The area of $S^n$ the unit $n$ -sphere (or hypersphere) is the same as the total solid angle it subtends at the origin. To calculate it, consider the following integral $$ I(n) = \int_{\R^{n+1}} e^{-\sum_{i=1}^{n+1} x_i^2}\, d^{n+1} x. $$ Switching to polar coordinates we let $r^2=\sum_{i=1}^{n+1} x_i^2$ and the integral becomes $$ I(n) = \int_{S^n} d\Omega \int_{0}^{\infty} r^{n} e^{-r^2}\, dr. $$ The first integral is the integral over all solid angles and is exactly what we want to evaluate. Let us denote it by $A(n)$ . With the change of variable $t=r^2$ , the second integral can be evaluated in terms of the gamma function $\Gamma(x)$ : $$ I(n)/A(n) = \frac{1}{2}\int_0^\infty t^{\frac{n-1}{2}} e^{-t}\, dt = \frac{1}{2}\Gamma\left(\frac{n+1}{2}\right). $$ We can also evaluate $I(n)$ directly in Cartesian coordinates: $$ I(n) = \left[ \int_{-\infty}^\infty e^{-x^2}\, dx \right]^{n+1} = \pi^{\frac{n+1}{2}}, $$ where we have used the standard Gaussian integral $\int_{-\infty}^\infty e^{-x^2}\, dx = \sqrt{\pi}$ .

Finally, we can solve for the area $$ A(n) = \frac{2\pi^{\frac{n+1}{2}}}{\Gamma\left(\frac{n+1}{2}\right)}. $$ If the radius of the sphere is $R$ and not $1$ , the correct area is $A(n)R^{n}$ .

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

$n=0$

$\Gamma(1/2)=\sqrt{\pi}$ , hence $A(0)=2$ (in this case, the area just counts the number of points in $S^0=\{+1,-1\}$ );

$n=1$

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

$n=2$

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

$n=3$

$\Gamma(2)=1$ , hence $A(3)=2\pi^2$ ;

$n=4$

$\Gamma(5/2)=3\sqrt{\pi}/4$ , hence $A(4)=8\pi^2/3$ .