Double integrals

At a Glance

• Frequency: 10 sub-parts across 8 of 13 years (2013, 2015, 2016, 2017, 2018, 2022, 2024, 2025)
• Priority tier: T1
• Marks (count): 10 (3), 12 (2), 13 (1), 15 (4)
• Average solve time: ~10 min
• Difficulty mix: medium 9, easy 1
• Section: A | Dominant type: computation

Why This Chapter Matters

Double integrals appear in 8 of the last 13 years, almost exclusively in Section A. The questions test one skill above all others: identifying the region correctly and setting up limits. Every computation mistake flows from a wrong region, and every region mistake flows from not sketching the curves. Five archetypes cover all appearances; the most common (5 of 10) is integration over a described region in Cartesian coordinates.

Minimum Theory

Iterated integrals. For $f$ continuous on a region $D$: $\iint_D f\,dA=\int_a^b\!\int_{g(x)}^{h(x)}f(x,y)\,dy\,dx=\int_c^d\!\int_{p(y)}^{q(y)}f(x,y)\,dx\,dy.$ Choose the variable order so that the region boundaries are cleanest to express.

Change of variables. Under $x=x(u,v)$, $y=y(u,v)$: $\iint_D f\,dx\,dy=\iint_{D'} f(x(u,v),y(u,v))\,|J|\,du\,dv,\qquad J=\frac{\partial(x,y)}{\partial(u,v)}.$ Alternatively, compute $|\partial(u,v)/\partial(x,y)|$ and take its reciprocal for $|J|$.

Polar coordinates. $x=r\cos\theta$, $y=r\sin\theta$, $dA=r\,dr\,d\theta$.

Bounding a double integral. If $m\le f(x,y)\le M$ on $D$, then $m\cdot\text{Area}(D)\le\iint_D f\le M\cdot\text{Area}(D)$.

Question Archetypes

Five patterns cover all double integral questions.

double-integral-region (5 question(s); 2013, 2015, 2016, 2018, 2025)

Recognition Cues

• “Evaluate $\iint_D f\,dA$ where $D$ is bounded by [curves].”
• Region described by a line and parabola, two parabolas, or a triangle.
• Often the integrand has absolute value — split the region where the sign changes.

Solution Template

1. Sketch the region: find intersection points; determine which curve is the upper/lower boundary.
2. Choose variable order: prefer $y$ outer (integrate $x$ first) when boundaries are expressible as $x=g(y)$; this avoids splitting in many cases.
3. Set up and evaluate the inner integral; simplify the result.
4. Evaluate the outer integral.

Worked Example(s)

2025 Paper 1, 2025-P1-Q3c-i (10 marks)

Evaluate $\iint_R y\,dA$ over the region bounded by $y=x$ and $y=4x-x^2$.

Intersections. $x=4x-x^2\Rightarrow x^2-3x=0\Rightarrow x=0$ or $x=3$. At $x=1$: parabola $=3>1=$ line, so parabola is above the line on $[0,3]$.

Inner integral (vertical strip, $y$ from $x$ to $4x-x^2$): $\int_x^{4x-x^2}y\,dy=\frac12[(4x-x^2)^2-x^2]=\frac12(15x^2-8x^3+x^4).$

$I=\frac12\int_0^3(15x^2-8x^3+x^4)\,dx=\frac12\left[5x^3-2x^4+\frac{x^5}{5}\right]_0^3=\frac12\cdot\frac{108}{5}=\boxed{\;\frac{54}{5}.\;}$

2013 Paper 1, 2013-P1-Q3c (15 marks)

$\iint_D xy\,dA$ where $D$ is bounded by the line $y=x-1$ and parabola $y^2=2x+6$.

Outer variable $y$ (avoids splitting): boundaries $x=(y^2-6)/2$ (parabola, left) and $x=y+1$ (line, right). Intersections at $y=-2$ and $y=4$.

$I=\int_{-2}^4\int_{(y^2-6)/2}^{y+1}xy\,dx\,dy=\int_{-2}^4\frac y2\!\left[(y+1)^2-\!\left(\frac{y^2-6}{2}\right)^2\right]dy.$

Bracket $=(4(y^2+2y+1)-(y^2-6)^2/4)$… expanding and collecting: integrand $=\frac18(-y^5+16y^3+8y^2-32y)$.

