Solve for volume of ellipsoid $(\mathbb x- \mathbf{\mu})^T \Sigma^{-1}(\mathbb x- \mathbf{\mu})=1/a$ For following ellipsoid:
$$(\mathbb x- \mathbf{\mu})^T \Sigma^{-1}(\mathbb x- \mathbf{\mu})=1/a \text{, where } x \in \mathbb R^n, \Sigma \succ0 \text{ and } a>0.$$
How to prove its volume is $a\pi\sqrt{\det \Sigma}$?
 A: The determinant follows directly by realizing that the volume element under coordinate changes transforms as  the Jacobian of the coordinate change, see here: https://en.wikipedia.org/wiki/Volume_element
Other than that, the formula is not correct for $x \in \mathbb R^n$. I give an explanation by doing it "manually" as follows:
The volume of
$$
(\mathbb x- \mathbf{\mu})^T \Sigma^{-1}(\mathbb x- \mathbf{\mu})=1/a
$$
remains constant w.r.t. changes of $\mu$. Further, represent the matrix  $\Sigma$ by its diagonalization   $\Sigma = \mathbb U^T \mathbb \Lambda \mathbb U $ where $ \mathbb U$ is a unitary transformation and $\mathbb \Lambda $ is the diagonal matrix of eigenvalues with elements $\lambda_i$. Then again, the volume remains constant under unitary transformations. So we may equivalently consider 
$$
\mathbb x^T \mathbb \Lambda^{-1} \mathbb x=1/a
$$
Taking square roots of the diagonal elements in $\mathbb \Lambda$ allows to write $\mathbb \Lambda = \mathbb \Gamma \mathbb \Gamma^T$, with $\mathbb \Gamma $ a diagonal matrix with elements $\gamma_i$. Letting  $\mathbb \Gamma^{-1} \mathbb x = \mathbb y$ we have that each coordinate is modified by a factor $\gamma_i$, i.e. the total volume is modified by a factor $\prod_i \gamma_i$, multiplied by the volume in $y$-space which is then the volume of a ball with radius $r$ given by 
$$
\mathbb y^T \mathbb y=1/a = r^2
$$
Now since $y \in \mathbb R^n$, this volume is not given as easily as in your formula. Rather, the volume of the ball is $\frac{1}{a^{n/2}} V_n$ where $V_n$ is the volume of the unit ball in $n$ dimensions which can be found e.g. here: 
https://en.wikipedia.org/wiki/Volume_of_an_n-ball
As to the other factor, the determinant, it can easily be seen that this derives indeed from the determinant since 
$$
\sqrt{\det \Sigma} = \sqrt{\det \mathbb U^T} \sqrt{\det \mathbb \Lambda} \sqrt{\det \mathbb U } = \sqrt{\det \mathbb \Lambda} = \sqrt{\prod_i \lambda_i} = \prod_i \gamma_i
$$
which we already established.
