x^2$ of a square with side $x$ is its circumference $2x$. The derivative of the volume $\pi r^2h$ of a cylinder with radius $r$ and height $h$ is its lateral surface area $2\pi rh$. How about a cube with volume $V=x^3$? In this case the derivative is $3x^2$ so it is not the surface area. Why is this? It is due to symmetry i.e. it depends on whether the volume increases symmetrically when you increase your variable such as length of side, radius, or height, etc. In case of a cube, increasing length from $x$ to $x+\Delta x$ results in the increase of volume from only three faces of the cube i.e. the volume of cube does not increase symmetrically in that case. When $\Delta x\to 0$, the volume increment becomes three faces resulting the derivative the area of those three faces $3x^2$.

How about a box with volume $V=xyz$? In this case, divergence would give a similar relationship. In fact, $\nabla\cdot V=\frac{\partial V}{\partial x}+\frac{\partial V}{\partial y}+\frac{\partial V}{\partial z}=yz+xz+xy$ which is the surface area of box with length $x$, width $y$ and height $z$. Imagine that a box is filled with fluid and assume that volume increase amounts to the fluid flowing into the box through its faces. $\frac{\partial V}{\partial x}$ would measure the rate of fluid flowing into the $yz$ face per unit time. That would indeed be the same as the area of the face $yz$.

## 3 thoughts on “The Derivative of Volume and Surface Area”

Thanks! You are right. The derivative of the area $x^2$ of a square does not coincide with its circumference $4x$ for the same reason why the derivative of the volume $x^3$ of a cube does not coincide with its surface area $4x^2$.