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11Double integrals over rectangles
You hang a rectangular awning over a patio. Its cloth is not flat: over every spot on the rectangular floor the cloth sits at a different height, sagging low in the middle and pulled tight at the poles. You want the volume of air trapped underneath. Measuring the height at the four corners is not enough, and there is no single height to multiply the floor by.
By the end you can turn a height that changes across a whole rectangle into one number — the trapped volume — and evaluate it as two ordinary integrals done one after another.
In 60 seconds
Over a rectangle, cut the base into a grid, multiply each cell's area by the height above it, add, and refine: that limit is the double integral, and it is computed as two single integrals nested inside each other.
the integrand factors into a function of x times a function of y
Three most common mistakes
Treating the inner limits as if they belonged to the outer variable. In dy dx the numbers c and d go with the inner dy and a and b with the outer dx; swap them and the answer is garbage.
Carrying the inner variable out of the inner integral. When you integrate over y with x held fixed, x is a constant, so the antiderivative in y may still contain x.
Reading the Midpoint estimate as the exact value. It is exact only for a flat plane; for anything curved it is an approximation that improves as the grid refines.
On this course only quizzes (10 percent) and homework (10 percent) carry a stated weight, so treat set-ups from this section as quiz-sized: one rectangle, one integrand, one order chosen for ease. The published weights say nothing finer than that, so no claim about how often a particular type appears is made here.
How much time do you have?
10 minutes
You leave able to evaluate a double integral over a rectangle by writing it as dy dx and doing the inner integral first — which is the whole mechanical skill.
In 60 seconds card, Iterated integrals: integrate one variable at a time, Fubini's Theorem: the double integral is either iterated integral
45 minutes
Add the meaning behind the symbol, the numerical estimate, the product shortcut and the average value, so a worded or set-up-only question does not surprise you.
everything in the 10 minute path, The double integral as a limit of sums over a rectangle, Estimating a double integral: the Midpoint Rule, Product integrands over a rectangle, Average value and the basic properties, the faded ladder
full read
Every worked example, the exam-shaped question, and the mixed practice, including the questions that fold in partial derivatives from last week.
all concepts in order, worked exam question, practice A through D
By the end of this section
Interpret the double integral over a rectangle as the limit of , and as a under the surface when the integrand is nonnegative.
Estimate a double integral over a rectangle with the Midpoint Rule and say in which direction the estimate is likely to err.
Evaluate an iterated integral over a rectangle by integrating the inner variable first with the outer variable held constant, then the outer.
Apply Fubini's Theorem to swap the order of integration on a rectangle and choose the order that makes the inner integral elementary.
Factor a product integrand over a rectangle into two single integrals and evaluate them separately.
Compute the average value of a function over a rectangle and bound a double integral using the comparison and linearity properties.
Syllabus coverage
15.1
Double integrals over rectangles: double Riemann sums, the definition and volume meaning, the Midpoint Rule, iterated integrals, Fubini's Theorem, the product rule for a rectangle, and the average value.
The single syllabus token is developed across all six blocks below, in the order the book uses: meaning first, then the numerical estimate, then the mechanical iterated-integral skill, then Fubini that justifies it, then the two shortcuts.
covered
general regions
Double integrals over non-rectangular regions
Not on this week's line; it is the next section, 15.2, where the inner limits stop being constants and start depending on the outer variable.
deferred
Recall first
The single definite integral as a limit of sums
$\int_a^b f(x)\,dx=\lim_{n\to\infty}\sum_{i=1}^{n} f(x_i^{*})\,\Delta x$, the limit of areas of thin rectangles as they get thinner.
The double integral copies this construction one dimension up: cells instead of intervals, and a height above each cell instead of above each point.
Evaluating a single definite integral
If $F'=f$ on $[a,b]$ then $\int_a^b f(x)\,dx=F(b)-F(a)$.
Every double integral here is finished by doing this twice, once for the inner variable and once for the outer.
