The volume contained inside $S^n$ , the $n$ -sphere (or hypersphere), is given by the integral $$ V(n) = \int_{\sum_{i=1}^{n+1}x_i^2\le1} d^{n+1} x. $$ Going to polar coordinates ($r^2=\sum_{i=1}^{n+1}x_i^2$ ) this becomes $$ V(n) = \int_{S^n} d\Omega \int_0^1 r^{n}\, dr. $$ The first integral is the integral over all solid angles subtended by the sphere and is equal to its area $A(n)=\frac{2\pi^{\frac{n+1}{2}}}{\Gamma\left(\frac{n+1}{2}\right)}$ , where $\Gamma(x)$ is the gamma function. The second integral is elementary and evaluates to $\int_0^1 r^{n}\, dr = 1/(n+1)$ .

Finally, the volume is $$ V(n) = \frac{\pi^{\frac{n+1}{2}}}{\frac{n+1}{2}\Gamma\left(\frac{n+1}{2}\right)} = \frac{\pi^{\frac{n+1}{2}}}{\Gamma\left(\frac{n+3}{2}\right)}. $$ If the sphere has radius $R$ instead of $1$ , then the correct volume is $V(n)R^{n+1}$ .

Note that this formula works for $n\ge0$ . The first few cases are

$n=0$

$\Gamma(3/2)=\sqrt{\pi}/2$ , hence $V(0)=2$ (this is the length of the interval $[-1,1]$ in $\R$ );

$n=1$

$\Gamma(2)=1$ , hence $V(1) = \pi$ (this is the familiar result for the area of the unit circle);

$n=2$

$\Gamma(5/2)=3\sqrt{\pi}/4$ , hence $V(2) = 4\pi/3$ (this is the familiar result for the volume of the unit sphere);

$n=3$

$\Gamma(3)=2$ , hence $V(3) = \pi^2/2$ .