$I=\frac18\int_{-2}^4(-y^5+16y^3+8y^2-32y)\,dy=\frac{288}{8}=\boxed{\;36.\;}$

2015 Paper 1, 2015-P1-Q4a (13 marks)

$\iint_R\sqrt{|y-x^2|}\,dA$ over $R=[-1,1]\times[0,2]$.

Absolute value. $|y-x^2|=x^2-y$ below the parabola $y=x^2$, $y-x^2$ above. Split the $y$-integral at $y=x^2$ for each $x$: $I_1(x)=\int_0^{x^2}\sqrt{x^2-y}\,dy=\frac23|x|^3,\qquad I_2(x)=\int_{x^2}^2\sqrt{y-x^2}\,dy=\frac23(2-x^2)^{3/2}.$ Integrate over $x\in[-1,1]$ (both integrands even in $x$): $I=\frac23\!\underbrace{\int_{-1}^1|x|^3\,dx}_{=1/2}+\frac23\!\underbrace{\int_{-1}^1(2-x^2)^{3/2}\,dx}_{=3\pi/4+2}=\frac13+\frac\pi2+\frac43=\boxed{\;\frac53+\frac\pi2.\;}$ (The $(2-x^2)^{3/2}$ piece uses substitution $x=\sqrt2\sin\theta$ and $\cos^4$ reduction.)

2016 Paper 1, 2016-P1-Q4c (15 marks)

$\iint_{[0,1]^2} f\,dA$ where $f=x+y$ for $x^2<y<2x^2$, $f=0$ elsewhere.

Capped region. For $x\le1/\sqrt2$: band $x^2<y<2x^2$ fits inside the unit square. For $x>1/\sqrt2$: $2x^2>1$ so the upper limit is capped at $y=1$.

$I=\underbrace{\int_0^{1/\sqrt2}\int_{x^2}^{2x^2}(x+y)\,dy\,dx}_{I_1}+\underbrace{\int_{1/\sqrt2}^{1}\int_{x^2}^{1}(x+y)\,dy\,dx}_{I_2}.$

$I_1=\int_0^{1/\sqrt2}(x^3+\frac32x^4)\,dx=\frac1{16}+\frac{3\sqrt2}{80}$. $I_2=\int_{1/\sqrt2}^1(x+\frac12-x^3-\frac{x^4}{2})\,dx=\frac{37}{80}-\frac{19\sqrt2}{80}$.

$I=\frac{21}{40}-\frac{\sqrt2}{5}\approx0.242.$

2018 Paper 1, 2018-P1-Q4b (12 marks)

Evaluate $\int_0^a\int_{x/a}^x\frac{x\,dy\,dx}{x^2+y^2}$.

Inner integral ($x$ fixed): $\int_{x/a}^x\frac{x\,dy}{x^2+y^2}=x\cdot\frac1x\!\left[\arctan\frac yx\right]_{x/a}^x=\arctan1-\arctan\frac1a=\frac\pi4-\arctan\frac1a$ (constant in $x$!).

$I=\int_0^a\!\left(\frac\pi4-\arctan\frac1a\right)dx=a\!\left(\frac\pi4-\arctan\frac1a\right)=\boxed{\;\frac{\pi a}4-a\arctan\frac1a.\;}$

Common Traps

• Always sketch first. The most common error is setting up limits for the wrong region (e.g., swapping which curve is upper vs lower).
• For $D$ bounded by a parabola and a line, using $y$ as the outer variable often avoids splitting; using $x$ outer may require two separate integrals.
• For the absolute-value integral (2015): unpack $|y-x^2|$ into two cases. Do not try to evaluate $\sqrt{|f|}$ directly.
• For the 2016 two-parabola integral: the split point is $x=1/\sqrt2$ (where $2x^2=1$ hits the top of the square), not $x=1$. Missing this cap gives an incorrect answer.
• For the 2018 arctan integral: the factor $x$ in the numerator cancels the $1/x$ from the arctan antiderivative, leaving a constant inner integral — a telltale sign that the problem is simpler than it looks.

change-of-variables-integral (2 question(s); 2015, 2017)

Recognition Cues

• The region is a rhombus or parallelogram with sides $x\pm y=$ const or $xy=$ const.
• The integrand is a function of $x+y$ and $x-y$, or of $x^2-y^2$ and $xy$.
• Substitution makes both the region (a rectangle) and the integrand simpler.