Partial evaluation: holding one variable fixed
From last week, $\partial/\partial y$ treats $x$ as a constant. The inner integral does the same thing with an antiderivative instead of a derivative.
The inner integral integrates in $y$ while $x$ sits still, so any $x$ in the integrand rides along as a constant.
Try it yourself first (2 questions)
1§11.0 — a single integral with a spectator constant●●○○○
One question before anything is defined. Treat $x$ as a fixed number and integrate in $y$ only; this is the single move the whole section leans on.
Given
$\displaystyle\int_0^2 (x+y)\,dy$, with $x$ held constant
Find
Evaluate the integral; your answer will still contain $x$.
Hint 1/4
You are integrating in $y$. Ask what plays the role of the constant here and what plays the role of the variable.
Hint 2/4
$\int (x+y)\,dy$ has antiderivative $xy+\tfrac{y^2}{2}$ because $x$ is a constant multiplier of $y$.
Hint 3/4
Put in the limits $y=0$ and $y=2$ with $x$ still a constant: evaluate $xy+\tfrac{y^2}{2}$ from $0$ to $2$.
evaluating the $y$-antiderivative at the two limits
Answer $$4x^{2}$$
Check
Independent check at $x=1$: the integrand is $y$, and $\int_1^3 y\,dy=4$, which equals $4\cdot 1^{2}$.
Notation
symbol
reads as
means
watch out
$\iint_R f\,dA$
the double integral of f over R d A
the limit of the double Riemann sums of $f$ over the rectangle $R$; when $f\ge 0$ it is the volume under the surface.
The $dA$ is a single , not two separate differentials to cancel. It records that we are adding up area times height.
$R=[a,b]\times[c,d]$
R equals the box a b cross c d
the rectangle of points $(x,y)$ with $a\le x\le b$ and $c\le y\le d$.
All four numbers are constants here. The moment a limit depends on the other variable you have left this section.
$\Delta A$
delta A
the area $\Delta x\,\Delta y$ of one cell of the grid.
It is an area, the product of two lengths, so multiplying it by a height gives a volume.
$\bar x_i,\ \bar y_j$
x bar i and y bar j
the midpoints of the $i$-th column interval and the $j$-th row interval, used by the Midpoint Rule.
Midpoints, not endpoints. Using a corner instead changes the estimate, though not the exact limit.
$\int_a^b\!\int_c^d f\,dy\,dx$
the iterated integral, d y then d x
do the inner integral in $y$ from $c$ to $d$ first, then the outer in $x$ from $a$ to $b$.
Differentials pair inside out: $dy$ with the inner limits $c,d$ and $dx$ with the outer limits $a,b$.
Conventions used here
The area element does not care about order
We write $dA=dx\,dy=dy\,dx$; the order in an iterated integral records only which single integral is done first, and on a rectangle both orders carry constant limits.
It stops the differentials from looking like quantities to be cancelled or reordered by algebra, and keeps the focus on which limits pair with which differential.
Inner limits pair with the inner differential
In $\int_a^b\int_c^d f\,dy\,dx$ the numbers $c,d$ belong to $dy$ and $a,b$ to $dx$, reading from the inside out. The innermost integral is always the one evaluated first.
The single most common wreck in the section is attaching the wrong pair of numbers to a differential; naming the rule once removes the guesswork.
Nonnegative means volume, otherwise signed
When $f\ge 0$ on $R$ the double integral is the volume under the surface. When $f$ takes negative values the integral is a signed volume, counting the part below the plane $z=0$ as negative.
It keeps a negative answer from looking like an error: a double integral can be negative, exactly as a single one can.
Any works in the limit
The definition allows the sample point $(x_{ij}^{*},y_{ij}^{*})$ to be any point of its cell — a corner, the centre, anywhere. For a continuous $f$ the limit is the same regardless of the choice.
It explains why the Midpoint Rule and a corner rule disagree for a coarse grid yet agree in the limit, so the two are not in conflict.
