Why the degree of $\pi$ does not increase 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?
 A: 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. :)
A: 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.