Solution Template

1. Set $u,v$ to be the natural level-set functions.
2. Compute the forward Jacobian $\partial(u,v)/\partial(x,y)$ and invert: $dx\,dy=|J^{-1}|\,du\,dv$ where $|J^{-1}|=1/|\partial(u,v)/\partial(x,y)|$.
3. Transform the region to a rectangle in $(u,v)$.
4. Rewrite the integrand in $(u,v)$ — often it simplifies drastically.
5. Evaluate as a product of single integrals if separable.

Worked Example(s)

2015 Paper 1, 2015-P1-Q3d (12 marks)

$\iint_R(x-y)^2\cos^2(x+y)\,dA$ over rhombus with vertices $(\pi,0),(2\pi,\pi),(\pi,2\pi),(0,\pi)$.

Substitution: $u=x+y$, $v=x-y$. Rhombus sides: $x+y=\pi$, $3\pi$ and $x-y=\pm\pi$ — so the rhombus maps to the rectangle $[\pi,3\pi]\times[-\pi,\pi]$. Forward Jacobian $\partial(u,v)/\partial(x,y)=\det\begin{pmatrix}1&1\\1&-1\end{pmatrix}=-2$, so $dx\,dy=\frac12\,du\,dv$.

$I=\frac12\int_\pi^{3\pi}\cos^2u\,du\int_{-\pi}^\pi v^2\,dv=\frac12\cdot\pi\cdot\frac{2\pi^3}{3}=\boxed{\;\frac{\pi^4}{3}.\;}$

2017 Paper 1, 2017-P1-Q1c (10 marks)

$\iint_R xy(x^2+y^2)\,dA$ over $R:\{-3\le x^2-y^2\le3,\;1\le xy\le4\}$.

Substitution: $u=x^2-y^2$, $v=xy$. Region maps to $[-3,3]\times[1,4]$. Forward Jacobian $\partial(u,v)/\partial(x,y)=2(x^2+y^2)$, so $dx\,dy=du\,dv/[2(x^2+y^2)]$.

Integrand $\times\,dx\,dy=xy(x^2+y^2)\cdot du\,dv/[2(x^2+y^2)]=v\,du\,dv/2$ — the $(x^2+y^2)$ cancels perfectly.

$I=\int_{-3}^3\!\int_1^4\frac{v}{2}\,dv\,du=\frac{15}{4}\cdot6=\boxed{\;\frac{45}{2}.\;}$

Common Traps

• Compute the forward Jacobian $|\partial(u,v)/\partial(x,y)|$ and use $dx\,dy=\frac{1}{|J|}\,du\,dv$ — NOT the other way around.
• For the rhombus: the Jacobian of $(u,v)=(x+y,x-y)$ is $-2$, giving $|J|=2$ and $dx\,dy=\frac12\,du\,dv$. Dropping the $\frac12$ halves the answer.
• Verify the region maps to a rectangle (check all four side equations).

area-between-curves (1 question; 2022)

Recognition Cues

• “Use double integration to find the area common to [circle] and [parabola].”
• Area $=\iint_D dA$ where $D$ is the common region.

Worked Example(s)

2022 Paper 1, 2022-P1-Q3b (15 marks)

Area common to circle $x^2+y^2=4$ and parabola $y^2=3x$.

Intersection: $x^2+3x=4\Rightarrow x=1$ (so $y=\pm\sqrt3$). By symmetry, double the area for $y\ge0$.

$y$ outer (parabola $x=y^2/3$ left, circle $x=\sqrt{4-y^2}$ right): $A_{\text{upper}}=\int_0^{\sqrt3}\!\left(\sqrt{4-y^2}-\frac{y^2}{3}\right)dy.$ $\int_0^{\sqrt3}\sqrt{4-y^2}\,dy$ via $y=2\sin\theta$ ($\theta\in[0,\pi/3]$) $=4\int_0^{\pi/3}\cos^2\theta\,d\theta=\frac{2\pi}{3}+\frac{\sqrt3}{2}$. $\int_0^{\sqrt3}y^2/3\,dy=\sqrt3/3$.

