Changing order of integration: Worked examples

This page is the drill ground for Changing order of integration. The parent taught you the 5-step recipe; here we run it against every kind of region the topic can hand you — and we do not skip a single case. Before any example, we lay out a matrix of scenarios so you can see the whole battlefield.

The scenario matrix

Cell	Region shape	The twist you must survive	Example
C1	Triangle (line + axes)	inner has no elementary antiderivative — reorder to save it	Ex 1
C2	Lens (two curves cross: line vs parabola)	must re-solve both boundaries for the new variable, pick correct branch	Ex 2
C3	Region needing a split one way	one orientation = 2 integrals, other = 1 clean integral	Ex 3
C4	Strip that exits onto different curves at different heights	the exit boundary switches — you MUST split	Ex 4
C5	Degenerate corner / zero-width tip	boundaries meet at a point; check the limit is fine, not undefined	Ex 5
C6	Circle / disk piece	reorder AND recognise polar is even better	Ex 6
C7	Word problem (real-world total)	translate physical region into limits, then reorder to compute	Ex 7
C8	Exam twist: unknown/general $f$ , symbolic answer	reorder without ever evaluating $f$	Ex 8

Every cell C1–C8 is hit below. Numeric answers are all machine-checked in the verify block.

Ex 1 — Cell C1: triangle, impossible inner integral

Worked example Statement

Evaluate $I = \int_{0}^{1} \int_{x}^{1} e^{y^{2}} d y d x .$

Forecast: Guess before reading — will the answer contain $e$ ? Will reordering make the $y$ -integral doable? (Hold that thought.)

Step 1 — Read the limits. Outer $0 \leq x \leq 1$ ; inner $x \leq y \leq 1$ . Why this step? The limits ARE the region. Reading them out loud gives us the boundary curves: the bottom of each vertical strip rides on the line $y = x$ , the top on the line $y = 1$ .

Step 2 — Sketch (look at Figure s01). Above $y = x$ , below $y = 1$ , right of $x = 0$ : a triangle with corners $(0, 0), (0, 1), (1, 1)$ .

Why this step?

\int e^{y^{2}} d y

has no elementary antiderivative, so the given order is a dead end. The sketch tells us the alternative order.

Step 3 — New outer variable $y$ . From the picture the whole region spans $0 \leq y \leq 1$ . Why this step? The lowest tile sits at $y = 0$ (the corner), the highest at $y = 1$ (the top edge). Outer limits must be pure numbers.

Step 4 — New inner variable $x$ . Fix a height $y$ and slide a horizontal strip (red in the figure). It enters at the left wall $x = 0$ and exits at the slanted boundary $y = x$ , i.e. $x = y$ . So $0 \leq x \leq y$ . Why this step? We must re-express the boundary $y = x$ as a formula for the new inner variable: solve $y = x$ for $x \Rightarrow x = y$ .

Step 5 — Rewrite & evaluate. $I = \int_{0}^{1} \int_{0}^{y} e^{y^{2}} d x d y = \int_{0}^{1} e^{y^{2}} [x]_{0}^{y} d y = \int_{0}^{1} y e^{y^{2}} d y .$ Substitute $u = y^{2}, d u = 2 y d y$ : $I = \frac{1}{2} \int_{0}^{1} e^{u} d u = \frac{1}{2} (e - 1) .$

Verify: Numerically $\frac{1}{2} (e - 1) \approx 0.8591$ . Sanity: the integrand $e^{y^{2}} \geq 1$ over a triangle of area $\frac{1}{2}$ , so $I \geq \frac{1}{2}$ — consistent. ✓

Ex 2 — Cell C2: lens between a line and a parabola

Worked example Statement

Evaluate $J = \int_{0}^{1} \int_{x^{2}}^{x} (x + y) d y d x .$

Forecast: The region is the sliver between $y = x$ and $y = x^{2}$ . Guess: after reordering, will the left edge be the line or the parabola?

Step 1 — Read. $0 \leq x \leq 1$ , $x^{2} \leq y \leq x$ . Why this step? Bottom of each vertical strip rides the parabola $y = x^{2}$ ; top rides the line $y = x$ . They coincide at $x = 0$ and $x = 1$ .

