# How to explain this relation between surface area and 1st derivative of volume(circle, sphere, square and cube)?

I have recently realized that the first derivative of the area of a circle $A = \pi r^2$ is the circumference (perimeter) $C = 2 \pi r$ and the first derivative of the volume of a sphere $V = 4 \pi r^3 / 3$ is the surface area of a sphere $SA = 4 \pi r^2$. (So basically 3D to 2D, 2D to 1D).

Then I tried with cube and square and found out that the first derivative of the area of a square $A = a^2$ is half the perimeter of a square, and the first derivative of the volume of a cube $V = a^3$ is half the surface area of a cube.

Is there any mathematical explantion to this?

For the sphere and circle there are wonderful explanations. Think of it this way - when the volume of a sphere increases, it is like you are adding infinitely many infinitely thin (flat) layers (or shells) to the outside of the sphere, each with an area of $4\pi r^2$.
You can even think about it in one dimension. If the length of a line segment is given by $d$, then its first derivative is $1$, and you are adding infinitely many points each time the length increases.
Sorry to add another answer, but this should be separate. Using these analogies, you should be able to find the "surface volume" and "hypervolume" of a hypersphere (4 dimensional sphere). Let us denote $V_n(r)$ the "interior space" (area, volume, hypervolume) of a sphere in $n$ dimensions, and $A_n(r)$ the "exterior space" (circumference, surface area, surface volume). Since the length of a chord of a circe that is some distance $d$ from the center is given by $2\sqrt{r^2-d^2}$, the area of a circle can be derived by the integral of the lengths of these chords from $d=-r$ to $r$: $$2\int_{-r}^r \sqrt{r^2-d^2} \, dd$$ Which gives us $\pi r^2$. You can use a similar formula for the volume of a sphere, since the area of each circular cross section of the sphere a distance $d$ from the center is given by $\pi(r^2-d^2)$, the volume of the sphere is $$\int_{-r}^r \pi(r^2-d^2) \, dd$$ also giving us the correct formula, $\frac{4}{3}\pi r^3$. Now you can find the analogous formula for the volume of a hypersphere: $$\int_{-r}^r \frac{4}{3}\pi(r^2-d^2)^\frac{3}{2} \, dd$$ and the "surface volume" is the derivative of this: $$\frac{4}{3}\pi(r^2-d^2)^\frac{3}{2}$$