Double integrals over rectangles — Fubini's theorem: Visual walkthrough

Step 1 — What are we even measuring?

WHAT. We have a flat rectangular floor and a curved roof floating above it. We want the amount of space trapped between floor and roof.

WHY. Everything that follows is just a clever way to measure this one blob of space. If you can see the blob clearly, every later step is bookkeeping.

PICTURE. Look at the figure. The black rectangle on the ground is our floor. Its corners are fixed by four plain numbers:

$R = how far x runs [a, b] \times how far y runs [c, d]$

$x$ runs left-to-right, from the number $a$ to the number $b$ .
$y$ runs front-to-back, from the number $c$ to the number $d$ .
The red surface floating above is the roof, written $z = f (x, y)$ : for every floor point $(x, y)$ , the roof height there is the number $f (x, y)$ .

The whole quantity we want is the volume $V$ of the red-capped solid.

Figure — Double integrals over rectangles — Fubini's theorem

Step 2 — Approximate the volume with tiny towers

WHAT. Before slicing cleanly, we build the volume out of little rectangular columns — the honest, brute-force definition.

WHY. This is where the volume comes from. We chop the floor into a grid of small tiles, stand a tower on each tile as tall as the roof above it, and add up the towers. This is the same trick as Riemann sums, just with tiles instead of intervals.

PICTURE. Each floor tile has width $Δ x$ and depth $Δ y$ , so its area is

$Δ A = tile width Δ x \cdot tile depth Δ y .$

Pick any point $(x_{i}^{*}, y_{j}^{*})$ inside tile number $(i, j)$ — the star just means "some sample point in that tile." The tower on it has

$tower volume = height at the sample f (x_{i}^{*}, y_{j}^{*}) \cdot base area Δ A .$

Add every tower:

$S = \sum_{i} \sum_{j} f (x_{i}^{*}, y_{j}^{*}) Δ A .$

The red tower in the figure is one such term; the grey towers are its neighbours.

The problem: this is a double limit — brutal to evaluate directly. Steps 3–6 turn it into two easy one-dimensional integrals.

Step 3 — Freeze one direction: make a single slice

WHAT. Instead of tiny towers, take a big knife and cut the solid with one flat vertical wall. Freeze $x$ at some value $x_{0}$ and slice straight back through the solid.

WHY. On that frozen wall, $x$ is no longer a variable — it is a locked number. So the roof, seen only on this wall, is a plain 1-variable curve $y \mapsto f (x_{0}, y)$ . A curve like that bounds an ordinary flat area, and flat areas we already know how to get from single-variable calculus.

PICTURE. The red shaded face in the figure is the cut. Its height at back-position $y$ is $f (x_{0}, y)$ ; it stretches from $y = c$ to $y = d$ . Its area — call it $A (x_{0})$ — is the ordinary integral of that height:

$A (x_{0}) = \int_{c}^{d} roof height along the frozen wall f (x_{0}, y) d y .$

Read $d y$ as "sweep $y$ from $c$ to $d$ , adding thin vertical strips of the red face."

Step 4 — Slide the slice: area becomes a function of position

WHAT. Now unfreeze $x$ and let the cutting wall slide from the left edge $x = a$ to the right edge $x = b$ . At each stopping point the face has a (possibly different) area.

WHY. The solid isn't one slice — it's a continuous stack of them. To describe the whole stack we need to know how the slice-area changes as the wall slides. That gives us a single new function of one variable.

PICTURE. The figure plots the sliding wall at three positions. Below, the black curve is

$A (x) = \int_{c}^{d} f (x, y) d y,$

the slice-area as a function of where the wall sits. The red dot marks $A (x_{0})$ from Step 3 — just one point on this curve. Where the roof is tall and wide, $A (x)$ is large; where the solid pinches, $A (x)$ dips.

Step 5 — Stack the slices back into a volume

WHAT. Give each slice a tiny thickness $d x$ and glue them side by side to rebuild the solid.

WHY. A slice with face-area $A (x)$ and thickness $d x$ is a thin slab of volume $A (x) d x$ . Adding all slabs from $x = a$ to $x = b$ reassembles the exact solid — this is precisely Volume by slicing (single-variable), the "loaf of bread" idea, one dimension up.