Step 2 — Sketch (Figure s02). A lens-shaped region between line (upper) and parabola (lower), corners $(0, 0), (1, 1)$ .

Why this step? We need to know, for a fixed height

y

, which curve is on the left and which on the right.

Step 3 — New outer $y$ : $0 \leq y \leq 1$ . Why this step? Lowest and highest points of the lens are the two corners.

Step 4 — New inner $x$ . For fixed $y \in (0, 1)$ , solve each boundary for $x$ :

line $y = x \Rightarrow x = y$ ,
parabola $y = x^{2} \Rightarrow x = y$ (positive branch — the region lives in the first quadrant).

Since $0 < y < 1$ gives $y < y$ , the strip runs $y \leq x \leq y$ . Why this step? We MUST re-solve both boundaries and pick the branch that lies in $R$ . Test $y = \frac{1}{4}$ : strip $[0.25, 0.5]$ — both inside the lens. ✓

Step 5 — Evaluate both ways (they must match). Original order: $J = \int_{0}^{1} \int_{x^{2}}^{x} (x + y) d y d x = \int_{0}^{1} [x y + \frac{y ^{2}}{2}]_{x^{2}}^{x} d x = \int_{0}^{1} (\frac{3}{2} x^{2} - x^{3} - \frac{x ^{4}}{2}) d x .$ $= \frac{1}{2} - \frac{1}{4} - \frac{1}{10} = \frac{3}{20} .$ Reordered check: $J = \int_{0}^{1} \int_{y}^{y} (x + y) d x d y = \frac{3}{20}$ too.

Verify: $J = \frac{3}{20} = 0.15$ , both orders agree. ✓

Ex 3 — Cell C3: one order = two integrals, other = one

Worked example Statement

Reverse the order and evaluate $S = \int_{0}^{2} \int_{0}^{x} y d y d x + \int_{2}^{4} \int_{0}^{4 - x} y d y d x .$ This is the total over the triangle with vertices $(0, 0), (4, 0), (2, 2)$ , written as two vertical-strip integrals (it needed splitting at $x = 2$ where the top edge changes from $y = x$ to $y = 4 - x$ ).

Forecast: Guess: can horizontal strips describe this triangle in ONE integral (no split)?

Step 1 — Read. Left piece: $0 \leq x \leq 2, 0 \leq y \leq x$ . Right piece: $2 \leq x \leq 4, 0 \leq y \leq 4 - x$ . Why this step? Two pieces because the top boundary of a vertical strip is $y = x$ for $x < 2$ but $y = 4 - x$ for $x > 2$ — the exit curve switches at the apex.

Step 2 — Sketch (Figure s03). A triangle: base on the $x$ -axis from $0$ to $4$ , apex at $(2, 2)$ .

Why this step? The picture shows that a horizontal strip at height

y

enters on the left edge

y = x \Rightarrow x = y

and exits on the right edge

y = 4 - x \Rightarrow x = 4 - y

, for ALL

y

— no split needed.

Step 3 — New outer $y$ : $0 \leq y \leq 2$ . Why this step? Apex height is $2$ , base at $0$ .

Step 4 — New inner $x$ : $y \leq x \leq 4 - y$ . Why this step? Both edges solved for $x$ ; one single description works.

Step 5 — Evaluate the single integral. $S = \int_{0}^{2} \int_{y}^{4 - y} y d x d y = \int_{0}^{2} y [(4 - y) - y] d y = \int_{0}^{2} y (4 - 2 y) d y = \int_{0}^{2} (4 y - 2 y^{2}) d y .$ $= [2 y^{2} - \frac{2}{3} y^{3}]_{0}^{2} = 8 - \frac{16}{3} = \frac{8}{3} .$

Verify: $S = \frac{8}{3} \approx 2.667$ . Cross-check by summing the two original pieces: $\int_{0}^{2} \frac{x ^{2}}{2} d x + \int_{2}^{4} \frac{( 4 - x ) ^{2}}{2} d x = \frac{4}{3} + \frac{4}{3} = \frac{8}{3}$ . ✓ (Reordering turned 2 integrals into 1.)