$A=2(A_{\text{upper}})=2\!\left(\frac{2\pi}3+\frac{\sqrt3}2-\frac{\sqrt3}3\right)=\boxed{\;\frac{4\pi}{3}+\frac{\sqrt3}{3}.\;}$

Common Traps

• “Common to” both curves means inside both: inside the disc ($x^2+y^2\le4$) AND inside the parabola ($y^2\le3x$).
• Intersection $x=1$, not $x=2$ — solve $x^2+3x=4$ carefully.
• The trig substitution gives $\theta\in[0,\pi/3]$ (since $\sin^{-1}(\sqrt3/2)=\pi/3$), not $[0,\pi/2]$.

area-polar (1 question; 2024)

Recognition Cues

• “Find the area inside/outside [cardioid/circle/rose].”
• Region described using $r=f(\theta)$; double integration in polar coordinates.

Worked Example(s)

2024 Paper 1, 2024-P1-Q4b (15 marks)

Area inside cardioid $r=a(1+\cos\theta)$ and outside circle $r=a$.

Intersections: $a(1+\cos\theta)=a\Rightarrow\theta=\pm\pi/2$. For $|\theta|<\pi/2$, cardioid $\ge$ circle.

$A=\int_{-\pi/2}^{\pi/2}\!\int_a^{a(1+\cos\theta)}r\,dr\,d\theta=\frac{a^2}{2}\int_{-\pi/2}^{\pi/2}[(1+\cos\theta)^2-1]\,d\theta=\frac{a^2}{2}\int_{-\pi/2}^{\pi/2}(2\cos\theta+\cos^2\theta)\,d\theta.$ Using symmetry and $\int_0^{\pi/2}\cos^2\theta\,d\theta=\pi/4$: $A=a^2(2+\pi/4)\cdot\frac{1}{1}=a^2\!\left(2+\frac\pi4\right)=\boxed{\;\frac{a^2(\pi+8)}{4}.\;}$

Common Traps

• Integration range is $[-\pi/2,\pi/2]$ only — for $|\theta|>\pi/2$ the cardioid is inside the circle, so there is no required region there.
• The half-angle identity $\cos^2\theta=(1+\cos2\theta)/2$ is needed for the $\cos^2$ term.

integral-bounds (1 question; 2017)

Recognition Cues

• “Prove $A\le\iint_D f\le B$” where $f$ involves a distance function.

Worked Example(s)

2017 Paper 1, 2017-P1-Q4d (10 marks)

Prove $\pi/3\le\iint_D dx\,dy/\sqrt{x^2+(y-2)^2}\le\pi$ over unit disc $D$.

Key: integrand $=1/\text{dist}((x,y),Q)$ where $Q=(0,2)$ is outside $D$. For any $P\in D$: $\text{dist}(P,Q)\in[2-1,2+1]=[1,3]$, so $1/\text{dist}\in[1/3,1]$.

$\frac13\cdot\underbrace{\iint_D dA}_{\pi}\le\iint_D\frac{dA}{\text{dist}}\le 1\cdot\pi,\qquad\text{i.e.}\quad\frac\pi3\le I\le\pi.\qquad\blacksquare$

Common Traps

• $Q=(0,2)$ is outside the disc — distance is bounded below by $1$ (the integral is proper, not improper).
• Reciprocating: large distance $\to$ small integrand; $\min$ distance at $(0,1)$ gives $\max$ integrand $=1$.

Marks-Aware Writing

10-mark questions (2017-Q1c, 2017-Q4d, 2025): Sketch the region (one line description), set up the iterated integral, evaluate. For change-of-variables: state the substitution, Jacobian, new region, integral.

12-13-mark questions (2015-Q3d, 2018): Change-of-variables or inner-then-outer integration. Show the substitution and Jacobian computation explicitly. For 2018: show the cancellation of the $x$ factor.

15-mark questions (2013, 2015-Q4a, 2016, 2022, 2024): Full working: identify region (with sketch/description), set up limits with explanation of which curve is upper/lower, inner integral, outer integral. For absolute-value integrands: show the region split explicitly.

Practice Set