Surface area of the unit ball in $R^{n}$

Question

I want to compute the surface area of the unit sphere in $R^{n}$, the parametrization is as follows $$x_{i}=\prod_{j=1}^{i-1}\sin{\theta_{j}}\cos{\theta_{i}}$$ where $1\leq i \leq n-1$ and $$x_{n}=\prod_{i=1}^{n-1}\sin{\theta_{j}}$$ But I am not sure how to proceed next since the integral seems to be so complicated. Could someone help me with that?

Answer 1

Here's a better approach, which you can work out for yourself.

Find a formula for the volume of the ball of radius $1$ in $\Bbb R^n$. This can be done recursively, relating to the ball in $\Bbb R^{n-2}$. Then adjust for the volume $V_n(r)$ of the ball of radius $r$.

Now, the area $A_{n-1}(r)$ of the boundary sphere is, not surprisingly, given by differentiating: $A_{n-1}(r)=V_n'(r)$. So, in other words, $A_{n-1}(1) = V_n'(1) = nV_n(1)$.

Answer 2

I like the intuitive approach Ted Shifrin suugested, however when I saw the differetiation trick I felt it lacked rigour. I bring here a more technical method, which assumed you already know the jacobian of ball coordinates (with variable $r$) $$\det J= r^{n-1}\prod_{i=1}^{n-1}\sin^{n-1-i}(\theta_i)$$ The parameterization $\psi$ for the sphere $R \mathbb S^{n-1}$ is the same but with constant $R$. It's derivative $D\psi$ is the same as $J$ but without the first column $J_1$ (the derivative w.r.t $r$) and plugging in constant $R$. If so, we can write

$$J = [J_1 \ \ D\psi]$$ Hence we get $$J^T J = \begin{bmatrix} J_1^T J_1 & J_1^T D\psi \\ D\psi^T J_1& D\psi^T D\psi \end{bmatrix}$$ Now I claim $J_1^T D\psi = 0 = D\psi^T J_1$. Why? the columns of $D\psi$ are tangent to the sphere, while $J_1$ is perpendicular to the sphere at every point, hence every dot product vanishes. If so, we get $$(\det J)^2 = \det J^T J = \left \| J_1 \right \|^2 \det D\psi^T D\psi = \det D\psi^T D\psi $$ Then we get $$\sqrt{\det D\psi^T D\psi} = \det J = R^{n-1}\prod_{i=1}^{n-1}\sin^{n-1-i}(\theta_i)$$ Note $R$ is now constant. Integrating we have $$\text{Area}(R\mathbb S^{n-1})= \int_{\text{angle domain}}\sqrt{\det D\psi^T D\psi} = \int_{\text{angle domain}} R^{n-1}\prod_{i=1}^{n-1}\sin^{n-1-i}(\theta_i) \text{d}^{n-1}\vec{\theta} $$ This is the same integral as the one of $\text{vol}(B_n)$ but without integration on $r$ which yielded a factor of $\int_0^1 r^{n-1}\text{d}r = \frac{1}{n}$. so we have $$\cdots = nR^{n-1}\text{vol}(B_n)$$ Very messy but I hoped it helped, at least for someone who sees this in the future.