Double integrals over general regions — Type I and II: Worked examples

Before we start, remember the two engines from the parent note:

$\iint_{D} f d A = \int_{a}^{b} \int_{g_{1} (x)}^{g_{2} (x)} f d y d x (Type I, vertical strips)$ $\iint_{D} f d A = \int_{c}^{d} \int_{h_{1} (y)}^{h_{2} (y)} f d x d y (Type II, horizontal strips)$

Everything below is just these two, applied to harder-and-harder regions.

The scenario matrix

Every double integral over a general region falls into one of these case classes. The table lists what makes each one special, and which worked example below handles it.

#	Case class	What makes it tricky	Example
A	Region needs both boundaries as functions of $x$	top and bottom are curves	Ex 1
B	Boundaries cross / swap — must find intersection	which curve is "top" flips	Ex 2
C	Region not Type I as a whole — must split	one description won't cover it	Ex 3
D	Only Type II works (Type I would need pieces)	left/right boundary is a single curve, top/bottom is not	Ex 4
E	Forced order-swap (no antiderivative the naïve way)	$\int e^{y^{2}}$ -style dead ends	Ex 5
F	Degenerate / zero-area limiting case	region collapses to a line	Ex 6
G	Sign of the integrand — negative $f$ , signed volume	answer can be negative	Ex 7
H	Word problem (mass / average value)	translate physics → limits	Ex 8
I	Exam twist — region defined by inequalities only	you must draw it yourself	Ex 9

We now clear the whole table.

Cell A — both boundaries are curves of $x$

Worked example Ex 1 — under a parabola cap

Compute $\iint_{D} (x + y) d A$ where $D$ lies between $y = x^{2}$ (bottom) and $y = 2 - x^{2}$ (top).

Forecast: the two parabolas make a lens shape. Guess: is the answer positive? (Both curves live in the region $y > 0$ near the middle, and $x + y$ is mostly positive there — so yes.)

Figure s01 below draws the lens: the coral curve is the bottom $y = x^{2}$ , the lavender curve is the top $y = 2 - x^{2}$ , the mint fill is $D$ , and the dashed vertical lines are the strips we sweep. Notice the corners at $x = \pm 1$ where the two curves meet.

Find where the strips start and stop (the $x$ -limits). Set the curves equal: $x^{2} = 2 - x^{2} \Rightarrow 2 x^{2} = 2 \Rightarrow x = \pm 1$ . Why this step? Vertical strips only exist where a bottom curve sits below a top curve; outside $[- 1, 1]$ the "cap" curve is below the parabola, so there is no region. In the figure these are exactly where the coral and lavender curves cross.
Identify top vs bottom. At $x = 0$ : bottom $= 0$ , top $= 2$ . So $g_{1} (x) = x^{2}$ , $g_{2} (x) = 2 - x^{2}$ . Why? Always test a sample point — never trust the order the curves were named in. The middle dashed strip in the figure confirms lavender sits above coral.
Set up Type I: $\int_{- 1}^{1} \int_{x^{2}}^{2 - x^{2}} (x + y) d y d x .$
Inner integral. $\int_{x^{2}}^{2 - x^{2}} (x + y) d y = [x y + \frac{y ^{2}}{2}]_{x^{2}}^{2 - x^{2}}$ . Top value: $x (2 - x^{2}) + \frac{( 2 - x ^{2} ) ^{2}}{2}$ . Bottom value: $x \cdot x^{2} + \frac{x ^{4}}{2}$ . Why this step? We integrate the inner variable $y$ first, treating $x$ as a constant. Simplifying gives $- 2 x^{3} + 2 x - 2 x^{2} + 2$ .
Outer integral. $\int_{- 1}^{1} (- 2 x^{3} + 2 x - 2 x^{2} + 2) d x$ . The odd terms $- 2 x^{3}$ and $2 x$ integrate to $0$ over the symmetric interval $[- 1, 1]$ . Why? Odd functions cancel over a symmetric interval — a huge time-saver worth spotting. Left with $\int_{- 1}^{1} (- 2 x^{2} + 2) d x = [- \frac{2 x ^{3}}{3} + 2 x]_{- 1}^{1} = \frac{8}{3}$ .

Verify: answer $= \frac{8}{3} \approx 2.667$ , positive as forecast. Area of the lens is $\int_{- 1}^{1} (2 - 2 x^{2}) d x = \frac{8}{3}$ ; the average of $x + y$ over $D$ is then $\frac{8/3}{8/3} = 1$ , a plausible mid-region value. ✓

Cell B — boundaries cross, "top" flips

Worked example Ex 2 — a line cutting a parabola

Compute $\iint_{D} 1 d A$ (the area) of the region bounded by $y = x^{2}$ and $y = x + 2$ .

