= Solution
For general coordinates $q=(a,b,c)$ expressed as functions of $u=(u,v,w)$, the <chain rule> gives $dq=J\,du$, where
$$
J=\frac{\partial(a,b,c)}{\partial(u,v,w)}
=\begin{pmatrix}a_u&a_v&a_w\\b_u&b_v&b_w\\c_u&c_v&c_w\end{pmatrix}.
$$
The <Jacobian determinant> is the oriented volume scaling of this linearized map. Thus the unsigned <volume element> transforms as $da\,db\,dc=|\det J|\,du\,dv\,dw$. Where the coordinate change is invertible, the inverse <Jacobian determinant> is $1/\det J$, by the <Jacobian chain rule>.
For <cylindrical coordinates>, use $x=r\cos\phi$, $y=r\sin\phi$, and unchanged height $z$. Direct differentiation gives
$$
\frac{\partial(x,y,z)}{\partial(r,\phi,z)}
=\begin{pmatrix}\cos\phi&-r\sin\phi&0\\\sin\phi&r\cos\phi&0\\0&0&1\end{pmatrix},
$$
whose <determinant> is $r$. Therefore
$$
\boxed{dV=r\,dr\,d\phi\,dz,\qquad
\det\frac{\partial(r,\phi,z)}{\partial(x,y,z)}=\frac1r\quad(r>0).}
$$
Both directions are displayed because “from Cartesian to cylindrical” can refer either to the derivative of the new coordinates or to the factor used in transforming a Cartesian volume integral. They are reciprocals; the cylindrical axis is a coordinate degeneracy.
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