PICTURE. The figure shows the loaf rebuilt from slabs; the red slab is the one at position $x$ , with thickness $d x$ drawn thick for visibility.

$V = \int_{a}^{b} slab face area A (x) slab thickness d x = \int_{a}^{b} (\int_{c}^{d} f (x, y) d y) d x .$

Reading it inside-out: first the inner $\int_{c}^{d} \dots d y$ builds one slab's face (Step 3), then the outer $\int_{a}^{b} \dots d x$ sweeps that slab across (Step 5). This nested form is the iterated integral — the "FREEZE, FILL, FILE" from the parent note.

Step 6 — Slice the OTHER way: nothing forced our choice

WHAT. Repeat Steps 3–5, but freeze $y$ first and slide the wall front-to-back instead of left-to-right.

WHY. In Step 3 we chose to freeze $x$ . That choice was arbitrary — the solid doesn't care which way we point the knife. Freezing $y$ gives a slice-area function of $y$ :

$B (y) = \int_{a}^{b} f (x, y) d x,$

and stacking those slabs gives the same solid, hence the same volume.

PICTURE. The figure shows the two knife directions side by side over the same solid: the Step-3 slice (grey, cuts along $y$ ) and the new slice (red, cuts along $x$ ). Same loaf, two grains.

$V = \int_{c}^{d} face area at fixed y B (y) d y = \int_{c}^{d} (\int_{a}^{b} f (x, y) d x) d y .$

Step 7 — The edge cases: where does this reasoning break?

WHAT. We check the corners: zero heights, negative heights, and roofs so wild that slicing lies to us.

WHY. Contract rule: the reader must never meet a case we skipped. Three matter.

PICTURE. Three panels.

Panel A — $f = 0$ (degenerate). Flat floor, no roof: every slice-area is $0$ , so $V = 0$ . Sanity holds.
Panel B — $f < 0$ (below the floor). If the roof dips under the $x y$ -plane, that slab counts as negative volume. The double integral is therefore a signed volume: space above the floor minus space below. Slicing still works verbatim — areas just carry a sign.
Panel C — $f$ not integrable. If $f$ blows up (unbounded, non-integrable), a slice-area $A (x)$ may not even be a finite number, or the two orders can disagree. Fubini's guarantee vanishes. This is why the hypothesis " $f$ continuous / integrable on $R$ " is not decoration.

The one-picture summary

Everything in one frame: the solid (Step 1–2), one red slice (Step 3), the slice-area profile $A (x)$ (Step 4), the slabs restacked (Step 5), and the twin knife direction (Step 6) — all labelled so you can read the derivation off the picture alone.

Recall Feynman retelling — the whole walkthrough in plain words

You've got a weird blob sitting on a rectangular table, and you want how much stuff is in it. First you guess by covering the table with tiny squares and standing a little tower on each — add the towers, that's roughly the volume, and shrinking the squares makes it exact. But adding zillions of towers is horrible. So instead you grab a knife. Freeze one direction and slice: each cut shows a flat face, and a flat face has an ordinary area you already know how to measure. Now slide the knife along and watch how the face-area grows and shrinks — that's a single humble curve $A (x)$ . Give every slice a sliver of thickness and glue them back: the areas, added up, rebuild the exact volume. The punchline: you could've sliced the loaf the other grain and gotten the same loaf — so the same number. That "same loaf, either grain" is Fubini. The only fine print: it must be a real, tame loaf — if the roof shoots off to infinity in a nasty way, slicing can lie, and you have to be more careful.

Connections

Double integrals over rectangles — Fubini's theorem — the parent result these pictures prove.
Riemann sums — Step 2's tower-sum is a 2-D Riemann sum.
Volume by slicing (single-variable) — Steps 3–5 ARE this idea, lifted up a dimension.
Change of order of integration — uses Step 6's freedom when one order is hard.
Fubini–Tonelli theorem — repairs Step 7 Panel C with an absolute-value hypothesis.
Double integrals over general regions — replaces the flat floor with a curvy one.
Triple integrals — slice a 4-D "solid" in three directions.