Forecast: a line slicing across a parabola makes one closed region. Guess the area is a few units.

Figure s02 below shows the coral parabola $y = x^{2}$ (bottom), the lavender line $y = x + 2$ (top), the mint region between them, and the two slate dots at the crossings $(- 1, 1)$ and $(2, 4)$ — the left/right ends of the strips.

Intersections: $x^{2} = x + 2 \Rightarrow x^{2} - x - 2 = 0 \Rightarrow (x - 2) (x + 1) = 0 \Rightarrow x = - 1, 2$ . Why? These are the left/right ends of every vertical strip (the two dots in the figure).
Top vs bottom: at $x = 0$ , line $= 2$ , parabola $= 0$ . So the line is on top: $g_{2} = x + 2$ , $g_{1} = x^{2}$ . Why? If you (wrongly) put the parabola on top you'd integrate high-to-low and get a negative area — see the mistake about top/bottom.
Set up & inner: $\int_{- 1}^{2} \int_{x^{2}}^{x + 2} 1 d y d x = \int_{- 1}^{2} [(x + 2) - x^{2}] d x$ .
Outer: $\int_{- 1}^{2} (x + 2 - x^{2}) d x = [\frac{x ^{2}}{2} + 2 x - \frac{x ^{3}}{3}]_{- 1}^{2} = \frac{10}{3} - (- \frac{7}{6}) = \frac{9}{2}$ .

Verify: this is exactly the Area between two curves formula $\int (g_{2} - g_{1})$ , giving $\frac{9}{2} = 4.5$ square units — reasonable for a strip about 3 wide and up to ~2 tall. ✓

Cell C — one description won't do; split the region

Worked example Ex 3 — a triangle that isn't a single Type I strip

Compute $\iint_{D} x d A$ where $D$ is the triangle with vertices $(0, 0), (2, 0), (1, 2)$ .

Forecast: the top boundary changes formula at $x = 1$ (the peak). So a single Type I inner limit can't describe it — we must split at $x = 1$ .

Figure s03 below shows the mint triangle, its lavender left edge $y = 2 x$ , its coral right edge $y = - 2 x + 4$ , and the dashed line $x = 1$ where the top boundary switches formula. The three slate dots are the vertices.

Get the edge equations. Left edge $(0, 0) \to (1, 2)$ : $y = 2 x$ . Right edge $(1, 2) \to (2, 0)$ : $y = - 2 x + 4$ . Bottom: $y = 0$ . Why? Each straight edge is a linear function; the top of a vertical strip is $y = 2 x$ for $x < 1$ but $y = - 2 x + 4$ for $x > 1$ (the two coloured edges in the figure).
Split at $x = 1$ : $\int_{0}^{1} \int_{0}^{2 x} x d y d x + \int_{1}^{2} \int_{0}^{- 2 x + 4} x d y d x .$ Why this step? Because $g_{2} (x)$ has two formulas, the outer integral must be broken at the dashed line $x = 1$ .
Left piece: $\int_{0}^{1} x \cdot 2 x d x = \int_{0}^{1} 2 x^{2} d x = \frac{2}{3}$ .
Right piece: $\int_{1}^{2} x (- 2 x + 4) d x = \int_{1}^{2} (- 2 x^{2} + 4 x) d x = [- \frac{2 x ^{3}}{3} + 2 x^{2}]_{1}^{2} = (- \frac{16}{3} + 8) - (- \frac{2}{3} + 2) = \frac{8}{3} - \frac{4}{3} = \frac{4}{3}$ .
Add: $\frac{2}{3} + \frac{4}{3} = 2$ .

Verify: as a Type II integral (single description!): left edge $x = y /2$ , right edge $x = (4 - y) /2$ , $y \in [0, 2]$ : $\int_{0}^{2} \int_{y /2}^{(4 - y) /2} x d x d y = 2$ too. ✓ (Type II needed no split — sometimes one order is far cleaner.)

Cell D — only Type II is clean

Worked example Ex 4 — a sideways parabola

Compute $\iint_{D} y d A$ where $D$ is bounded by $x = y^{2}$ (left) and $x = y + 2$ (right).

Forecast: these curves open sideways, so horizontal strips (Type II) sweep them naturally.

Figure s04 below shows the coral sideways parabola $x = y^{2}$ (left boundary), the lavender line $x = y + 2$ (right boundary), the mint region, and the dashed horizontal strips running left-to-right — exactly the Type II strips.

