Why the degree of $\pi$ does not increase

Question

What is the underlying reason, that $\pi$ is of degree $1$ in the volume formula for an $n$-ball of radius $r$,

i.e. the perimeter of circle is $$2\pi \cdot r$$ its area is $$\pi \cdot r^2$$ the volume of a ball is $$\frac{4}{3} \pi \cdot r^3$$

$r$ gets raised to an appropriate power, but $\pi$ remains $\pi$, with only a constant in front of it changing?

Answer 1

I don't know if this qualifies as an "underlying" reason, but various rearrangement proofs like this one can be very convincing as to "why" $\pi$ remains just $\pi$ while going from circumference to area.

Pretty much the same idea, but using integration instead of rearrangement, applies in higher dimensions.

IMHO the question isn't why $\pi$ does not become $\pi^2$ etc, but why $\pi$ does not become some totally unrelated constant at all. :)

Answer 2

I'm not sure what answer is expected here, but if you think of the area of a circle as being the limit of the sum of the areas of lots of concentric circles, then $$ A=\sum_{r=0}^R 2\pi r \delta $$ where $\delta$ is the line width of the circle, so $$ A=\int_{r=0}^{R}2\pi r dr $$ it is quite clear that $\pi$ is a constant and you get $A=\pi r^2$. Similar to that, if you think of the volume of a sphere as being the sum of the areas of lots of circles with the same axes, you are integrating $$ V=\int_{r=-R}^R \pi (R^2-r^2) dr \\ =\frac43\pi r^3 $$ and again the $\pi$ is just a constant which goes outside the integral. If the $\pi$ was in the limits of integration you might get $\pi^2$ or something, but it doesn't appear there.