Chapter 10 · Double integrals: setting up, changing order, polar

The hook

The integral school can't touch

Here is an integral over a triangle. The function is harmless; the region is harmless. And done one way it is impossible:

I = \int_0^1 \int_x^1 e^{y^2}\,dy\,dx.

One region, one function. Vertical strips (left) ask you to integrate

e^{y^2}

in

y

first — and no elementary antiderivative exists. Horizontal strips (right) integrate in

x

first, which is trivial, and the rest falls out.

Slice it the other way. Holding $y$ fixed and integrating across in $x$ first, the inner integral is just $e^{y^2}$ times the width of the strip:

\int_0^1 \int_0^y e^{y^2}\,dx\,dy = \int_0^1 y\,e^{y^2}\,dy = \frac{e-1}{2} \approx 0.8591.

Same region, same function. The only thing that changed is the direction you slice. That is the entire engine of this chapter.

Why this matters

In this unit the marks live in the first two lines of the answer — the sketch and the limits — not in the integration that follows. Set the region up correctly and the calculus is ordinary single-variable work you already own. Set it up wrong and no amount of cleverness downstream can save you. Double integrals are also the gateway to everything left in the course: triple integrals, line and surface integrals, and the three great theorems all rest on this one skill.

Intuition before formulas

A camera already does this twelve times a second

A grayscale photo is a function $f(x, y)$ — a brightness value at each point of the frame. The total light the sensor collects is a single number, $\iint_R f\,dA$ . And the sensor reaches that number the only way anything can: it samples $f$ on a grid of pixels and adds up millions of tiny pillars — one standing over each pixel, its height the brightness $f$ there. Keep the three roles apart: the photo is the function, the total light is the one number, the pixels are how the sum gets computed. Every photo on her phone is a double integral, evaluated millions of tiny pillars at a time.

The double integral as a pillar sum

\iint_R f\,dA = \lim_{\Delta A \to 0} \sum_k f(x_k, y_k)\,\Delta A.

Each term is a pillar: base $\Delta A$ , height $f(x_k,y_k)$ , so volume $f \cdot \Delta A$ . Add the pillars and shrink the bases to zero. When $f \ge 0$ the sum is exactly the volume under the surface $z = f(x,y)$ sitting above the region $R$ ; and taking $f = 1$ makes every pillar one unit tall, so $\iint_R 1\,dA$ is just the area of $R$ .

Warm-up · a rectangle, both orders, the same number

Integrate $f = 4 - x - y$ over the rectangle $[0,2]\times[0,1]$ . Over a rectangle the limits are four plain constants, so nothing can go wrong yet — slice it either way:

\int_0^2\!\int_0^1 (4 - x - y)\,dy\,dx = \int_0^2 \Big(4 - x - \tfrac12\Big)\,dx = 7 - 2 = 5,

\int_0^1\!\int_0^2 (4 - x - y)\,dx\,dy = \int_0^1 \big(8 - 2 - 2y\big)\,dy = 6 - 1 = 5.

Both give $\mathbf{5}$ . That both orders agree is no accident — it has a name, Fubini's theorem, and the next beat leans on it hard.

The honest truth about "seeing" it

Nobody — not your lecturer, not the people who invented this subject — visualizes a 3D solid directly. Everyone reduces it to a flat shadow and a sweep of slices. This is not a talent you lack; it is a thing no one does. The procedure below is the whole skill.

The slicing ritual — five moves, every single time

sketch → shadow → slice → limits → integrate. Draw the region. Find its shadow — the interval the outer variable sweeps across (in 2D the shadow is just that interval; name it now, because in the triple-integral chapter the shadow becomes a flat region and the word carries over). Choose a slice direction. Read the limits off the sketch. Only then integrate. Skip the sketch and you are reading limits off the algebra — the single most expensive mistake in this unit.

