Volume of ellipsoid via determinant From Convex Optimization by Boyd & Vanderberge:  

Let $$E = \{v : \|Av + b\|_2 \le 1\}$$ be the equation of a general
   ellipsoid for $A \in S^n_{++}$ and $b \in \Bbb R^n$, where $S^n_{++}$
   means positive definite symmetric matrices of size $n$. Then, the volume of $E$ is proportional to $\det \left( A^{-1} \right)$.

Can someone explain or prove why this is true?  I have no idea how to show this and this is mentioned in passing.
 A: This is a consequence of the multidimensional change of variables formula which states that (under suitable hypotheses) the following formula holds:

\begin{equation}
\int_{g(B)} f = \int_B (f \circ g) \cdot |\det g'|,
\end{equation}
where $g'$ is the gradient of $g$.

For your case, choose $f = 1$ (the constant function), let $B = \{\xi \in \mathbb{R}^n : \lVert \xi \rVert_2 \leq 1\}$ be the unit ball in $\mathbb{R}^n$, and define $g: \mathbb{R}^n \to \mathbb{R^n}$ by
\begin{equation}
g(\xi) = A^{-1} \cdot(\xi - b)
\end{equation}
Then, you can easily verify that $E = g(B)$, so that by the formula above,
\begin{align}
\text{vol}(E) &= \int_E 1 \\
&= \int_{g(V)} 1 \\
&= \int_B (1 \circ g) \cdot |\det g'| \\
&= \int_B 1 \cdot |\det A^{-1}| \\
&= \left( \det A^{-1} \right) \cdot \text{vol}(B)
\end{align}
In the last line, I got rid of the absolute values because of positive-definiteness. So, the volume of $E$ is indeed proportional as claimed, with the constant of proportionality being the volume of the unit ball in $\mathbb{R}^n$.
