Integrate using polar coordinates first, then integrate with respect to z. Thus the integrand has an extra factor of r, as it did above. You can also prove this by taking the determinant of the jacobian. Here is a matrix of partials as r θ and z map to x y and z. Verify that the determinant is r.
cos(θ) sin(θ) 0 -r×sin(θ) r×cos(θ) 0 0 0 1
As an exercise, find the volume of a slice of cylinder bounded by r < 1, z > 0, and z < y. In cylindrical coordinates, the latter constraint is written z < r×sin(θ). The integrand that defines volume becomes r, the determinant of the jacobian.
Integrate with respect to z first, giving r2×sin(θ). Integrate with respect to r, giving sin(θ)/3. Finally integrate with respect to θ, giving a volume of 2/3.