Triple and double integrals seem very logical; however, switching from a rectangular integral to a cylindrical or spherical one seems a bit messy. The derivations for switching $dV$ from $dxdydz$ to $rdrdθdz$ (cylindrical) or $ρ^2sin(φ)dρdθdφ$ (spherical) are completely logical. But going from the triple integral to an iterated integral is where I start to feel confused. I'll start with an example that I understand:
Say there is a region Q defined by $a\le{x}\le{b}, h_{1}(x)\le{y}\le{h_{2}(x)}, g_{1}(x,y)\le{z}\le{g_{2}(x,y)}$, or, equivalently, by $ρ_{1}\le{ρ}\le{ρ_{2}}, θ_{1}\le{θ}\le{θ_{2}}, \phi_{1}\le{\phi}\le{\phi_{2}}$. Notice that in this specific example, the bounds of the spherical dimensions are all constants. Thus, because of constant bounds, the triple Reimann sum definition of a triple integral is usable:
$\iiint_Q{f(x,y,z)dV}= \lim_{max(\Delta{ρ})\to 0, max(\Delta{θ})\to 0, max(\Delta{\phi})\to 0}\sum_{i=1}^n\sum_{j=1}^m\sum_{k=1}^p f(\rho_i\sin(\phi_k)cos(\theta_j), \rho_i\sin(\phi_k)\sin(\theta_j)\rho_i\cos(\phi_k))\rho^2_i\sin(\phi_k)\Delta\rho_i\Delta\theta_j\Delta\phi_k$
This definition makes it clear that the volume of the region will be found through physically multiplying out every one-dimensional portion of any $dV$, and thus it makes sense that the region Q's volume will be found. This can only be used, to the extent of my knowledge (which could absolutely be wrong), because the bounds of the spherical variables are constants. However, now imagine the bounds set above for the spherical coordinates were variable. The triple sum definition wouldn't be used anymore, but now we would use the definition with any $\Delta{V_i}$:
$\iiint_Q{f(x,y,z)dV}= \lim_{max\Delta{V_i}\to 0}\sum_{i=1}^nf(\rho_i\sin(\phi_i)cos(\theta_i), \rho_i\sin(\phi_i)\sin(\theta_i)\rho_i\cos(\phi_i))\Delta{V_i}$
$\iiint_Q{f(x,y,z)dV}= \iiint_Q{f(\rho\sin(\phi)cos(\theta), \rho\sin(\phi)\sin(\theta)\rho\cos(\phi))}dV$
$\iiint_Q{f(x,y,z)dV}= \iiint_Q{f(\rho\sin(\phi)cos(\theta), \rho\sin(\phi)\sin(\theta)\rho\cos(\phi))}\rho^2\sin(\phi)d{\rho}d{\theta}d{\phi}$
$\iiint_Q{f(x,y,z)dV}= \int_{\phi_{1}}^{\phi_{2}}\int_{\theta_{1}}^{\theta_{2}}\int_{\rho_1}^{\rho_2} {f(\rho\sin(\phi)cos(\theta), \rho\sin(\phi)\sin(\theta)\rho\cos(\phi))}\rho^2\sin(\phi)d{\rho}d{\theta}d{\phi}$ (remember, bounds are now variable)
That last step is where my confusion lies. Textbooks and math resources will immediately jump from the triple integral to an iterated integral with the given bounds, just as I have done above. The reason why this irks me is because it seems like the text is treating $d{\rho}d{\theta}d{\phi}$ as a pseudo-$dV$, and then applying Fubini's theorem with ${f(\rho\sin(\phi)cos(\theta),\rho\sin(\phi)\sin(\theta)\rho\cos(\phi))}\rho^2\sin(\phi)$ being the entire integrand. This would make sense if you got rid of the geometric meaning of $\rho, \theta$, and $\phi$ and instead put those variables as axes in a rectangular grid. So my question sums up to be: is intentionally getting rid of the geometric meaning of the spherical variables being used? Does the Fubini Theorem give reasoning to convert to an iterated integral as long as 3 $\Delta$(variable)'s are present (3 in this case, anyway)? Or do the bounds of a spherical or cylindrical triple integral need to be constant (which I doubt)?
I may be missing something large, and if so, I appreciate any help you can offer. Thank you for taking the time to read this!