When solving some exercises, I forgot the formula for the surface area of the unit sphere. However, I remember that the length of the perimeter of a circle of radius $r$ is $2 \pi r$. So I figured I'll just compute
$$2 \int_0^1 2 \pi r \, \mathrm dr,$$
starting to integrate from the north pole of a unit sphere up to its center and as that area occurs twice, I added a factor of $2$. However, this integral yields $2 \pi$ and not the desired $4 \pi$. Where did I forget another factor of $2$?
I tried doing the same computation with the volume of the sphere since its surface area will just be the derivative of the volume. However, again
$$2 \int_0^1 \pi r^2 \, \mathrm dr = \frac{2}{3} \pi,$$
missing a factor of $2$. What am I doing wrong?
Best Answer
Let $V_n(R)$ be the ($n$)-volume of the Euclidean ball $\mathbb B^n_R$ with radius $R$ in $\mathbb R^n$and let $ S_n(R)$ be the $n$ ("area") of the sphere of radius $R$ in $\mathbb R^{n+1}$. We have $$ \frac{d}{dr} V_n(r)=S_{n-1}(r), e.g. \text{for $n=2, V_2(r)=π r^2, S_1(r)=2π r$,} $$ and for $n=3, V_3(r)=\frac{4}{3}π r^3, S_2(r)=4π r^2$. To recover all formulas, we note that $$ 1=\int_{\mathbb R^n}e^{-π\vert x\vert ^2} dx=\int_0^{+\infty}e^{-π r^2} r^{n-1} dr S_{n-1}(1), \quad\text{so that } $$ $\quad$ $$\boxed{ S_{n-1}(R)=R^{n-1}\frac{2\pi^{n/2}}{\Gamma(n/2)},\qquad V_{n}(R)=R^n\frac{\pi^{n/2}}{\Gamma(1+\frac n2)}.} $$ We need only $\Gamma(1/2)=π^{1/2},\ \text{for $x>0$:}\Gamma (x+1)=x\Gamma (x),\ \text{for $n\in \mathbb N: \Gamma(n+1)=n!$} $