Angles in trig integrands are radians
Wherever a trigonometric function appears in an integrand, its argument is in radians, so $\int_0^{\pi}\sin x\,dx=2$ and never a degree-based value.
Only in radians do the antiderivatives $\int\sin=-\cos$ and $\int\cos=\sin$ hold without an extra scaling factor.
11.1The double integral as a limit of sums over a rectangle
Defines the symbol: a limit of box-volume sums over a grid, equal to the volume under the surface when $f\ge 0$.
Last week the definite integral of one variable added up thin strips to get an area. Keep the recipe, add a second direction: thin cells instead of thin strips, a height above each cell.
Solvable with what we have
The area under $y=x^{2}$ on $[0,2]$: $\int_0^2 x^{2}\,dx=\tfrac{8}{3}$, one integral, one variable.
The volume of a box: base area times a single constant height.
The volume of a solid whose slice area $A(x)$ is known: $\int_a^b A(x)\,dx$, from single-variable calculus.
Not solvable yet
The air under the awning of the opening, where the height changes as you move in $x$ and as you move in $y$ at once.
The volume under $z=x^{2}+y^{2}$ over a square, where no single $A(x)$ is handed to you.
Reach for the tool in hand: slicing, $V=\int_a^b A(x)\,dx$. Fix $x$, and the slice is the region under the curve $z=f(x,y)$ as $y$ runs across the rectangle. Its area is
$$A(x)=\int_c^d f(x,y)\,dy.$$
So $V=\int_a^b\Big(\int_c^d f(x,y)\,dy\Big)dx$.
Why it fails
Nothing is wrong with it — that is the point. The slice area is itself an integral, so a single integral will not finish the job; you are forced to integrate twice. The rest of the section is about doing that honestly, and about why the two integrations can be done in either order.
DefinitionDefinition: the double integral over a rectangle
Conditions
$R=[a,b]\times[c,d]$ is a rectangle
$[a,b]$ is cut into $m$ equal pieces and $[c,d]$ into $n$ equal pieces, giving $mn$ cells of area $\Delta A=\Delta x\,\Delta y$
$(x_{ij}^{*},y_{ij}^{*})$ is any sample point in cell $(i,j)$
Chop the rectangle into a grid of cells. Over each cell raise a box whose height is the value of $f$ at a sample point in that cell; its volume is height times cell area. Add all the box volumes, then let the cells shrink. The number they approach is the double integral.
Why the sum approaches a volume
Each term $f(x_{ij}^{*},y_{ij}^{*})\,\Delta A$ is the volume of a rectangular column: base $\Delta A$, height the surface value at the sample point. The columns together are a blocky staircase model of the solid under the surface. Refining the grid shrinks the mismatch between the staircase and the smooth solid to zero whenever $f$ is continuous, so the sums converge and their limit is the volume when $f\ge 0$.
The rectangle is cut into a grid; over the shaded cell of area $\textcolor{#1f6feb}{\Delta A}$ the box has height $\textcolor{#6f42c1}{f}$ at the marked sample point.
Looks like this, but is not
$V=(\text{base area})\times(\text{one height})=(b-a)(d-c)\cdot f(\text{some point})$. It has the right units and it is quick.
That is the box formula, and it is exact only when the height never changes across $R$. For a curved surface it is a single Riemann term with $m=n=1$: an estimate, not the integral. It becomes correct only in the limit of many cells.
A constant height really is base times height
The definition must reproduce what you already know when the surface is flat.
Given
$f(x,y)=5$ on $R=[0,3]\times[0,2]$
Find
the double integral, from the definition
SolutionEvery sample value is the same
$$f(x_{ij}^{*},y_{ij}^{*})=5\ \text{for every cell}$$
the height is constant, so the sample point does not matter
the cell areas add back to the whole rectangle's area
$$=5\cdot(3)(2)=30$$
the rectangle is $3$ by $2$, so its area is $6$
Answer $$\iint_R 5\,dA=30$$
Check
Independent check by geometry: this is a box $3$ by $2$ by $5$, volume $30$.
