volume of tilted ellipsoid

Question

I'm supposed to calculate the volume of $$(2 x+y+z)^2 + (x+2 y+z)^2 + (x+y+2 z)^2 \leq 1$$ simplifying it gives $$6 (x^2 + y^2 + z^2) + 10 (x y + y z + x z) \leq 1$$

After drawing it using GeoGebra, I saw that it's a tilted ellipsoid inside the unit sphere, but I'm unable to think of how to solve this. I tried replacing the coordinates and I tried using spherical coordinates but I was unable to go anywhere with them.

The final answer is $\frac{\pi}{3}$ meaning it's $\frac{1}{4}$ of a unit sphere's volume, and still, I wasn't able to conclude anything useful from it. Any hints would be greatly appreciated.

Answer 1

Hint. Let $X=2x+y+z$, $Y=x+2y+z$, $Z=x+y+2z$ then the volume of the given ellipsoid $E$ is $$V=\iiint_E 1\, dxdydz=\iiint_{X^2+Y^2+Z^2\leq 1} f(X,Y,Z)\, dXdYdZ$$ where $f(X,Y,Z)$ is a suitable function. Since here the substitution is linear, the function $f$ is constant and the integral on the right reduces to this constant multiplied by the volume of the unit ball. Can you take it from here?

Answer 2

Consider the linear map $$A = \begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2\end{bmatrix}$$ and notice that your ellipsoid $E$ is the preimage of unit ball in $\Bbb{R}^3$ via $A$ (or image via $A^{-1}$).

Therefore

$$\operatorname{vol}(E) = \operatorname{vol}(A^{-1}(\operatorname{Ball}(\Bbb{R}^3))) = \left|\det A^{-1}\right| \operatorname{vol}(\operatorname{Ball}(\Bbb{R}^3)) = \frac1{\left|\det A\right|} \cdot \frac43\pi = \frac13\pi.$$