I recently made a post on triple integrals with cylindrical/spherical coordinates, and that lead me to a breakthrough in my own understanding of double and triple integrals with variable bounds of integration in the iterated integrals. I wanted to check to make sure if this logic is correct, and if it’s not, please correct me!

Define a function $ M(x, y, z) $ on region Q in space with bounds $alphale{x}le{beta}, h_1(x)le{y}le{h_2(x)}, g_1(x,y)le{z}le{g_2(x,y)}$ upon which we would like to integrate. Now construct a rectangular prism that completely surrounds Q, which we will call P, with bounds $ale{x}le{b}, cle{y}le{d}, ele{z}le{f}$ where $$ale{alpha}le{beta}le{b}$$ $$cle{h_1(x)}le{h_2(x)}le{d}$$ $$ele{g_1(x,y)}le{g_2(x,y)}le{f}$$ Define the following function:

$$N(x, y, z)=

begin{cases}

M(x, y, z), &text{(x, y, z) β Q} \

0, &text{(x, y, z) β P and (x, y, z) β Q}

end{cases}$$

Now, partition P and construct the triple integral on the region P:

$$lim_{maxDelta{x} to 0, maxDelta{y} to 0, maxDelta{z} to 0}sum_{i=1}^thetasum_{j=1}^{phi}sum_{k=1}^{rho}N(x_i, y_j, z_k)Delta{z_k}Delta{y_j}Delta{x_i} = intint_Pint{N(x, y, z)dzdydx}$$

Use Fubini’s Theorem:

$$intint_Pint{N(x, y, z)dzdydx} = int_a^bint_c^dint_e^fN(x, y, z)dzdydz$$

Notice, however, that when integrating upon any fixed $(x, y)$, the inside z-integral would become:

$$int_e^fN(x, y, z)dz = int_e^{g_1(x, y)}N(x, y, z)dz + int_{g_1(x,y)}^{g_2(x,y)}N(x, y, z)dz + int_{g_2(x,y)}^fN(x, y, z)dz$$

$$int_e^fN(x, y, z)dz = 0 + int_{g_1(x,y)}^{g_2(x,y)}N(x, y, z)dz + 0$$

$$int_e^fN(x, y, z)dz =int_{g_1(x,y)}^{g_2(x,y)}M(x, y, z)dz$$

The change from $N$ to $M$ is significant but can be done because on $g_1(x,y)le{z}le{g_2(x,y)}$, $N=M$.

All the steps up to this point seem logical. However, these next parts are where I feel slightly iffy, so it could be wrong:

Now apply a somewhat similar logic to y: use any fixed x and break up the iterated integrals as such:

$$int_c^dint_e^fN(x, y, z)dzdy = int_c^{h_1(x)}int_e^fN(x,y,z)dzdy + int_{h_1(x)}^{h_2(x)}int_e^fN(x,y,z)dydz+int_{h_2(x)}^dint_e^fN(x,y,z)dydz$$

With the outer two iterated integrals, only y values that lie in P would be evaluated, which would thus make them equal 0. The middle iterated integral stays and the variable bounds of the z-integral can be added using the previous analysis:

$$int_c^dint_e^fN(x,y,z)dzdy = int_{h_1(x)}^{h_2(x)}int_{g_1(x,y)}^{g_2(x,y)}M(x,y,z)dzdy$$

Again, a similar logic can be applied to x (except no variables need to be fixed now): break the iterated integral up into 3 iterated integrals (with constants as bounds on the inner integrals), realize that the outer 2 iterated integral expressions are 0 because only x-values in P are evaluated, and thus only the middle integral remains with the bounds of $alpha$ and $beta$. Plug in the variable-bounded, doubly-iterated integral discussed earlier, and we finally have our solution:

$$int_a^bint_c^dint_e^fN(x,y,z)dzdydx = int_{alpha}^{beta}int_{h_1(x)}^{h_2(x)}int_{g_1(x,y)}^{g_2(x,y)}M(x,y,z)dzdydx = intint_Qint M(x,y,z)dzdydx$$

I feel very strong about the work up until after the discussion of the inner-most iterated integral. Is all of this logic sound? If not, what should I change to make this valid? I mainly want to stick with the triple Reimann sum definition of a triple integral here, so if you see an error, try to keep the correction within that definition (but if that cannot be done, that is perfectly ok!) Thank you for reading!