No limit needed; the sum was already exact.
A constant integrand pulls straight out: $\iint_R k\,dA=k\cdot\text{area}(R)$. Keep this as a sanity anchor for every later answer.
A four-cell estimate of a curved volume
A first, rough number for a genuinely curved surface, using the definition with a coarse grid.
Given
$f(x,y)=x+y$ on $R=[0,2]\times[0,2]$
A $2\times 2$ grid, sample point the upper-right corner of each cell
Find
an estimate of $\iint_R (x+y)\,dA$
SolutionList the four corner values
$$\Delta A=(1)(1)=1$$
each cell is $1$ by $1$
$$f(1,1)=2,\ f(2,1)=3,\ f(1,2)=3,\ f(2,2)=4$$
the upper-right corners of the four cells
Add heights times cell area
$$\sum f\,\Delta A=(2+3+3+4)\cdot 1=12$$
each height multiplies the same cell area $1$
Answer $$\text{estimate}=12$$
Check
Independent check on direction: $f$ increases toward the upper-right, so upper-right corners overshoot. The exact value found later is $8$, and indeed $12>8$.
One grid, four evaluations, no antiderivatives.
A coarse corner sum is only a first look. The sample point and the grid both matter until the cells are small; the next block chooses the sample point that behaves best.
Checkpoint
§11.1 — what the symbol adds up●○○○○
Thirty seconds, no computing. A double integral $\iint_R f\,dA$ over a rectangle adds up which product, over a grid of cells?
Given
$R$ is a rectangle cut into small cells of area $\Delta A$
Find
Which product is being summed?
Hint 1/4
Look at the definition and read off the two things multiplied in each term.
Hint 2/4
Each term is a height at a sample point times a cell area, $f(x^{*},y^{*})\,\Delta A$.
Hint 3/4
So the pieces are a value of $f$ and an area $\Delta A$, not a length or a slope.
Estimate the integral by sampling $f$ at the centre of every cell, multiplying each value by the cell area, and adding. The centres average out the rise and fall across a cell better than a corner does.
Four equal cells, each sampled at its $\textcolor{#6f42c1}{\text{centre}}$ rather than at a corner.
Looks like this, but is not
It looks like it should be an exact formula, since it is a clean weighted sum of $f$-values.
It is exact only when $f$ is a plane, $f=\alpha x+\beta y+\gamma$, because then the centre value equals the cell's true average. For anything curved it is an estimate whose error shrinks as the grid refines.
Midpoint estimate of a bowl over a square
A concrete estimate, with the exact value computed later so the error is visible.
Given
$f(x,y)=x^{2}+y^{2}$ on $R=[0,2]\times[0,2]$
A $2\times 2$ grid, so $\Delta A=1$
Find
the Midpoint Rule estimate of $\iint_R (x^{2}+y^{2})\,dA$
SolutionCell centres
$$\bar x\in\{0.5,1.5\},\ \bar y\in\{0.5,1.5\}$$
each interval $[0,1]$ and $[1,2]$ has midpoint $0.5$ and $1.5$
Evaluate at the four centres
$$f(0.5,0.5)=0.5,\quad f(0.5,1.5)=2.5$$
$0.25+0.25$ and $0.25+2.25$
$$f(1.5,0.5)=2.5,\quad f(1.5,1.5)=4.5$$
$2.25+0.25$ and $2.25+2.25$
Sum times cell area
$$(0.5+2.5+2.5+4.5)\cdot 1=10$$
the four values add to $10$, and $\Delta A=1$
Answer $$\iint_R (x^{2}+y^{2})\,dA\approx 10$$
Check
Independent check against the exact value: $\int_0^2\int_0^2 (x^{2}+y^{2})\,dy\,dx=\tfrac{32}{3}\approx 10.67$, so the estimate $10$ is low by about six percent, as a coarse grid on a bowl should be.