Formal statement

Reading limits off the sketch

Two ways to slice a region $R$ . Vertical slices fix $x$ and run $y$ up the strip; horizontal slices fix $y$ and run $x$ across it:

The two iterated forms

\iint_R f\,dA = \int_a^b \!\int_{g_1(x)}^{g_2(x)} f\,dy\,dx \qquad\text{or}\qquad \int_c^d \!\int_{h_1(y)}^{h_2(y)} f\,dx\,dy.

(You may meet these called Type I and Type II regions elsewhere; we will say vertical and horizontal.) One rule governs every setup, and it is worth saying once, cleanly:

The one rule for limits

The outer limits are constants. The inner limits may depend only on the outer variable. If an outer limit still contains a variable, or an inner limit mentions the inner variable itself, the setup is broken — re-read it off the sketch.

Worked example · WE-2 · the ritual, walked in full

Compute $\iint_R xy\,dA$ , where $R$ is the sliver between $y = x^2$ and $y = x$ . Sketch first: the two curves meet at $x = 0$ and $x = 1$ , and on that interval $x \ge x^2$ , so $y = x$ is the roof and $y = x^2$ the floor.

Vertical slices. Fix $x$ in $[0,1]$ ; $y$ runs from the floor $x^2$ to the roof $x$ :

\int_0^1\!\int_{x^2}^{x} xy\,dy\,dx = \int_0^1 x\cdot\frac{x^2 - x^4}{2}\,dx = \frac12\Big(\tfrac14 - \tfrac16\Big) = \frac{1}{24}.

Horizontal slices. Re-describe the same region: fix $y$ in $[0,1]$ ; $x$ runs from the line $x = y$ to the curve $x = \sqrt{y}$ (because $y = x^2$ means $x = \sqrt{y}$ ):

\int_0^1\!\int_{y}^{\sqrt{y}} xy\,dx\,dy = \int_0^1 y\cdot\frac{y - y^2}{2}\,dy = \frac12\Big(\tfrac13 - \tfrac14\Big) = \frac{1}{24}.

Moral (Fubini, stated correctly). When $f$ is continuous on $R$ , both orders always give the same number — full stop. The hook's vertical-slice integral equals $\tfrac{e-1}{2}$ too; it is not wrong, it simply cannot be finished by hand because $e^{y^2}$ has no elementary antiderivative. Slice direction never changes the answer — only whether you can reach it.

Changing the order is a re-sketch, never a relabel

To swap $dy\,dx$ for $dx\,dy$ : re-sketch the region, then re-describe it with the slices turned ninety degrees. Never just swap the differentials and keep the old numbers — that describes a different region. The swap is the move that rescues the impossible-looking integral.

Worked example · WE-3 · a swap that makes it doable

Evaluate $\displaystyle\int_0^1\!\int_x^1 \sin(y^2)\,dy\,dx$ . The inner integral $\int \sin(y^2)\,dy$ has no elementary form — so swap. The region is $0 \le x \le y \le 1$ ; re-described with horizontal slices, $y$ runs $0 \to 1$ and $x$ runs $0 \to y$ :

\int_0^1\!\int_0^y \sin(y^2)\,dx\,dy = \int_0^1 y\,\sin(y^2)\,dy = \frac{1 - \cos 1}{2} \approx 0.2298.

The $x$ -integral handed us the factor $y$ that makes the substitution $u = y^2$ work. Reaching for the swap the moment the inner integral looks impossible is the reflex this chapter is built to install.

End of sitting 1 — you can now sketch a region, read off the limits in either direction, and rescue an integral by changing the order. After the break: the polar recipe for round regions, a short mass-and-centroid detour, and the Slice Lab to drill all of it.

Formal statement · polar

Round regions: the polar recipe

When the region is a disk, a wedge, or a ring, Cartesian limits drag in ugly square roots. Switch to polar — $x = r\cos\theta$ , $y = r\sin\theta$ , $x^2 + y^2 = r^2$ — and the limits become constants. There is one thing you must never drop:

The polar area element

dA = r\,dr\,d\theta.

