Volume of an $n$-dimensional ellipsoid

Question

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?

Answer 1

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 $m(L(A)) = \inf \{ \sum_k m(L(R_k)) | \{R_k\}_k \text{ is a cover of }A \text{ by rectangles}\} = |\lambda| m(A)$.

Answer 2

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 $\phi(y) = A^{-1}y$.