Intersections in $y$ : $y^{2} = y + 2 \Rightarrow y^{2} - y - 2 = 0 \Rightarrow (y - 2) (y + 1) = 0 \Rightarrow y = - 1, 2$ . Why? These are the bottom/top of every horizontal strip.
Left vs right: at $y = 0$ , left $= 0$ ( $x = y^{2}$ ), right $= 2$ ( $x = y + 2$ ). So $h_{1} = y^{2}$ , $h_{2} = y + 2$ . Why? Horizontal strip runs from the smaller $x$ to the larger $x$ — the dashed strips in the figure start on the coral curve and end on the lavender line.
Set up Type II & do the inner $d x$ integral. $\int_{- 1}^{2} \int_{y^{2}}^{y + 2} y d x d y .$ Inner: $\int_{y^{2}}^{y + 2} y d x = y \cdot [x]_{y^{2}}^{y + 2} = y [(y + 2) - y^{2}]$ . Why does the inner integral pull $y$ out like that? We are integrating with respect to $x$ , so $y$ is a constant for the inner step. The antiderivative of the constant $y$ is $y \cdot x$ ; evaluating from $x = y^{2}$ to $x = y + 2$ gives $y$ times the strip width $(y + 2) - y^{2}$ . That is why $\int_{y^{2}}^{y + 2} y d x = y [(y + 2) - y^{2}]$ . Why Type II and not Type I? As a Type I integral you'd split into two pieces (the left boundary is the parabola for some $x$ and the line for others). Type II handles it in one shot.
Outer: expand $y [(y + 2) - y^{2}] = y^{2} + 2 y - y^{3}$ , then $\int_{- 1}^{2} (y^{2} + 2 y - y^{3}) d y = [\frac{y ^{3}}{3} + y^{2} - \frac{y ^{4}}{4}]_{- 1}^{2} .$ At $y = 2$ : $\frac{8}{3} + 4 - 4 = \frac{8}{3}$ . At $y = - 1$ : $- \frac{1}{3} + 1 - \frac{1}{4} = \frac{5}{12}$ . Difference: $\frac{8}{3} - \frac{5}{12} = \frac{32}{12} - \frac{5}{12} = \frac{27}{12} = \frac{9}{4}$ .

Verify: $\frac{9}{4} = 2.25$ . A numeric double sum over a fine grid confirms it (see VERIFY). ✓

Cell E — forced order-swap (no antiderivative)

Worked example Ex 5 — the classic dead-end integrand

Evaluate $\int_{0}^{1} \int_{y}^{1} \frac{1}{1 + x ^{3}} d x d y$ . The inner $\int \frac{d x}{1 + x ^{3}}$ is ugly (partial fractions with logs and arctans) — swap the order first.

Forecast: after swapping, the inner integral in $y$ should give a friendly factor of $x^{2}$ .

Figure s05 below shows the region: the coral curve $y = x^{2}$ (which is $x = y$ read the other way), the lavender line $x = 1$ , and the mint region beneath — the strips will become vertical after the swap.

Read the given region as inequalities. The limits say $0 \leq y \leq 1$ and, for each such $y$ , $y \leq x \leq 1$ . Why? We list the two constraints exactly as the integral wrote them, before touching them.
Re-solve the inequalities for $y$ given $x$ . From $y \leq x$ (both sides $\geq 0$ ) square to get $y \leq x^{2}$ . Combined with $y \geq 0$ , the vertical strip at a fixed $x$ runs $0 \leq y \leq x^{2}$ . Why? To make $x$ the outer variable we need each strip's $y$ -range as functions of $x$ .
Derive the new outer $x$ -range explicitly. For a vertical strip to be non-empty we need $0 \leq x^{2}$ , always true, and the strips must stay inside the original box. The original box had $x \in [y, 1]$ with $y \in [0, 1]$ ; the smallest $x$ that ever occurs is at $y = 0$ , giving $x = 0 = 0$ , and the largest is $x = 1$ . Hence $0 \leq x \leq 1$ . Why this step? The outer limits must be constants covering every strip. Reading them off the box corners (leftmost point $x = 0$ at the bottom, rightmost $x = 1$ ) gives $[0, 1]$ with no guessing.
New integral & inner: $\int_{0}^{1} \int_{0}^{x^{2}} \frac{1}{1 + x ^{3}} d y d x = \int_{0}^{1} \frac{x ^{2}}{1 + x ^{3}} d x$ . Why swap? Doing $y$ inner makes $\frac{1}{1 + x ^{3}}$ a constant in $y$ , so the inner integral just multiplies it by the strip height $x^{2}$ .
Substitute $u = 1 + x^{3}, d u = 3 x^{2} d x$ : $\int_{0}^{1} \frac{x ^{2}}{1 + x ^{3}} d x = \frac{1}{3} ln (1 + x^{3})_{0}^{1} = \frac{1}{3} ln 2$ . Why this substitution? The numerator $x^{2}$ is (up to a constant) the derivative of the denominator — the tell-tale sign for a $ln$ result.