Zoom in on a tiny polar cell. Its radial side is

\Delta r

; its arc side is an arc of radius

r

subtending

\Delta\theta

, so its length is

r\,\Delta\theta

. Area

\approx \Delta r \cdot r\,\Delta\theta

— and that stray

r

is the whole reason

dA = r\,dr\,d\theta

.

Worked example · WE-4 · a quarter disk in polar

Compute $\iint_R (x^2 + y^2)\,dA$ over the quarter disk $r \le 2$ in the first quadrant. In polar the integrand is $x^2 + y^2 = r^2$ , the wedge is $0 \le \theta \le \tfrac{\pi}{2}$ , $0 \le r \le 2$ , and $dA = r\,dr\,d\theta$ :

\int_0^{\pi/2}\!\int_0^{2} r^2\cdot r\,dr\,d\theta = \int_0^{\pi/2}\!\int_0^{2} r^3\,dr\,d\theta = \frac{\pi}{2}\cdot 4 = 2\pi.

The same integral in Cartesian form would carry $\sqrt{4 - x^2}$ through every line. The $r$ in $dA$ is doing real work — leave it out and the answer is simply wrong.

A short detour: mass and centroid

Give the region a density $\rho(x,y)$ and the same machinery weighs it. The mass is the integral of the density; the centroid (the balance point) is the density-weighted average position:

Mass and centroid

m = \iint_R \rho\,dA, \qquad \bar{x} = \frac{1}{m}\iint_R x\,\rho\,dA, \qquad \bar{y} = \frac{1}{m}\iint_R y\,\rho\,dA.

One worked example fixes the pattern. Take the unit-density triangle with vertices $(0,0)$ , $(1,0)$ , $(0,1)$ , so $\rho = 1$ and the region is $0 \le x \le 1$ , $0 \le y \le 1 - x$ :

m = \int_0^1\!\int_0^{1-x} 1\,dy\,dx = \frac12, \qquad \bar{x} = \frac{1}{m}\int_0^1\!\int_0^{1-x} x\,dy\,dx = \frac13, \qquad \bar{y} = \frac13.

The centroid $\left(\tfrac13, \tfrac13\right)$ sits one-third of the way in from each leg, as the symmetry of the triangle demands.

Scope — read the hedge

Mass and centroid appear on some Indian engineering papers; unless your class notes emphasize them, you need only the three formulas above and this one worked example. Do not over-invest here — the slicing and the order-change are where the marks concentrate.

Interactive playground

The Slice Lab

Three drills, one rule: commit your limits before anything renders. Build them, check them, and only then does the region light up and the slices sweep. Committing — even to a wrong setup — is what builds the skill; recognising the right limits once you see them is not the same thing.

The Slice Lab

Region:

Slice direction:

Region:

Your flipped order needs:

Region:

Area element dA =

A slot lights green when it matches the sketch, amber when it does not — and the verdict tells you why. The region only draws after you check.

What to notice

The outer limits are always constants. Every correct setup in Mode 1 has two numbers on the outside; a variable there is the checklist's first red flag.
Flipping is re-describing, not relabelling. In Mode 2 the square roots appear and disappear as the slices turn — because you are reading a genuinely different picture, not swapping symbols.
R4 forces a split. One region, but in the vertical direction the roof changes formula at $x = 1$ , so it needs two sets of limits. No single pair can describe it.
The $r$ is not optional. In Mode 3, dropping the $r$ from $dA$ is flagged immediately — and an offset circle swept through a full turn double-counts itself.

Retrieval gauntlet

Test yourself before moving on

Commit to an answer before reading the feedback. Sketch each region first — these are setup questions, and the sketch is the answer key.

Quiz 1 · read the limits off the sketch

$R$ is bounded by $y = x^2$ and $y = 4$ . Which vertical-slice setup is correct?

