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!

  • 2
    $\begingroup$ Does the answer in this or the answer to this answer your question? $\endgroup$ – André Armatowski Aug 2 '20 at 13:33
  • $\begingroup$ So what I've gathered is: I can use the double/triple Reimann sum definition of a double/triple integral (which must have constants as the limits of integration) for a region that does not have constant bounds, but instead of summing the original integrand function, I can simply defined another function that equals my original function's values in the oddly shaped region and 0 outside the region? If this is correct, how does this translate to the actual algebra where I have variable limits on my integrals? $\endgroup$ – Lightbulb Aug 2 '20 at 14:08
  • $\begingroup$ Yes,If you are talking about evalutaing something like $$\int_{a}^{b}\int_{g(x)}^{h(x)}f(x,y)\,dy\,dx$$ then suppose $\alpha\leq \ g(x) \leq h(x) \leq \beta$ for every $x\in [a,b]$ where $\alpha,\beta\in\mathbb{R}$ then for any $t\in[a,b]$ $$\int_{\alpha}^{\beta}f(t,y)\,dy=\int_{\alpha}^{g(t)}0\,dy+\int_{g(t)}^{h(t)}f(t,y)\,dy+\int_{h(t)}^{\beta}0\,dy$$. Is this what you meant? $\endgroup$ – André Armatowski Aug 2 '20 at 15:47
  • $\begingroup$ Absolutely!! That clears up double integration perfectly, thank you so much. Now in extending to triple integrals: the inside integral converting to variable bounds follows easily, but what about the bounds of the second integral? Is this correct?: let's say you were integrating with respect to z first, and its variable bounds are $g_1(x,y)$ and $g_2(x,y)$. If you were to change to these bounds like you have done, the inside integral with respect to z would not be 0 if the next integral has bounds (that could be variable) that lie in the oddly-shaped region within the constant bounds. Correct? $\endgroup$ – Lightbulb Aug 2 '20 at 17:19
  • $\begingroup$ I am not entirely sure that works since we only integrate in the "$z$" direction once, we must make sure to not leave any extra room in this direction. I would instead consider applying the same idea as that for double integrals: Assuming that we want to evaluate $$\int_{a}^{b}\int_{h_{1}(x)}^{h_{2}(x)}\int_{g_{1}(x,y)}^{g_{2}(x,y)}f(x,y,z)\,dz\,dy\,dx.$$ We split it up as an double inner integral by fixing any $t\in [a,b]$. We are then left with $$\int_{h_{1}(t)}^{h_{2}(t)}\int_{g_{1}(t,y)}^{g_{2}(t,y)}f(t,y,z)\,dz\,dy$$ Here since $t$ is fixed we have reduced it to my above comment. $\endgroup$ – André Armatowski Aug 2 '20 at 17:39

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