I have the following problem. We are in $\mathbb R^{2n+1}$ and we want to compute $$\int_S d\mathcal H^{2n}$$ where $S:=\{(x_1,\dots,x_{2n},t): \sqrt{x_1^2+ \dots+x_{2n}^2}=r, |t|\leq r^2\}.$
We have a parametrization of $S$ given by $\phi: \mathbb S\to \mathbb R^{2n+1}, \, (x_1,\dots,x_{2n})\mapsto (x_1,\dots,x_{2n}, x^2_1+\dots + x^2_{2n}),$ where $\mathbb S$ is the $2n-1$-dimensional sphere in $\mathbb R^{2n}$ of radius $r$.
So $$\int_S d\mathcal H^{2n}=\int_\mathbb S J_\phi dx_1\dots dx_{2n},\\ J_\phi:=\sqrt {det(\mathcal J^t\phi \cdot \mathcal J\phi) }; $$ $\mathcal J\phi$ is the jacobian matrix of $\phi$.
Now, $\mathcal J^t\phi \cdot \mathcal J\phi= I_{2n}+4A,$ where $I_{2n}$ is the identity matrix $2n\times 2n$ and $A=(a_{ij})=x_i x_j$, so $\mathcal J^t\phi \cdot \mathcal J\phi$ is symmetric. My problem is: how can I compute the determinant? Is there an easy way? Is it correct to calculate the integral in this way?
Thanks in advance, any kind of help is really appreciated.
Edit:
Ok, the parametrization is clearly wrong, so I think a correct one is simply the identity: $$\sigma: \mathbb S \times [-r^2,r^2] \to \mathbb R^{2n+1}, \quad (x,\rho)\mapsto (x,\rho). $$ In this case $J\sigma=1,$ so
$$\int_S d\mathcal H^{2n}=\int_{-r^2}^{r^2}\int_\mathbb S dx_1\dots dx_{2n} d\rho=|\mathbb S|2r^2=2n \omega_{2n} r^{2n-1} \, 2r^2=2n \omega_{2n}2 r^{2n+1}.$$
But this seems to be wrong, since on my notes I've written that $$\int_S d\mathcal H^{2n}=(2n-1) \omega_{2n}2 r^{2n+1}.$$