Changing order of integration: Question bank — Maths deep dive

Before we start, a shared vocabulary — and a picture to anchor it. Throughout, we write $R$ for the region of integration: the flat 2D patch in the $x y$ -plane over which the double integral $\iint_{R} f d A$ is taken. Every "strip" lives inside this $R$ .

Outer = "how far do I sweep"; inner = "where does this one strip start and stop". The figure shows exactly that: the outer range is the flat interval on the axis, the inner limits are the two curves the red strip touches.

True or false — justify

TF1. Swapping the order of a double integral always changes its numerical value.

False — by Fubini's Theorem both iterated integrals equal the same area-weighted total; only the bookkeeping (the limits) changes, never the number.

TF2. If the inner integral has no elementary antiderivative, the double integral cannot be evaluated at all.

False — you may be stuck in one order, but reordering can turn the inner integral into something trivial (e.g.

e^{y^{2}}

inner becomes

y e^{y^{2}}

after swap), so the double integral still has a clean value.

TF3. After swapping, the new outer limits are just the min and max of the four numbers in the old limits.

False — the outer limits are the overall range of the new outer variable, which you read off the sketch of

R

, not by shuffling the old numbers around.

TF4. For any region, both a vertical-strip description and a horizontal-strip description exist as a single integral (no splitting).

False — many regions are "simple" one way but need splitting into two or more integrals the other way; only convex-ish, well-behaved regions are single-integral both ways.

TF5. When you flip the order, the boundary curves of the region stay the same.

True as curves, but you must re-solve each equation for the new inner variable — e.g.

y = x^{2}

stays the same parabola but is written

x = y

when

x

becomes inner.

TF6. Fubini's theorem needs

f

to be positive to guarantee the two orders agree.

False — continuity of

f

on the region

R

(or integrability with a finite integral of

∣ f ∣

over

R

) is what's required; the sign of

f

is irrelevant to the swap being valid.

TF7. If a boundary is

x = y

on a region, then swapping could legitimately introduce

x = - y

as well.

Only if the region actually includes the negative branch — you pick the branch that lies in

R

by testing an interior point, never both automatically.

TF8. The order

d x d y

always corresponds to horizontal strips.

True — the innermost differential names the variable that moves within a single strip, and

d x

moving means the strip is horizontal (fixed

y

TF9. Reordering can change a proper integral into an improper one.

True — for some Improper and Non-elementary Integrals the region is unbounded or

f

blows up, and one orientation may hide the singularity as a finite inner integral while the other exposes it; the swap is still valid where Fubini's hypotheses hold.

Spot the error

SE1. "Reverse

\int_{0}^{1} \int_{0}^{x} f d y d x

to get

\int_{0}^{x} \int_{0}^{1} f d x d y

." — What's wrong?

The outer limit

\int_{0}^{x}

still contains

x

(Blunder A); outer limits must be constants. Correct swap over that triangle is

\int_{0}^{1} \int_{y}^{1} f d x d y

SE2. "The region is

x^{2} \leq y \leq x

, so after swapping the inner limits are

\int_{x^{2}}^{x} f d x

— i.e. still written in terms of

x

." — What's wrong?

The boundaries were never re-solved for

x

(Blunder B). Solve

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

and

y = x \Rightarrow x = y

, giving inner limits

\int_{y}^{y} f d x

with outer

y \in [0, 1]

SE3. "For the strip in the lens between

y = x

and

y = x^{2}

at fixed

y \in (0, 1)

x

runs from

y

y

." — What's wrong?

The order is backwards: for

0 < y < 1

we have

y < y

, so the strip enters at

x = y

and exits at

x = y

; the lower limit must be the smaller value,

x = y

SE4. "I copied the min of all limits (0) and max (1) for the outer, so the outer is

0

1

— no sketch needed." — What's wrong?

Skipping the sketch (Blunder C) — the true outer range and whether the region needs splitting can only be seen geometrically; a lucky number match doesn't validate it.

SE5. "In

\int_{0}^{2} \int_{0}^{4 - x^{2}} f d y d x

the region is a full circle, so I swap to

\int_{- 2}^{2} \int_{0}^{4 - y^{2}} f d x d y

." — What's wrong?

The original region is only the quarter disk in the first quadrant (

x, y \geq 0

), not a full or half circle. New limits:

0 \leq y \leq 2

0 \leq x \leq 4 - y^{2}

SE6. "The inner limits became constants and the outer became a function of

y

— that's a valid swapped integral." — What's wrong?

It's inside-out. Inner limits may be functions of the outer variable; the outer must be constant. This describes an ill-formed integral.

SE7. "Swapping

\int_{0}^{1} \int_{x}^{1} e^{y^{2}} d y d x

I keep the boundary as

y = x

and write

\int_{0}^{1} \int_{0}^{y = x} \dots

." — What's wrong?

