Why isn't the volume of a sphere $ π^2r^3?$

The problem is that the points closer to the center line of the circle don't move a distance of $2\pi r$ when the circle is rotated $2\pi$ radians. Rather a point at distance $h$ from the center line moves a distance of $2\pi h$. Now at distance $h$ the length of the rotating segment parallel to the center line is $2\sqrt{r^2 - h^2}$. Thus the volume is $$\int_0^r 2\sqrt{r^2 - h^2} \cdot 2\pi h \, dh = \frac{4 \pi r^3}{3}.$$

It is only the top of the semicircle that travels $2\pi$, That gives a contribution to the volume of $2\pi$ times the area of the top.

Other parts of the semicircle - near the diameter - travel much shorter distances, so contribute smaller multiples of their area to the volume.

For a more general view, any dimension, it turns out that the easy way to find the volume of higher dimensional spheres, including the one in $\mathbb R^3,$ is by adding two to the dimension and using polar coordinates. Given that the $n$-volume of the sphere of radius $R$ in $\mathbb R^n$ is $$ \omega_n R^n, $$ the easy calculation is that the $(n+2)$-volume of the sphere of radius $R$ in $\mathbb R^{n+2}$ is $$ \frac{2 \pi \omega_n}{n+2} R^{n+2}, $$ or $$ \omega_{n+2} = \frac{2 \pi \omega_n}{n+2}. $$

Your example is $\omega_1 = 2,$ a segment of "radius" $1$ so length $2,$ so $\omega_3 = \frac{2 \pi 2}{3} = \frac{4 \pi }{3}.$

The odd ones are $$ \omega_1 = 2, \; \omega_3 = \frac{4 \pi }{3}, \; \omega_5 = \frac{8 \pi^2 }{15}, \ \omega_7 = \frac{16 \pi^3 }{105},..., \; \omega_{2k +1} = \frac{ 2^{k+1} \; \; \pi^k}{(2k+1)!!} $$

The even ones are $$ \omega_2 = \pi, \; \omega_4 = \frac{ \pi^2 }{2}, \; \omega_6 = \frac{ \pi^3 }{6}, ..., \; \omega_{2k} = \frac{\pi^k}{k!} $$

Compare http://en.wikipedia.org/wiki/N-sphere#Closed_forms