Four evaluations, one multiplication.
The Midpoint value came in under the true value because the bowl curves upward and the centre sits below the cell's average height. Refining the grid closes the gap.
Checkpoint
§11.2 — when the estimate is exact●●○○○
One sentence to accept or reject; it fixes what the rule can and cannot promise.
Given
Claim: the Midpoint Rule gives the exact value of $\iint_R f\,dA$ whenever $f(x,y)=2x+3y+1$.
Find
True or false, and say why in one sentence.
Hint 1/4
Ask what shape the surface $z=2x+3y+1$ is.
Hint 2/4
The Midpoint Rule is exact precisely for a flat plane, because a centre value equals the average over a cell there.
Hint 3/4
Here $f=2x+3y+1$ is a plane, so the midpoint value is the true cell average.
Hint 4/4
So the claim is true.
Show solutionCentre equals average for a plane
$$\text{average of a linear } f \text{ over a cell}=f(\text{centre})$$
a plane rises as much above the centre on one side as it falls on the other
Answer $$\text{True}$$
Check
Independent check on a cell: over $[0,1]\times[0,1]$, $\iint (2x+3y+1)\,dA=1+1.5+1=3.5$, and $f(0.5,0.5)\cdot 1=1+1.5+1=3.5$.
⚠ Reporting the Midpoint estimate as the exact answer
the formula looks closed-form and produces a tidy number
11.3Iterated integrals: integrate one variable at a time
Computes a double integral as inner-then-outer single integrals; the inner one freezes the outer variable.
The motivating slice already wrote the answer as one integral inside another. Here is how to actually compute that, and the notation the book uses for it.
The inner integral integrates $f$ in $y$ from $c$ to $d$ while $x$ is frozen, leaving a function of $x$ alone. The outer integral then integrates that function in $x$ from $a$ to $b$, leaving a number.
The $\textcolor{#d1690a}{\text{inner}}$ integral adds $f$ up one vertical strip at a frozen $x$; the $\textcolor{#1f6feb}{\text{outer}}$ integral then slides that strip across $x$.
Looks like this, but is not
You might integrate both variables in one sweep: $\int (x+2y)\,d(x,y)$ as if $d(x,y)$ were a single step.
There is no one-step antiderivative in two variables. The only legal move is one integral at a time; the inner one turns the two-variable integrand into a one-variable function first.
Evaluate the integral of x plus 2y over a rectangle
The full inner-then-outer routine on a clean integrand.
Independent check by the other order: $\int_0^1\int_0^2 (x+2y)\,dx\,dy=\int_0^1 (2+4y)\,dy=2+2=4$, the same.
One inner integral, one outer integral.
The inner integral collapsed a function of two variables to $x+1$, a function of one. That collapse is the whole trick; everything after it is single-variable calculus.
An x-squared rides through the inner integral
Here the frozen variable genuinely survives the inner step.
Independent check by the differentials: $dy$ is innermost, so its integral is innermost and runs first.
⚠ Attaching the inner limits to the outer variable
the four numbers are written left to right and it is tempting to pair them left to right too
wrong$$\int_0^2\!\int_0^1 f\,dy\,dx\to \text{use }0,2\text{ for }dy$$
right$$dy\ \text{takes the inner }0,1;\ dx\ \text{takes the outer }0,2$$
⚠ Dropping the frozen variable in the inner integral
with $x$ held constant it is easy to forget it is still in the integrand
wrong$$\int_0^1 (x^{2}+y)\,dy=\tfrac12$$
right$$\int_0^1 (x^{2}+y)\,dy=x^{2}+\tfrac12$$
11.4Fubini's Theorem: the double integral is either iterated integral
Guarantees the double integral equals either iterated integral on a rectangle, so you may pick the easier order.
Two things still need justifying: that the iterated integral equals the double integral at all, and that the two orders agree. Fubini says yes to both.
