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?
 A: 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 $4\pi r^2$.
Now imagine, when you add area to a square, you are adding a corner in a v-shape to the square with a length of half of the perimeter. The 3D equivalent of this is adding a pyramid-like shape covering three  mutually adjacent faces of the cube to the corner of a cube each time the volume increases.
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.
A: 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:
$$\frac{4}{3}\pi(r^2-d^2)^\frac{3}{2}$$
