I'm currently studying Analysis on Manifolds by James Munkres and I'm trying to determine the surface area of the $n$-sphere $S^n(a)$ in terms of the $n-1$ ball $B^{n-1}(a)$ however I have trouble with one part of the computation. I proceeded as follows:
We can parametrize $S^n(a)=\{x\in \mathbb{R}^{n+1}\mid x_1^2+\dotsc +x_{n+1}^2 = a^2\}$ via the coordinate chart $\alpha:(0,2\pi)\times B^{n-1}(a)\rightarrow S^n(a)$ where
$$\alpha(t,x_3,\dotsc,x_{n+1}) = (\sqrt{a^2-x_3^2-\cdots-x_{n+1}^2}\cos t,\sqrt{a^2-x_3^2-\cdots-x_{n+1}^2}\sin t,x_3,\dotsc,x_{n+1}).$$
The surface area of $S^{n}(a)$ (in Munkres called volume) is then given by
$$v(S^n(a)) = \int_{(0,2\pi)\times B^{n-1}(a)}V(D\alpha)$$
where $V(D\alpha) = (\det D\alpha^{tr}D\alpha)^{1/2}$. So all that remains is to determine $V(D\alpha)$. This is however where I run into problems. I find that if we denote $r = \sqrt{a^2-x_3^2-\cdots-x_{n+1}^2}$ then
$$D\alpha = \begin{bmatrix} -r \sin t& \frac{-x_3}{r}\cos t & \frac{-x_4}{r}\cos t & \cdots & \frac{-x_{n+1}}{r}\cos t \\ r \cos t & \frac{-x_3}{r}\sin t & \frac{-x_4}{r}\sin t & \cdots & \frac{-x_{n+1}}{r}\sin t \\ 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1 \end{bmatrix}.$$
And $$D\alpha^{tr}D\alpha = \begin{bmatrix} r^2 & 0 & 0 & \cdots & 0 \\ 0 & \frac{x_3^2}{r^2}+1 & \frac{x_3x_4}{r^2} & \cdots & \frac{x_3x_{n+1}}{r^2} \\ 0 & \frac{x_4x_3}{r^2} & \frac{x_4^2}{r^2}+1 & \cdots & \frac{x_4x_{n+1}}{r^2} \\ \vdots & \vdots & \vdots & \ddots &\vdots \\ 0 & \frac{x_{n+1}x_3}{r^2} & \frac{x_{n+1}x_4}{r^2} & \cdots & \frac{x_{n+1}^2}{r^2}+1 \end{bmatrix}.$$
Now I have no idea how to compute the determinant of $D\alpha^{tr}D\alpha$ even though I strongly suspect its value to be $a^2$ based on computations for $n=2$ and $n=3$. Is there any way to see that this determinant is indeed $a^2$?
The lower right block can be written using $$ \pmatrix{x_3^2 & x_3x_4 & \dots &x_3x_{n+1} \\ \vdots &&&\vdots\\ x_{n+1}x_3 & x_{n+1}x_4 & \dots& x_{n+1}^2} = \pmatrix{x_3\\\vdots\\x_{n+1}}\pmatrix{x_3&\dots&x_{n+1}} $$ as $$ I + r^{-2} \pmatrix{x_3\\\vdots\\x_{n+1}}\pmatrix{x_3&\dots&x_{n+1}} $$ whose determinant can be computed by Weinstein-Aronszajn identity to be equal to $$ 1 + r^{-2}\pmatrix{x_3&\dots&x_{n+1}} \pmatrix{x_3\\\vdots\\x_{n+1}} = 1 + r^{-2}(a^2-r^2) = r^{-2}a^2 $$