Ex 4 — Cell C4: exit boundary switches (must split the OTHER way)

Worked example Statement

Reverse the order in $T = \int_{0}^{2} \int_{x^{2}}^{2 x} f (x, y) d y d x$ (region between $y = x^{2}$ and $y = 2 x$ ).

Forecast: The curves cross at $(0, 0)$ and $(2, 4)$ . Guess: when we use horizontal strips, does one exit curve serve the whole height, or does it switch?

Step 1 — Read. $0 \leq x \leq 2$ , $x^{2} \leq y \leq 2 x$ . On $[0, 2]$ , $x^{2} \leq 2 x$ , so the parabola is below the line. Why this step? Establishes the region: below the line $y = 2 x$ , above the parabola $y = x^{2}$ .

Step 2 — Sketch (Figure s04). A curved sliver from $(0, 0)$ to $(2, 4)$ .

Why this step? We check where a horizontal strip enters/exits. Solve for

x

: line

y = 2 x \Rightarrow x = y /2

; parabola

y = x^{2} \Rightarrow x = y

(positive branch). For

0 < y < 4

y /2 < y

— the strip runs

y /2 \leq x \leq y

, and this holds for the WHOLE range

0 \leq y \leq 4

Step 3 — New outer $y$ : $0 \leq y \leq 4$ (apex at $(2, 4)$ ). Step 4 — New inner $x$ : $\frac{y}{2} \leq x \leq y$ . Why this step? Both boundaries re-solved for $x$ ; correct branch of $\cdot$ chosen (region in first quadrant). Test $y = 1$ : strip $[0.5, 1]$ ; check the midpoint $(0.7, 1)$ satisfies $0.49 \leq 1 \leq 1.4$ ✓.

Step 5 — Result. $T = \int_{0}^{4} \int_{y /2}^{y} f (x, y) d x d y .$ To make it concrete, take $f = 1$ (area). Original: $\int_{0}^{2} (2 x - x^{2}) d x = 4 - \frac{8}{3} = \frac{4}{3}$ . Reordered: $\int_{0}^{4} (y - \frac{y}{2}) d y = \frac{2}{3} \cdot 8 - \frac{1}{4} \cdot 16 = \frac{16}{3} - 4 = \frac{4}{3}$ . ✓

Verify: area $= \frac{4}{3}$ both orders. ✓ (Here the exit curve did NOT switch — contrast Ex 3 where it did. Reading the picture is what tells you which.)

Ex 5 — Cell C5: degenerate tip, check the limit is fine

Worked example Statement

Reverse the order in $U = \int_{0}^{1} \int_{0}^{1 - x^{2}} x d y d x$ — a quarter disk. Watch the degenerate corners where the strip width shrinks to zero.

Forecast: At $x = 1$ the top boundary $1 - x^{2} = 0$ : the vertical strip has zero height. Guess: does that cause any trouble in either order?

Step 1 — Read. $0 \leq x \leq 1$ , $0 \leq y \leq 1 - x^{2}$ : the quarter of the unit circle in the first quadrant. Why this step? $y = 1 - x^{2}$ is the upper arc $x^{2} + y^{2} = 1$ .

Step 2 — Sketch (Figure s05). Quarter disk, corners $(0, 0), (1, 0), (0, 1)$ .

Why this step? At

x \to 1

the strip height

\to 0

(a degenerate tip at

(1, 0)

). Integrands stay finite, so the integral is a proper (non-improper) integral — no blow-up.

Step 3 — New outer $y$ : $0 \leq y \leq 1$ . Step 4 — New inner $x$ : solve arc $y = 1 - x^{2} \Rightarrow x = 1 - y^{2}$ (positive branch). So $0 \leq x \leq 1 - y^{2}$ . Why this step? Horizontal strip enters at wall $x = 0$ , exits at arc $x = 1 - y^{2}$ . At $y \to 1$ this width $\to 0$ — the SAME degenerate tip, now at the top. The formula is still finite there, confirming no problem.

