Why isn't the volume of a sphere $ π^2r^3?$ 
Imagine two similiar circles. We rotate one on half of the other circumference. It gives a sphere with volume of $1/2(2πr) \cdot πr^2 = π^2r^3$
 A: 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.
A: 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
A: 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
$$\int_0^r 2\sqrt{r^2 - h^2} \cdot 2\pi h \, dh = \frac{4 \pi r^3}{3}.$$