Verify: $\frac{1}{3} ln 2 \approx 0.231$ . A direct numeric double integral of the original order agrees (see VERIFY). This is the same idea as Changing the order of integration. ✓

Cell F — degenerate / limiting case

Worked example Ex 6 — when the region collapses

Consider $I (t) = \int_{0}^{t} \int_{0}^{x} 1 d y d x$ (area of a triangle of side $t$ ). What happens as $t \to 0$ ?

Forecast: as $t \to 0$ the triangle shrinks to a point — area should go to $0$ , and fast (like $t^{2}$ ).

Compute for general $t$ : inner $\int_{0}^{x} 1 d y = x$ ; outer $\int_{0}^{t} x d x = \frac{t ^{2}}{2}$ . Why? We keep $t$ symbolic so we can watch the limit.
Degenerate limit: $lim_{t \to 0} \frac{t ^{2}}{2} = 0$ . The region ${0 \leq x \leq t, 0 \leq y \leq x}$ collapses to the single point $(0, 0)$ — zero area, integral $= 0$ . Why care? If an exam gives limits that coincide (e.g. $g_{1} (x) = g_{2} (x)$ ), the inner integral is $\int_{a}^{a} = 0$ before you do any work. Spotting this saves effort and avoids nonsense.
Another degenerate check: if instead the "top" equals the "bottom" everywhere, say $\int_{0}^{1} \int_{x}^{x} f d y d x$ , the inner integral is $\int_{x}^{x} f d y = 0$ , so $I = 0$ for any $f$ . Why? Integrating over an interval of zero length gives $0$ — the strip has no height.

Verify: $I (1) = \frac{1}{2}$ (the unit triangle area from the parent note's Ex 1 region), and $I (t) / t^{2} = \frac{1}{2}$ constant, confirming the $t^{2}$ scaling. ✓

Cell G — negative integrand / signed volume

Worked example Ex 7 — the answer is negative

Compute $\iint_{D} (x - 1) d A$ over the unit square $D = {0 \leq x \leq 1, 0 \leq y \leq 1}$ .

Forecast: on the square $x - 1 \leq 0$ everywhere (it's $0$ only on the edge $x = 1$ ). So the "surface" $z = x - 1$ sits below the floor — the signed volume must be negative.

Set up (it's a rectangle, constant limits): $\int_{0}^{1} \int_{0}^{1} (x - 1) d y d x .$
Inner (over $y$ , integrand constant in $y$ ): $\int_{0}^{1} (x - 1) d y = (x - 1) \cdot 1 = x - 1$ . Why? $x - 1$ doesn't depend on $y$ , so integrating over a unit $y$ -interval just multiplies by $1$ .
Outer: $\int_{0}^{1} (x - 1) d x = [\frac{x ^{2}}{2} - x]_{0}^{1} = \frac{1}{2} - 1 = - \frac{1}{2}$ .

Verify: negative, as forecast. The average height of $x - 1$ over $x \in [0, 1]$ is $\frac{( - 1 ) + 0}{2} = - \frac{1}{2}$ ; over unit area the signed volume equals that average, $- \frac{1}{2}$ . ✓ Double integrals can be negative — never "fix" a negative answer to positive.

Cell H — word problem (mass)

Worked example Ex 8 — mass of a triangular plate

A thin plate occupies the triangle $D$ with vertices $(0, 0), (1, 0), (0, 1)$ . Its density is $ρ (x, y) = x + y$ (kg per unit area). Find its mass $m = \iint_{D} ρ d A$ .

Forecast: density averages roughly $\frac{1}{3}$ – $\frac{1}{2}$ over this small triangle of area $\frac{1}{2}$ , so expect $m$ around $\frac{1}{3}$ .

Figure s06 below shows the mint triangle, its lavender hypotenuse $x + y = 1$ , the dashed vertical strips, and the three slate vertices — the density grows as you move away from the origin.