Quiz 2 · reverse the order

Reverse the order of $\displaystyle\int_0^1\!\int_x^1 f\,dy\,dx$ .

Quiz 3 · polar, and the r that must not vanish

$\displaystyle\iint_R (x^2 + y^2)\,dA$ over the upper half disk $x^2 + y^2 \le 9$ , $y \ge 0$ , equals:

Predict the next step. You have sketched the upper half-disk of radius 3 and written

\iint (x^2+y^2)\,dA

. What does the integrand become in polar, and what extra factor rides in with

dA

?

Sketch. Half-disk, radius 3, sitting on the $x$ -axis: $\theta$ runs $0 \to \pi$ , $r$ runs $0 \to 3$ .

Convert. $x^2 + y^2 = r^2$ , and $dA = r\,dr\,d\theta$ , so the integrand is $r^2 \cdot r = r^3$ :

\int_0^{\pi}\!\int_0^{3} r^3\,dr\,d\theta = \int_0^{\pi}\frac{3^4}{4}\,d\theta = \frac{81}{4}\cdot\pi = \frac{81\pi}{4}.

The single most common slip here is writing $\int_0^\pi\!\int_0^3 r^2\,dr\,d\theta$ and getting $9\pi$ — the dropped $r$ .

Quiz 4 · which one is built for polar?

Which is best evaluated in polar? (i) $\iint xy\,dA$ over the triangle $(0,0),(1,0),(1,1)$ ; (ii) $\iint \sqrt{x^2+y^2}\,dA$ over the disk $x^2+y^2 \le 4$ ; (iii) $\iint (x+y)\,dA$ over $[0,1]\times[0,2]$ .

Quiz 5 · the swap is the whole point

Evaluate $\displaystyle\int_0^1\!\int_y^1 \cos(x^2)\,dx\,dy$ .

Predict the next step. You have sketched the region of

\int_0^1\!\int_y^1 \cos(x^2)\,dx\,dy

and seen the inner integral is impossible. What is the very next move — and what does the region become?

Sketch. The region is $0 \le y \le x \le 1$ — the triangle below $y = x$ . Swapping to vertical slices, $x$ runs $0 \to 1$ and $y$ runs $0 \to x$ .

Swap, then substitute. The inner $x$ -integral hands you a factor $x$ :

\int_0^1\!\int_0^{x}\cos(x^2)\,dy\,dx = \int_0^1 x\cos(x^2)\,dx = \frac{\sin(x^2)}{2}\Big|_0^1 = \frac{\sin 1}{2} \approx 0.4207,

with $u = x^2$ , $du = 2x\,dx$ . The swap is the whole solution.

Exam translator

How this appears on exams — and six to practise

Exam shapes

In standard Indian engineering-math exams, and likely in MCC201A unless your class notes differ, double integrals wear four recognisable costumes — each one you have now seen worked:

"Evaluate $\iint_R f\,dA$ over $\langle$ region $\rangle$ " → sketch, slice, read limits (WE-2). Most of the marks are the setup.
"Change the order of integration and evaluate" → re-sketch, never relabel (WE-3). The swap is usually there because the printed order is impossible.
"Evaluate using polar coordinates" → circular region or radial integrand; $dA = r\,dr\,d\theta$ (WE-4). The dropped $r$ is the classic lost mark.
"Find the area / volume / mass / centroid" → $\iint_R 1\,dA$ for area, $\iint_R f\,dA$ for volume, the density formulas for the rest.

Six cold problems

Worked solutions are complete; nothing points off this page. The order below is interleaved on purpose — do them as listed, do not block by method. Two of them, PR2 and PR6, open with a faded bridge: part of the work is done for you, the key step left for you to predict. Every solution opens with the sketch.

