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**ThePerfectHacker** I assume the volume you are trying to find is the set $\displaystyle V=\{ (x,y,z)| (x+y+z)^2+(x-2y-z)^2+(2x-y+z)^2 \leq h^2 \}$ (where $\displaystyle h>0$).

The Jacobian theorem says,

$\displaystyle \iiint_R f = \iiint_{\bold{g}^{-1}(R)} (f \circ \bold{g}) \cdot | \det D(\bold{g})| $

Here $\displaystyle \bold{g}:\mathbb{R}^3 \to \mathbb{R}^3$ is a differenciable vector function.

And $\displaystyle \det D (\bold g)$ is the determinant of the Frechet' derivative i.e. the Jacobian.

Of course, $\displaystyle \bold{g}^{-1}(R)$ is the set $\displaystyle \{ (x,y,z)\in \mathbb{R}^3 | \bold{g}(x,y,z) \in R \}$

It is conventient to have $\displaystyle \bold{g}$ be such a function so that $\displaystyle \bold{g}(x+y+z,x-2y-z,2x-y+z)=(x,y,z)$.

If we let $\displaystyle A = \left( \begin{array}{ccc}1&1&1\\1&-2&-1\\2&-1&1 \end{array} \right) $ then $\displaystyle A(x,y,z)^T = (x+y+z,x-2y-z,2x-y+z)^T$.

Thus, $\displaystyle A^{-1} (x+y+z,x-2y-z,2x-y+z)^T = (x,y,z)^T$.

This means, the function $\displaystyle \bold{g}(\bold{x}) = A^{-1}\bold{x}$ (thinking of vectors in $\displaystyle \mathbb{R}^3$ as coloums).

Now, this is convenient because $\displaystyle \bold{g}^{-1}(V) = \{ (x,y,z)| \bold{g}(x,y,z) \in V \} = \{ (x,y,z) | x^2+y^2+z^2 \leq h^2 \}$

The volume of $\displaystyle V$ is given by,

$\displaystyle \iiint_V 1 = \iiint_{\bold{g}^{-1}(V)} |\det D(\bold{g})| $

But, $\displaystyle \det D(\bold{g}) = \det A^{-1} = (\det A)^{-1} = 3^{-1}$

Thus the volume is $\displaystyle \tfrac{4}{9}\pi h^3$ because the volume of a sphere is $\displaystyle \tfrac{4}{3}\pi h^3$