TheoremFubini's Theorem
Conditions
$f$ is continuous on $R=[a,b]\times[c,d]$ (or bounded with a small discontinuity set)
For a continuous integrand on a rectangle, the double integral equals the iterated integral, and you may take the two single integrals in whichever order you like; both give the same number.
Proof
The idea is the slicing picture: integrating in $y$ first builds the slice area $A(x)$, and integrating that in $x$ sweeps the slices into the volume; doing it the other way sweeps the perpendicular slices. A full proof needs the theory of the limit of the double sums, deferred to the text.
Same rectangle, two sweeps: $\textcolor{#1f6feb}{dy\,dx}$ slides a vertical strip sideways, $\textcolor{#d1690a}{dx\,dy}$ slides a horizontal strip upward; the totals agree.
Looks like this, but is not
You might expect the freedom to swap order to survive on any region, not just a rectangle.
On a rectangle the limits are constants and swapping is harmless. On a general region the inner limits depend on the outer variable, and swapping order changes those limits — the subject of the next section, not this one.
Both orders give the same number
Evaluate one integral twice, once each way, and watch them agree.
Independent check by the product shortcut, valid because $6x^{2}y=(6x^{2})(y)$ factors: $\big(\int_0^1 6x^{2}dx\big)\big(\int_0^2 y\,dy\big)=(2)(2)=4$.
Two full set-ups on purpose, to show the agreement.
Since the orders always agree here, the only reason to prefer one is ease. The next block and the exam question are entirely about spotting the easy order.
Checkpoint
§11.4 — is swapping order allowed●●○○○
One sentence to accept or reject about what Fubini permits on a rectangle.
Given
Claim: for a continuous $f$ on a rectangle, $\int_a^b\int_c^d f\,dy\,dx=\int_c^d\int_a^b f\,dx\,dy$.
Find
True or false, and give the reason in one sentence.
Hint 1/4
Ask which theorem governs swapping the order of integration.
Hint 2/4
Fubini's Theorem states exactly this equality for continuous $f$ on a rectangle.
Hint 3/4
The integrand is assumed continuous and the region is a rectangle, so the hypotheses hold.
When the integrand splits into an $x$-only factor times a $y$-only factor, the double integral is just the product of the two separate single integrals over their own intervals.
Why the product splits
In the inner integral $\int_c^d g(x)h(y)\,dy$ the factor $g(x)$ is constant in $y$ and pulls out, leaving $g(x)\int_c^d h(y)\,dy$. The remaining $y$-integral is a plain number, call it $K$. The outer integral is then $\int_a^b g(x)\,K\,dx=K\int_a^b g(x)\,dx$, which is the product of the two single integrals.
Looks like this, but is not
$\displaystyle\iint_R (x+y)\,dA \stackrel{?}{=}\Big(\int x\,dx\Big)\Big(\int 1\,dy\Big)+\dots$ — trying to split a sum the same way.
The rule splits a product, not a sum. $x+y$ is not $g(x)h(y)$, so it does not factor; you either integrate it directly by iteration or split it as two separate integrals $\iint x\,dA+\iint y\,dA$.
A product that factors on sight
The integrand is already a product of an x-part and a y-part.
Independent check by iteration: $\int_0^1\int_0^1 e^{x}e^{y}\,dy\,dx=\int_0^1 e^{x}(e-1)\,dx=(e-1)(e-1)=(e-1)^{2}$.
A sum in an exponent is a product of exponentials. Spotting that turned a two-variable integral into two copies of the same one-variable integral.
Checkpoint
§11.5 — which integrand factors●●○○○
Thirty seconds, no computing. Which of the integrands below is a function of $x$ times a function of $y$?
Given
Four candidate integrands over a rectangle
Find
Which one factors as $g(x)\,h(y)$?
Hint 1/4
Ask whether you can write each as one bracket in $x$ times one bracket in $y$.
Hint 2/4
A product like $x^{a}y^{b}$ factors; a sum like $x+y$ or a coupled argument like $\cos(x+y)$ does not.