You must express the inner variable

x

explicitly (Blunder B): solve

y = x

for

x

to get

x = y

, so the inner is

\int_{0}^{y}

, giving

\int_{0}^{1} \int_{0}^{y} e^{y^{2}} d x d y

SE8. "The region for

\int_{1}^{e} \int_{0}^{l n x} f d y d x

swaps to

\int_{0}^{1} \int_{1}^{e^{y}} f d x d y

with no thought about which side of

y = ln x

." — What's right/wrong here?

This one is actually correct: solving

y = ln x

for

x

gives

x = e^{y}

(right edge), left edge is

x = 1

, and

y

ranges

0

1

. The trap is assuming it's wrong — always verify by testing an interior point rather than by suspicion.

Why questions

WQ1. Why must the outer limits be constants while inner limits may be curves?

The outer integral is the last one done — its result must be a plain number, so no free variable may survive; the inner integral is done first with the outer variable held fixed, so that variable legitimately appears in the inner bounds.

WQ2. Why does reordering sometimes make an "impossible" inner integral trivial?

Reversing puts the other variable inside; the strip-length factor it contributes (e.g. multiplying by

y

) can supply exactly the derivative factor needed, letting substitution or cancellation collapse the antiderivative.

WQ3. Why is sketching the region called "80% of the work"?

The picture is the only reliable way to read off the true outer range, spot required splittings, and choose the correct root branch — algebra alone silently produces wrong limits.

WQ4. Why does the swap rest on Fubini's theorem rather than being obvious?

The double integral is a single limit over the 2D region

R

; that this limit equals either iterated (nested 1D) integral is a genuine theorem requiring integrability, not a triviality of notation. See Fubini's Theorem.

WQ5. Why can a region need splitting in one orientation but not the other?

If a horizontal strip would exit onto different curves depending on its height (the boundary "changes formula" partway up), you must split; turning to vertical strips may let every strip exit on one single curve, needing no split.

WQ6. Why do we sometimes change to Polar Coordinates in Double Integrals instead of just swapping

d x d y

For circular/annular regions both Cartesian orders give ugly

r^{2} - x^{2}

limits; polar makes the region a simple rectangle in

(r, θ)

, a different (and often better) kind of "reordering" via Change of Variables (Jacobian).

WQ7. Why does the shared boundary curve play a special role in the swap?

It is the single equation touched by both descriptions; the entire order change is the algebraic act of solving that shared boundary for the opposite variable — everything else just relabels which slot is inner.

WQ8. Why doesn't the value depend on whether we sum rows-then-columns or columns-then-rows?

The double integral is a limit of a sum over the same set of tiles, and finite summation is commutative; the iterated integrals are just two orders of adding identical tiles, so the totals coincide.

Edge cases

EC1. What are the swapped limits when the region degenerates to a single point (the two boundary curves meet)?

The inner limits collapse (lower = upper), the strip has zero length there, and the integral contributes

0

at that value — a measure-zero edge that never affects the result.

EC2. What happens at

y = 0

and

y = 1

for the lens between

y = x

and

y = x^{2}

after swapping?

At both endpoints the curves meet (

x = y = y

y = 0

and

y = 1

), so the inner interval

[y, y]

shrinks to a point — the region pinches to the corners

(0, 0)

and

(1, 1)

EC3. If the region is a rectangle

a \leq x \leq b, c \leq y \leq d

, what changes when you swap?

Nothing but the differential order — both inner and outer limits are already constants, so

\int_{a}^{b} \int_{c}^{d} = \int_{c}^{d} \int_{a}^{b}

; this is the cleanest case of Fubini.

EC4. On a region where

y

appears, why must you check whether

0 < y < 1

y > 1

before ordering the inner limits?

Because

y

crosses the line

y

: for

0 < y < 1

y < y

, but for

y > 1

y < y

; the smaller value is always the lower limit, so the inequality can flip and mislabel enter/exit edges.

EC5. What if a horizontal strip exits the region, re-enters, and exits again (a non-simple region)?

The single inner integral is invalid; you must split into a sum of integrals, one per contiguous piece of the strip inside

R

— a case the sketch reveals immediately.

EC6. For an unbounded region (e.g.

x \geq 0, y \geq x

out to infinity), is the swap still legal?

Yes wherever the improper double integral converges absolutely; then Fubini/Tonelli let you swap, but you must keep the limits improper and confirm convergence — see Improper and Non-elementary Integrals.

EC7. Does the reordering idea extend to three variables?

Yes — for Triple Integrals — Order of Integration there are

3! = 6

possible orders of

d x, d y, d z

, all equal by Fubini, and the same "outer constants, inner surfaces/curves" discipline applies at each level.

Recall One-line survival rule

Every trap above is defeated by the same reflex: sketch $R$ , choose the new outer's full range as constants, re-solve each boundary for the new inner variable, and test an interior point. The single rule that fixes the most errors on this page ::: Outer limits must be constants; inner limits may be curves — never the reverse.