Volume of an $n$-dimensional ellipsoid Consider the convex body $S_r \subset \mathbb{R}^n$ defined as follows
$$S_r := \{ \mathbb{x} \in \mathbb{R}^n : \| A \mathbb{x}\| \leq r \}$$
for some invertible $n \times n$ matrix $A$ and some real constant $r > 0$. Here, $\|\cdot \|$ denotes the Euclidean norm. How is the volume of $S_r$ calculated?
 A: Use the substitution rule (where $\phi : B\to\phi(B)$ is a diffeomorphism)
$$
\int_{\phi(B)}f(x)\,dx = \int_Bf(\phi(y))|\det\phi'(y)|\,dy.
$$
Here, $B = B_r(0)$ (the ball with radius $r$ and center $0$) and $\phi(y) = A^{-1}y$.
A: Note that the Lebesgue measure and the Euclidean norm are rotation invariant.
Note that $x \in S_r$ iff $|x^T A^TA x| \le r$ and since $A^TA$ is symmetric
and positive definite there is a unitary $U$ and a diagonal $\Lambda$
such that $A^TA = U \Lambda U^T$.
Hence the volume of $S_r$ is the same as the volume of $\{x | \sum_k \lambda_k x_k^2 \le r^2 \} =  \{x | \sum_k ({ \sqrt{\lambda_k} x_k})^2 \le r^2 \} = \{({r x_1 \over \sqrt{\lambda_1}},...,{r x_n \over \sqrt{\lambda_n}}) | \|x\| \le 1 \}$.
It is straightforward to see that the volume of the last set is 
${r^n \over \sqrt{\lambda_1 \cdots \lambda_n}} = {r^n \over |\det A|}$ times the
volume of the unit ball $\overline{B}(0,1)$.
Hence the problem reduces to that of finding the volume of the unit ball in $\mathbb{R}^n$.
Addendum: Given a measurable set $A$ and a map $L(x) = (x_1,...,\lambda x_k, ...,x_n)$ (exactly one of the coordinates is multiplied by $\lambda$), it
is straightforward to show that $m(L(A)) = |\lambda| m(A)$. Since the map
$x \mapsto ({r x_1 \over \sqrt{\lambda_1}},...,{r x_n \over \sqrt{\lambda_n}})$
can be written as the composition of $n$ such operations, we see that
\begin{eqnarray}
m(\{({r x_1 \over \sqrt{\lambda_1}},...,{r x_n \over \sqrt{\lambda_n}}) | \|x\| \le 1 \}) &=& |{r \over \sqrt{\lambda_1}}|m(\{(x_1,...,{r x_n \over \sqrt{\lambda_n}}) | \|x\| \le 1 \}) \\
&\vdots& \\
&=& {r^n \over \sqrt{\lambda_1 \cdots \lambda_n}} m( \{ x | \|x\| \le 1 \}) 
\end{eqnarray}
And more:
To see why $m(L(A)) = |\lambda| m(A)$ for a measurable $A$, we use the definition of Lebesgue outer measure. We have
$m(A) = \inf \{ \sum_k m(R_k) | \{R_k\}_k \text{ is a cover of }A \text{ by rectangles}\}$.
It is not difficult to see that $m(L(R)) = |\lambda| m(R)$ for a rectangle.
If ${\cal R}_A= \{ \{R_k\}_k |  \{R_k\}_k \text{ is a cover of }A \text{ by rectangles}\} $, then it is straightforward to see that
${\cal R}_{L(A)} = \{ \{L(R_k)\}_k | \{R_k\}_k \in {\cal R}_A \}$.
Hence $m(L(A)) = \inf \{ \sum_k m(L(R_k)) | \{R_k\}_k \text{ is a cover of }A \text{ by rectangles}\} = |\lambda| m(A)$.