Hint 3/4
Check $x\,y^{2}$: it is $(x)(y^{2})$, a clean product.
Hint 4/4
So $x\,y^{2}$ is the one that factors.
Show solutionFactor test
$$x\,y^{2}=(x)(y^{2})\ \checkmark$$
one $x$-factor times one $y$-factor
$$x+y,\ x-y,\ \cos(x+y)\ \text{do not factor}$$
sums and coupled arguments are not products
Answer $$x\,y^{2}$$
Check
Independent check: $\iint_{[0,1]^2} xy^{2}\,dA=(\int_0^1 x\,dx)(\int_0^1 y^{2}dy)=\tfrac12\cdot\tfrac13=\tfrac16$, which the iterated integral also gives.
⚠ Splitting a sum as if it were a product
the product rule is fresh and the eye wants to apply it everywhere
Gives the typical height as total over area, and uses comparison and linearity to bound integrals you cannot evaluate.
One number often wanted from a surface is not its total but its typical height. And a few properties let you bound an integral you cannot, or would rather not, evaluate.
DefinitionAverage value over a rectangle
Conditions
$R=[a,b]\times[c,d]$ with area $A(R)=(b-a)(d-c)>0$
The average value is the total, divided by the area of the base. It is the single constant height whose flat box holds exactly the same volume as the true surface.
The $\textcolor{#6f42c1}{f_{\text{ave}}}$ box holds the same volume as the bumpy $\textcolor{#1f6feb}{\text{surface}}$ above the base.
Looks like this, but is not
The average of $f$ over $R$ might seem to be the average of its values at the four corners.
Corner values ignore everything in between. The average is the integral over the area, which weights every point of $R$ equally; four corners are just four points of infinitely many.
Average height of a tilted plane
A plane over a square, whose average you can also guess by symmetry.
Independent check by symmetry: the average of $x$ over $[0,2]$ is $1$ and likewise for $y$, so the average of $x+y$ is $1+1=2$.
One double integral, one division.
For a plane the average is just the value at the centre of the rectangle, here $f(1,1)=2$. That is the same fact that makes the Midpoint Rule exact for planes.
Bounding an integral without evaluating it
The comparison property gives a range for an integral with no elementary antiderivative.
Independent check on plausibility: the true value is about $0.64$, comfortably inside $[\tfrac13,1]$.
When an integrand has no clean antiderivative, the comparison property still pins the answer between $m\,A(R)$ and $M\,A(R)$ using only its largest and smallest values.
Checkpoint
§11.6 — average of a ramp●●○○○
Thirty seconds. The average of a function that depends on $x$ alone can be read off its one-variable average.
Given
$f(x,y)=x$ on $R=[0,4]\times[0,1]$
Find
Find the average value of $f$ over $R$.
Hint 1/4
The height does not depend on $y$, so the average is just the average of $x$ over $[0,4]$.
Hint 2/4
$f_{\text{ave}}=\tfrac{1}{A(R)}\iint_R x\,dA$, with $A(R)=4\cdot 1=4$.
Hint 3/4
$\iint_R x\,dA=\int_0^4\int_0^1 x\,dy\,dx=\int_0^4 x\,dx=8$, so divide $8$ by $4$.
the integrand is $\tfrac{d}{dy}\tfrac{e^{y}-1}{y}$, and the limit at $0$ is $1$
Answer $$e-2$$
Check
Independent check: the answer matches part a, exactly as Fubini promises; only the labour differs.
Same integrand, same rectangle, same answer $e-2$ — but dy first is two clean lines and dx first is an integration by parts followed by spotting an antiderivative.
How to tell them apart
Before choosing an order, glance at each inner integral. If a leading factor of $x$ or $y$ makes one direction an elementary antiderivative and the other a by-parts, take the elementary one.
Scaffolding comes off
The common skeleton
Read the rectangle: the four constant limits a, b, c, d.
Choose an order and write the iterated integral, dy dx or dx dy.
