<s>[INST] Find Volume V bounded by surface $\sqrt[3]{ x^2}+\sqrt[3]{ y^2}+\sqrt[3]{ z^2}=1$
I try this substitution $x={u}^3$,$y={v}^3$ and $z={k}^3$ ,and I get sphere with radius 1
 [/INST] Let $x=r\sin^3\alpha\cos^3\beta$, $y=r\sin^3\alpha\sin^3\beta$ 
 and $z=r\cos^3\alpha$.
Thus, $J=9r^2\sin^5\alpha\cos^2\alpha\sin^2\beta\cos^2\beta$ and the volume it's
$$9\int_{0}^1r^2dr\int_{0}^{\frac{\pi}{2}}\sin^5\alpha\cos^2\alpha d\alpha\int_{0}^{2\pi}\sin^2\beta\cos^2\beta d\beta=\frac{4\pi}{35}$$
About Jacobian see here: https://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant
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