Step 5 — Evaluate. $U = \int_{0}^{1} \int_{0}^{1 - y^{2}} x d x d y = \int_{0}^{1} \frac{1}{2} (1 - y^{2}) d y = \frac{1}{2} [y - \frac{y ^{3}}{3}]_{0}^{1} = \frac{1}{2} \cdot \frac{2}{3} = \frac{1}{3} .$

Verify: Original order: $\int_{0}^{1} x 1 - x^{2} d x = [- \frac{1}{3} (1 - x^{2})^{3/2}]_{0}^{1} = \frac{1}{3}$ . Both give $\frac{1}{3}$ ; the degenerate tip contributed no defect. ✓

Ex 6 — Cell C6: disk piece where polar wins

Worked example Statement

Consider $V = \int_{0}^{1} \int_{0}^{1 - x^{2}} (x^{2} + y^{2}) d y d x$ over the same quarter disk. Reorder — then note Polar Coordinates in Double Integrals is even cleaner.

Forecast: Guess whether the Cartesian reorder is pleasant, or whether you'd rather switch to $r, θ$ .

Step 1 — Read & sketch. Same quarter disk as Ex 5 (reuse Figure s05). Why this step? Region identical; only the integrand changed to $x^{2} + y^{2}$ .

Step 2 — Reorder (Cartesian). By the same solving as Ex 5: $V = \int_{0}^{1} \int_{0}^{1 - y^{2}} (x^{2} + y^{2}) d x d y .$ Why this step? Shows the reorder is valid, but the inner integral $\int (x^{2} + y^{2}) d x = \frac{x ^{3}}{3} + y^{2} x$ evaluated at $x = 1 - y^{2}$ is messy.

Step 3 — Switch to polar. On the quarter disk: $x^{2} + y^{2} = r^{2}$ , $d A = r d r d θ$ , with $0 \leq r \leq 1$ , $0 \leq θ \leq \frac{π}{2}$ . Why this step? Why polar and not more Cartesian gymnastics? Because the region is a circular arc and the integrand depends only on $r$ — polar makes the limits constants and separates the integral. This is exactly what Change of Variables (Jacobian) formalises (the Jacobian gives the $r$ factor).

Step 4 — Evaluate. $V = \int_{0}^{π /2} \int_{0}^{1} r^{2} \cdot r d r d θ = \int_{0}^{π /2} [\frac{r ^{4}}{4}]_{0}^{1} d θ = \frac{1}{4} \cdot \frac{π}{2} = \frac{π}{8} .$

Verify: $V = \frac{π}{8} \approx 0.3927$ . (Cartesian reorder gives the same after grinding.) ✓

Ex 7 — Cell C7: real-world word problem

Worked example Statement

A triangular metal plate has vertices $(0, 0), (2, 0), (0, 2)$ (units: cm). Its surface density is $ρ (x, y) = e^{y^{2}}$ g/cm². The mass is $M = \iint_{R} e^{y^{2}} d A$ . Set it up in the convenient order and compute.

Forecast: Guess: which variable must be the outer one so the inner integral is doable?

Step 1 — Translate the plate into a region. The triangle is bounded by $x = 0$ , $y = 0$ , and the line through $(2, 0), (0, 2)$ : $x + y = 2$ , i.e. $0 \leq x \leq 2 - y$ . So $R : 0 \leq y \leq 2, 0 \leq x \leq 2 - y$ . Why this step? A word problem is just a region + integrand in disguise; we must extract both.

Step 2 — Sketch (Figure s06).

Why this step? Confirms the hypotenuse

x + y = 2

and which order makes

e^{y^{2}}

integrable.

Step 3 — Choose $y$ as OUTER. If $y$ were inner we'd face $\int e^{y^{2}} d y$ — impossible. With $y$ outer, the inner $x$ -integral treats $e^{y^{2}}$ as a constant. Why this step? Same lesson as Ex 1: put the "impossible" variable on the OUTSIDE.

Step 4 — Inner limits. For fixed $y$ , $x$ runs $0$ to $2 - y$ . Why this step? Left wall $x = 0$ , right boundary the hypotenuse $x = 2 - y$ .

