Double integrals over general regions — Type I and II: Visual walkthrough

Before line one, three plain-word promises:

Step 1 — What "volume under a surface" actually means

WHAT. We chop the floor into tiny square patches. Over one patch sitting at $(x, y)$ with area $Δ A$ , the tent has roughly constant height $f (x, y)$ , so the little box of stuff above it has volume $f (x, y) Δ A$ . Add all boxes: total volume $\approx \sum f (x, y) Δ A$ .

WHY. We can only define the double integral as the limit of this sum as the patches shrink: $\iint_{D} f d A = lim_{patches \to 0} \sum_{patches in D} box height f (x, y) box floor Δ A .$ Here $\sum$ means "add up", and the subscript "patches in $D$ " is the crucial part — we only count patches that land inside $D$ .

PICTURE. The red boxes stand only on the shaded region $D$ ; the height of each is the tent value $f$ at that spot.

Figure — Double integrals over general regions — Type I and II

The problem this page solves: that " $\sum$ over patches in $D$ " is hard when $D$ has slanted edges. We need to turn it into ordinary one-variable integrals with clear limits.

Step 2 — The one thing we know how to do: rectangles

WHAT. Over a rectangle $R = [a, b] \times [c, d]$ (the notation means $a \leq x \leq b$ AND $c \leq y \leq d$ — a box with vertical sides at $x = a, b$ and horizontal sides at $y = c, d$ ), the sum is easy because both variables run between constants. We already know how to handle this.

WHY. With constant limits we can add the boxes in an organized way: first pile up one column, then slide across. That organized adding is the iterated integral.

PICTURE. The rectangle $R$ has straight sides; every $x$ from $a$ to $b$ meets the same top ( $d$ ) and bottom ( $c$ ). Nothing bends.

So our whole strategy will be: smuggle the curved region $D$ into a rectangle.

Step 3 — Enclose $D$ in a rectangle

WHAT. Draw the smallest rectangle $R = [a, b] \times [c, d]$ that completely contains $D$ . Choose:

$a, b$ = the leftmost and rightmost $x$ -values $D$ reaches,
$c, d$ = the lowest and highest $y$ -values $D$ reaches,

so that $c \leq g_{1} (x)$ and $g_{2} (x) \leq d$ hold everywhere (the curves stay inside the box).

WHY. We only know how to integrate over rectangles (Step 2). By trapping $D$ inside $R$ , we can work on $R$ — provided we can make the corners of $R$ that stick out beyond $D$ count for nothing.

PICTURE. The teal rectangle hugs the plum region $D$ . The four corner slivers (still cream) are inside $R$ but outside $D$ — those are the parts we must silence.

Step 4 — Extend by zero (silence the outside)

WHAT. Invent a new height $F$ that equals the real height on $D$ and is flat-zero everywhere else: $F (x, y) = {f (x, y) 0 (x, y) \in D (on the region, keep the real tent) (x, y) \in / D (off the region, floor level)$ Read term by term: the top line keeps the tent where we want it; the bottom line flattens the tent to height $0$ in the corner slivers.

WHY. A box of height $0$ has volume $0$ . So integrating $F$ over the whole rectangle gives exactly the same volume as integrating $f$ over just $D$ : $\iint_{D} f d A = \iint_{R} F d A .$ The left side is what we want; the right side lives on a rectangle where we can iterate.

PICTURE. Same rectangle, but now the corner slivers are pressed down to the floor (height $0$ ). Only the tent over $D$ pokes up.

Step 5 — Iterate on the rectangle (Fubini)

WHAT. On a rectangle, Fubini's Theorem says we may add the boxes column-by-column: $\iint_{R} F d A = \int_{a}^{b} (one full column, x fixed \int_{c}^{d} F (x, y) d y) d x .$ Term by term:

the inner $\int_{c}^{d} F (x, y) d y$ freezes $x$ and slides $y$ up the column from $c$ to $d$ , adding the box heights in that one vertical strip;
the outer $\int_{a}^{b} (\dots) d x$ then slides that finished strip from left ( $a$ ) to right ( $b$ ).

WHY. This is the "add a column, then slide across" idea of Step 2, now written as symbols. It only works because the rectangle's limits $a, b, c, d$ are all constants.

PICTURE. One vertical teal strip at a fixed $x$ is highlighted; the arrow shows $y$ sweeping from $c$ up to $d$ . A ghosted strip shows the outer sweep sliding from $a$ to $b$ .