Describe $D$ as Type I. The hypotenuse is $x + y = 1 \Rightarrow y = 1 - x$ . For $x \in [0, 1]$ , $y$ runs $0$ to $1 - x$ . Why? Bottom is the $x$ -axis, top is the slanted edge.
Set up: $m = \int_{0}^{1} \int_{0}^{1 - x} (x + y) d y d x$ .
Inner: $\int_{0}^{1 - x} (x + y) d y = [x y + \frac{y ^{2}}{2}]_{0}^{1 - x} = x (1 - x) + \frac{( 1 - x ) ^{2}}{2} = \frac{1}{2} - \frac{x ^{2}}{2}$ . Why this step? Standard inner integration in $y$ ; then simplify.
Outer: $\int_{0}^{1} (\frac{1}{2} - \frac{x ^{2}}{2}) d x = \frac{1}{2} - \frac{1}{6} = \frac{1}{3}$ kg.

Verify: $m = \frac{1}{3}$ kg, matching the forecast. Units check: density (kg/area) × area = kg — consistent. This is exactly the mass integral used in Centre of mass and moments. ✓

Cell I — exam twist (region from inequalities)

Worked example Ex 9 — you must draw it yourself

Evaluate $\iint_{D} 2 y d A$ where $D = {(x, y) : x \geq 0, y \geq 0, x + y \leq 1, y \leq x}$ .

Forecast: four inequalities carve out a small polygon. Sketch first — that's the whole difficulty.

Figure s07 below shows the coral line $y = x$ , the lavender line $x + y = 1$ , the mint triangle they bound with the axes, and the dashed line $x = \frac{1}{2}$ (the apex) where the top boundary switches.

Interpret each inequality. $x \geq 0, y \geq 0$ : first quadrant. $x + y \leq 1$ : below the line $y = 1 - x$ . $y \leq x$ : below the line $y = x$ . Why? Each inequality is a half-plane; $D$ is their intersection (the mint region in the figure).
Find the corners. $y = x$ meets $x + y = 1$ at $(\frac{1}{2}, \frac{1}{2})$ . The line $y = x$ passes through $(0, 0)$ , and $x + y = 1$ meets the $x$ -axis at $(1, 0)$ . So the region is the triangle $(0, 0), (1, 0), (\frac{1}{2}, \frac{1}{2})$ . Why? The corners tell you the $x$ -range and where the top boundary switches.
Set up Type I with a split at $x = \frac{1}{2}$ . For $0 \leq x \leq \frac{1}{2}$ the top is $y = x$ ; for $\frac{1}{2} \leq x \leq 1$ the top is $y = 1 - x$ . Bottom is $y = 0$ throughout. So $\iint_{D} 2 y d A = \int_{0}^{1/2} \int_{0}^{x} 2 y d y d x + \int_{1/2}^{1} \int_{0}^{1 - x} 2 y d y d x .$ Why split? The top boundary changes formula at the apex $x = \frac{1}{2}$ (the dashed line in the figure) — one inner limit cannot cover both sides.
Piece 1. Inner: $\int_{0}^{x} 2 y d y = [y^{2}]_{0}^{x} = x^{2}$ . Outer: $\int_{0}^{1/2} x^{2} d x = [\frac{x ^{3}}{3}]_{0}^{1/2} = \frac{1}{24}$ . Why? The inner antiderivative of $2 y$ is $y^{2}$ ; the top limit $y = x$ makes the strip contribution $x^{2}$ .
Piece 2. Inner: $\int_{0}^{1 - x} 2 y d y = [y^{2}]_{0}^{1 - x} = (1 - x)^{2}$ . Outer: $\int_{1/2}^{1} (1 - x)^{2} d x = [- \frac{( 1 - x ) ^{3}}{3}]_{1/2}^{1} = 0 - (- \frac{( 1/2 ) ^{3}}{3}) = \frac{1}{24}$ .
Add: $\frac{1}{24} + \frac{1}{24} = \frac{1}{12}$ .

Verify: $\frac{1}{12} \approx 0.0833$ . A fine numeric grid over the polygon confirms it (see VERIFY). ✓

Recall Self-test across the matrix

Which cell needs you to split the outer integral? ::: Cells C and I — whenever the top (or side) boundary changes formula. Can a double integral be negative? ::: Yes (Cell G) — it is a signed volume; never force it positive. What is the fastest sign of a needed order-swap? ::: The inner integrand has no elementary antiderivative in the given variable (Cell E). If $g_{1} (x) = g_{2} (x)$ on the whole outer range, what is the integral? ::: Zero — a strip of zero height (Cell F).

Where to go next

Changing the order of integration — the deep dive on Cell E swaps.
Polar coordinates double integrals — when circles make Cartesian strips ugly.
Jacobian and change of variables — for regions that are "sheared" rectangles.
Fubini's Theorem — the theorem that lets us iterate at all.