PR1 · between two parabolas

Shape · find the intersections, then vertical slices

PR1. Evaluate $\iint_R (x + 2y)\,dA$ , where $R$ is between $y = 2x^2$ and $y = 1 + x^2$ .

Sketch. Set $2x^2 = 1 + x^2\Rightarrow x^2 = 1\Rightarrow x = \pm1$ . On $[-1,1]$ , $1 + x^2 \ge 2x^2$ , so $y = 1 + x^2$ is the roof.

\int_{-1}^{1}\!\int_{2x^2}^{1+x^2}(x + 2y)\,dy\,dx = \frac{32}{15}.

The inner integral in $y$ is elementary; the outer is a polynomial in $x$ . The only real decision was reading roof and floor off the sketch.

PR3 · a polar annulus

Shape · ring ⇒ polar; watch the integrand's own r

PR3. Evaluate $\iint_R \dfrac{1}{x^2+y^2}\,dA$ over the annulus $1 \le r \le 2$ .

Sketch. A full ring: $\theta$ runs $0 \to 2\pi$ , $r$ runs $1 \to 2$ . The integrand $\dfrac{1}{x^2+y^2} = \dfrac{1}{r^2}$ , and $dA = r\,dr\,d\theta$ :

\int_0^{2\pi}\!\int_1^{2}\frac{1}{r^2}\cdot r\,dr\,d\theta = \int_0^{2\pi}\!\int_1^{2}\frac{1}{r}\,dr\,d\theta = 2\pi\,\ln 2.

The $r$ from $dA$ cancelled one power downstairs and turned the integrand into $1/r$ — which is exactly why the answer carries a logarithm.

PR2 · faded bridge · a forced swap

Shape · the inner integral is impossible ⇒ change the order

PR2. Evaluate $\displaystyle\int_0^4\!\int_{\sqrt{x}}^{2}\frac{1}{y^3+1}\,dy\,dx$ .

Bridge: $\int \frac{1}{y^3+1}\,dy$ is a mess, so swap. The region is $0 \le x \le y^2 \le 4$ . Before revealing, predict the next step: in the flipped order, what are the new limits on $x$ and on $y$ ?

Re-describe. The curve $y = \sqrt{x}$ is $x = y^2$ ; the region $\sqrt{x} \le y \le 2$ , $0 \le x \le 4$ becomes $0 \le x \le y^2$ , $0 \le y \le 2$ :

\int_0^{2}\!\int_0^{y^2}\frac{1}{y^3+1}\,dx\,dy = \int_0^{2}\frac{y^2}{y^3+1}\,dy.

Then substitute $u = y^3 + 1$ , $du = 3y^2\,dy$ :

\int_0^{2}\frac{y^2}{y^3+1}\,dy = \frac13\big[\ln(y^3+1)\big]_0^2 = \frac13\ln 9 = \frac{2}{3}\ln 3 \approx 0.7324.

The $x$ -integral handed you the $y^2$ that the substitution needed — the swap did all the work.

PR4 · decision-only · name the move

Shape · no integration — just choose the right tool and say why

PR4. For each, name the move and why — do not integrate. (a) $\iint \ln(x^2+y^2)\,dA$ over $1 \le r \le 3$ ; (b) $\int_0^1\!\int_{\sqrt{x}}^{1}\sqrt{y^3+1}\,dy\,dx$ ; (c) $\iint x^2 y\,dA$ over $[1,3]\times[0,2]$ .

(a) Polar. An annulus plus a radial integrand ( $\ln(x^2+y^2) = \ln r^2$ ) — both signals point polar.

(b) Swap. $\int \sqrt{y^3+1}\,dy$ has no elementary antiderivative; reversing the order integrates in $x$ first and produces the $y^2$ -type factor that rescues it.

(c) Nothing clever. A rectangle with a separable polynomial integrand — either order, four constant limits. Reaching for polar or a swap here would only make it worse. (That is the trap (c) trains against: not every integral wants a tool.)