Step 6 — Collapse the dead zones (the punchline)

WHAT. Look at one column at a fixed $x$ . As $y$ climbs from $c$ to $d$ , the height $F$ is zero until $y$ reaches the bottom curve $g_{1} (x)$ , then equals the real $f$ up to the top curve $g_{2} (x)$ , then is zero again up to $d$ . Splitting the inner integral at those two curves: $\int_{c}^{d} F d y = = 0 \int_{c}^{g_{1} (x)} 0 d y + \int_{g_{1} (x)}^{g_{2} (x)} f d y + = 0 \int_{g_{2} (x)}^{d} 0 d y = \int_{g_{1} (x)}^{g_{2} (x)} f (x, y) d y .$ The two zero pieces vanish, leaving the tent-part only.

WHY. The dead zones contributed literally nothing (height $0$ ), so the column's true content runs only between the curves. This is exactly why the inner limits are the curves $g_{1} (x)$ and $g_{2} (x)$ : they are where the tent starts and stops along that column.

PICTURE. The same column, now colour-coded: grey dead segments below $g_{1}$ and above $g_{2}$ , orange live segment between them.

Feed this back into the outer integral of Step 5: $\iint_{D} f d A = \int_{a}^{b} \int_{g_{1} (x)}^{g_{2} (x)} f (x, y) d y d x$ Read it: outer = slide constant-to-constant across $x$ ; inner = climb curve-to-curve up each column. $■$

Step 7 — Edge case: is $D$ always Type I?

WHAT. No. A column can only have one entry-point and one exit-point for the formula to work — i.e. every vertical line hits $D$ in a single unbroken segment. If a vertical line enters, exits, then re-enters $D$ (a dumbbell or crescent), no single pair $g_{1}, g_{2}$ describes the column.

WHY. Step 6 assumed exactly two crossings: zero → live → zero. Three or more crossings breaks that split. The fix: cut $D$ into sub-regions $D_{1}, D_{2}, \dots$ , each simple, and add: $\iint_{D} f d A = \iint_{D_{1}} f d A + \iint_{D_{2}} f d A .$

PICTURE. Left: a good region — every vertical line crosses once in, once out. Right: a bad crescent where one vertical line crosses four times, chopped by a dashed cut into two Type I pieces.

Step 8 — Degenerate case: curves that touch (area collapses)

WHAT. At an $x$ where $g_{1} (x) = g_{2} (x)$ , the column has zero height — the inner integral is $\int_{g_{1}}^{g_{1}} f d y = 0$ . This is not an error; it is the honest boundary of the region.

WHY. In the between-curves area the two curves meet at the endpoints. Setting $f \equiv 1$ : $Area (D) = \int_{a}^{b} (g_{2} (x) - g_{1} (x)) d x,$ and at a touching point $g_{2} - g_{1} = 0$ , so that column adds nothing — exactly as the geometry demands. If the curves cross inside $[a, b]$ (top and bottom swap), you must split at the crossing and reassign which curve is $g_{2}$ (top) on each piece; otherwise $g_{2} - g_{1}$ goes negative and the sign flips.

PICTURE. A region between $y = x^{2}$ and $y = x$ : the columns have full height in the middle and pinch to zero height at the meeting points $x = 0$ and $x = 1$ .

The one-picture summary

Everything on this page in a single frame: enclose ( $a, b, c, d$ ) → silence the corners ( $F = 0$ ) → one column climbs $g_{1} \to g_{2}$ → slide $a \to b$ .

Recall Feynman retelling — the whole walkthrough in plain words

You want the amount of stuff under a bumpy tent, but the tent's floor is a weird slanted shape, not a neat box (Step 1). The only shape you know how to handle is a box, so you draw the smallest box that fits the whole tent inside it (Step 3). The box has corners that stick out past the tent's real floor — so you press those corner-tents flat to the ground, making them count as nothing (Step 4). Now you're allowed to measure inside the whole box the easy way: pick a thin vertical slice, add up the tent height along it from bottom to top, then slide that slice across (Step 5). But along any slice, the flattened parts are zero — the tent only exists between the bottom curve and the top curve — so the slice really only runs curve-to-curve (Step 6). That's the formula: slide left-to-right between fixed numbers, climb bottom-curve to top-curve inside each slice. If a slice ever pokes in and out of the shape more than once, cut the shape into simpler pieces first (Step 7). And where the two curves kiss, the slice has no height and quietly adds zero — which is exactly the edge of the region (Step 8).