Step 5 — Evaluate. $M = \int_{0}^{2} \int_{0}^{2 - y} e^{y^{2}} d x d y = \int_{0}^{2} (2 - y) e^{y^{2}} d y .$ Split: $\int_{0}^{2} 2 e^{y^{2}} d y - \int_{0}^{2} y e^{y^{2}} d y$ . The second is elementary: $\frac{1}{2} (e^{4} - 1)$ . The first is a genuine non-elementary integral (see Improper and Non-elementary Integrals); write it as $2 \int_{0}^{2} e^{y^{2}} d y$ , a value $\approx 2 (16.4526) = 32.905$ . $M \approx 32.905 - \frac{1}{2} (e^{4} - 1) = 32.905 - 26.799 = 6.106 g .$ Why this step? Even here, reordering isolated the elementary chunk; the remaining term is honestly non-elementary and must be evaluated numerically.

Verify: $M \approx 6.11$ g. Sanity by units: g/cm² × cm² = g ✓. Sanity by bound: area $= 2$ cm², density between $1$ and $e^{4} \approx 54.6$ , and mass $6.1$ lies in $[2, 109]$ ✓.

Ex 8 — Cell C8: exam twist, general $f$ , symbolic reorder

Worked example Statement

Reverse the order of integration, without evaluating, for $W = \int_{0}^{2} \int_{0}^{y /2} f (x, y) d x d y + \int_{2}^{4} \int_{0}^{2 - y /2 + ?} \dots$ — actually the clean twist: reverse $W = \int_{0}^{1} \int_{- y}^{y} f (x, y) d x d y .$ This region is symmetric across $x = 0$ , so it spans quadrants left and right of the $y$ -axis.

Forecast: Both signs of $x$ appear. Guess: after reordering, does the region split at $x = 0$ , and what are the two branches?

Step 1 — Read. $0 \leq y \leq 1$ , $- y \leq x \leq y$ : the region under the parabola $y = x^{2}$ opening upward, capped at $y = 1$ , symmetric about the $y$ -axis. Why this step? $x = \pm y$ are the two branches of $y = x^{2}$ ; both signs of $x$ occur, so we must handle negative- $x$ and positive- $x$ tiles.

Step 2 — Sketch (Figure s07).

Why this step? Shows the region is a parabolic cap over

- 1 \leq x \leq 1

, bounded below by

y = x^{2}

and above by

y = 1

Step 3 — New outer $x$ : overall $- 1 \leq x \leq 1$ . Why this step? Widest tiles are at $y = 1$ where $x = \pm 1$ . Note the outer variable now takes negative values — a case the naive "min=0" habit would miss.

Step 4 — New inner $y$ : for a fixed $x$ , vertical strip enters at the parabola $y = x^{2}$ and exits at $y = 1$ . So $x^{2} \leq y \leq 1$ . This single description covers BOTH signs of $x$ because $x^{2}$ is even — no split needed. Why this step? Re-solving the boundary $x = \pm y$ for $y$ gives the single curve $y = x^{2}$ , which merges the two branches.

Step 5 — Result. $W = \int_{- 1}^{1} \int_{x^{2}}^{1} f (x, y) d y d x .$ Why this step? Fully symbolic; valid for any integrable $f$ by Fubini's Theorem over this general region.

Verify (with a concrete $f = 1$ , i.e. area): original $\int_{0}^{1} 2 y d y = 2 \cdot \frac{2}{3} = \frac{4}{3}$ ; reordered $\int_{- 1}^{1} (1 - x^{2}) d x = 2 - \frac{2}{3} = \frac{4}{3}$ . Both $= \frac{4}{3}$ , confirming the reorder (and that negative $x$ was correctly included). ✓

Recall Which cell does each trap live in?

"Left the outer limit as a curve" happens most in which cell type? ::: The word/general ones (C7, C8) — always reduce the OUTER limits to constants. When does reordering turn 2 integrals into 1? ::: When the exit curve switches in one orientation but not the other (C3). When is the negative branch of a square root the correct one? ::: When the region lies left of the $y$ -axis (C8) — pick the branch that lands in $R$ .

Where to go next

The reorder mechanism is licensed by Fubini's Theorem.
Circular/annular regions: prefer Polar Coordinates in Double Integrals and the general Change of Variables (Jacobian).
The three-variable analogue: Triple Integrals — Order of Integration.