PR5 · a volume

Shape · volume under a surface = ∬ f dA over the shadow

PR5. Find the volume under $z = 4 - x - y$ over the triangle $(0,0),(1,0),(0,1)$ .

Sketch. The triangle is $0 \le x \le 1$ , $0 \le y \le 1 - x$ , and on it $z = 4 - x - y > 0$ , so the volume is the double integral of the height:

\int_0^1\!\int_0^{1-x}(4 - x - y)\,dy\,dx = \frac{5}{3}.

"Volume under a surface" is just $\iint_R f\,dA$ with $f$ the height — the same setup, read as a solid.

PR6 · faded bridge · the split

Shape · one direction is one piece; the other needs a split

PR6. $R$ is bounded by $x = y^2$ and $x = 2 - y$ , for $0 \le y \le 1$ (the curves meet at $y = 1$ , the point $(1,1)$ ). Evaluate $\iint_R y\,dA$ with horizontal slices, then set up but do not evaluate the vertical-slice version.

Bridge: horizontally it is one clean piece. Before revealing, predict the next step: sweeping in the vertical direction, where does the roof of the region change formula — and what does that force you to do?

Horizontal slices (one piece). For $y$ in $[0,1]$ , $x$ runs from the parabola $x = y^2$ to the line $x = 2 - y$ :

\int_0^1\!\int_{y^2}^{2-y} y\,dx\,dy = \int_0^1 y\,(2 - y - y^2)\,dy = \frac{5}{12}.

Vertical slices (set up only). Sweeping in $x$ , the roof changes formula at $x = 1$ : to the left the cap is the parabola $y = \sqrt{x}$ ; to the right it is the line $y = 2 - x$ . So the region must be split at $x = 1$ :

\int_0^1\!\int_0^{\sqrt{x}} y\,dy\,dx \;+\; \int_1^2\!\int_0^{2-x} y\,dy\,dx \quad\big(= \tfrac{5}{12}\ \text{either way}\big).

Same region, same answer — but the vertical direction needs two integrals where the horizontal needed one. Spotting where the roof formula changes is the whole skill the split is testing.

Three things to carry out of this chapter

The marks are in the setup. Sketch, find the shadow, slice, read the limits off the picture — then integrate. Outer limits constant; inner limits depend only on the outer variable.
Changing order is a re-sketch, never a relabel — and both orders always give the same number (Fubini). The swap is the move that rescues an impossible inner integral.
Round region or radial integrand ⇒ polar, with $dA = r\,dr\,d\theta$ . The $r$ is never optional.

Pre-flight checklist — run it before you integrate

\textbf{1.}\ \text{Sketch drawn? Limits read off the sketch, not the statement?}

\textbf{2.}\ \text{Outer limits constant? Inner limits depend only on the outer variable?}

\textbf{3.}\ \text{Polar: is the } r \text{ in } dA = r\,dr\,d\theta \text{ present?}

Before you move on

The camera finished summing its millions of tiny pillars before she lowered the phone — and now she can set one up by hand. Re-do two of PR1–PR6 from a blank page tomorrow: not re-read, re-do. Setting up the region is muscle memory, and muscle memory is built by reaching for the pencil, not by recognising a finished solution.

What's next. There is a famous integral even sideways slicing cannot save — $\int_{-\infty}^{\infty} e^{-x^2}\,dx$ . Chapter 11 squares it, switches to polar, and the impossible becomes a one-liner. The tool that makes the switch rigorous — the Jacobian — is Chapter 11's headline.

Same material, another voice

If a different explanation would help, this one is worth your time — free, from MIT OpenCourseWare:

Read / watch: MIT 18.02SC, Unit 3 Part A — Double Integrals — sessions on the definition, examples, exchanging the order, and polar coordinates map one-to-one onto this chapter.

✓ Chapter complete. Your progress, and every quiz answer, is saved on this computer — revisit any time.