TITLE: Volume of hyperellipsoid
QUESTION [7 upvotes]: How can I compute the volume of the hyperellipsoid corresponding to a Mahalanobis distance $r^2 = (x-\mu)^{T}\Sigma^{-1}(x-\mu)$? 
I'm a bit confused because the answer involves $r$: 
$$V = V_{d} |\Sigma|^{1/2}r^{d}$$ with $V_{d}$ as the volume of a d-dimensional unit hypersphere. I have seen that some statements of this problem describe $V_{d}$ as:
$$V_{d} = \left\{ \begin{array}{ll}
\pi^{d/2}/(d/2)! \;\;\; \text{for d even}\\\
2^{d}\pi^{(d-1)/2}\left(\frac{d-1}{2}\right)!/(d)! \;\;\; \text{for d odd}
\end{array}\right.$$
I thought I was supposed to integrate $r^{2}$ over $r\in [0, 1]$ and the surface of a unit hypersphere, but that doesn't give the right answer. What is the right procedure? 
Thanks in advance

REPLY [4 votes]: You can put $r$ on the right hand side: $1 = (x-\mu)^T(r \Sigma^{1/2})^{-2} (x-\mu)$. Then the answer becomes clearer:
$V = V_d\ \det(r\Sigma^{1/2}) = V_d\ \det(\Sigma)^{1/2}r^d$.