volume of tilted ellipsoid

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?


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

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

Tags:

Volume

Integration

Vector Analysis

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