Let $A \in Mat_{n,n}(\mathbb{R})$ be symmetrical and positive definite.
The function $f: \mathbb{R}^n \to \mathbb{R}$ is defined as $f(\vec{x}) = \langle A\vec{x},\vec{x} \rangle$.
One has to find the Volume of the the "ellipsoid" described by $f(\vec{x}) \leq 1$. Or in other words, the volume of the folowing set: $$\{\vec{x} \in \mathbb{R}^n : \langle A\vec{x},\vec{x} \rangle \leq 1\}$$
One can assume we already know the volume of $\omega_n$, the $n$-dimensional Sphere.
I found it on an old analysis paper on and the question seems to be very interesting, but I have no idea on how this exercise can be solved, so your help is greatly appreciated.
Update Looking at the hint from the comment this seems like a good sketch of proving this but I would love if someone could "fill in the blanks" / find the errors.
Since A can be diagonalized $A = B^{-1}PB$ with $P$ being filled with the positive eigenvalues we can rewrite our equation:
$f(\vec{x}) = \langle A\vec{x},\vec{x} \rangle = \langle B^{-1}PB\vec{x},\vec{x} \rangle$
Let's now look at $g(\vec{x}) = \langle P\vec{x},\vec{x} \rangle = \lambda_1x_1^2 + \cdots + \lambda_nx_n^2$
So our set becomes $\{\lambda_1x_1^2 + \cdots + \lambda_nx_n^2 \leq 1\}$
We now use the substitution $v_i = \sqrt{\lambda_i} x_i$ which results in $C = \{v_1^2 + \cdots + v_n ^2 \leq 1\}$ we know that the volume of $C$ is $\omega_n$ Since we have substituted we must use the multidimensional substitution rule:
Since $v_i = \sqrt{\lambda_i} x_i \implies x_i = \frac{v_i}{\sqrt{\lambda_i}}$
We can now build the functional matrix of the change in variable:
$$\begin{bmatrix} \frac{\partial x_i}{\partial v_i} & \cdots \\ \cdots \end{bmatrix} = \begin{bmatrix}\frac{1}{\sqrt{\lambda_i}} & \cdots \\ \cdots \end{bmatrix}$$ building the determinant of the functional matrix gives me $\frac{1}{\sqrt{\det P}}$
Using the substitution we therefore get $\frac{1}{\sqrt{\det P}}\omega_n$
this however only works for a matrix $P$ which only has positive values on its diagonal?
Thank you for helping.