Do the inner integral, holding the outer variable constant; you are left with a function of one variable.
Do the outer integral of that function over its interval; you are left with a number.
Sanity-check: redo in the other order, or bound the answer.
1 · fully worked
Evaluate the integral of 2x plus y over a rectangle
Independent check by the other order: $\int_0^2\int_0^1 (2x+y)\,dx\,dy=\int_0^2 (1+y)\,dy=2+2=4$.
Each skeleton step produced exactly one line. On the next rung the lines are given and you supply the reasons.
2 · you write the reasoning
Easier on purpose: the integrand depends on $x$ only, so the inner integral is almost free. For each line, say why it is legal. Evaluate $\displaystyle\iint_R (x+1)\,dA$ over $R=[0,2]\times[0,1]$.
Set it up as dy dx over $0\le x\le 2$, $0\le y\le 1$.
reasoning
The order dy dx is chosen; on a rectangle either order works and the limits are constants.
Inner integral in $y$: the integrand has no $y$, so it is a constant in $y$.
reasoning
With no $y$ in the integrand, $\int_0^1 (x+1)\,dy=(x+1)\int_0^1 dy=(x+1)\cdot 1$; the interval has length one.
Outer integral in $x$ of the leftover $x+1$.
reasoning
The inner integral left a function of $x$ alone, so the outer step is single-variable calculus.
Evaluate at $2$ and $0$.
reasoning
Putting in the limits $2$ and $0$ gives $2+2=4$; the lower limit contributes nothing here.
3 · find the buried error
Harder than the two above, and the work is done for you — badly. Exactly two of the four steps contain an error. Find them. Problem: $\displaystyle\iint_R x\,y^{2}\,dA$ over $R=[1,2]\times[0,3]$.
Step 1. Set up $\int_1^2\int_0^3 x\,y^{2}\,dy\,dx$.
Independent check on (b): Fubini guarantees the hard order gives the same $e-2$; and $e-2\approx0.718$ is a plausible average height for a surface running from $x e^{0}=x$ up to $x e^{x}$ on the square.
One order comparison, one clean nesting, one division. The marks live in part (a): the wrong order costs an integration by parts.
Read the inner integral before committing to an order. A factor that makes the inner antiderivative elementary in one direction and forces parts in the other is the whole reason Fubini is worth having.
Practice
A · concept 4 questions
1§11.1 — sign of a double integral●●○○○
One claim to accept or reject. It decides whether a negative answer should alarm you.
Given
Claim: $\iint_R f\,dA\ge 0$ for every integrable $f$ on a rectangle $R$.
Find
True or false, with a one-sentence reason.
Hint 1/4
Ask whether the definition forces $f$ to be nonnegative.
Hint 2/4
The sum $\sum f(x^{*},y^{*})\,\Delta A$ inherits the sign of the sampled values of $f$.
Hint 3/4
If $f<0$ somewhere, those terms are negative and can make the total negative; take $f=-1$.
Hint 4/4
So the claim is false: $\iint_R (-1)\,dA=-\text{area}(R)<0$.
Show solutionConstant negative height
$$\iint_R (-1)\,dA=-\,\text{area}(R)<0$$
a constant pulls out; the area is positive
Answer $$\text{False}$$
Check
Independent check: a surface at height $-1$ encloses a signed volume equal to minus the base area.
2§11.4 — the hypothesis Fubini needs●●○○○
No computing. Fubini's Theorem lets you swap the order of integration on a rectangle under one condition below.
Given
Fubini's Theorem on a rectangle $R=[a,b]\times[c,d]$
Find
Which condition is the one Fubini requires?
Hint 1/4
Recall the exact hypothesis stated with the theorem.
Hint 2/4
The theorem asks for continuity of $f$ on the rectangle, not positivity or a special shape.
Hint 3/4
So look for the option that says $f$ is continuous on $R$.
Independent check by splitting: $\iint_R x\,dA=(\int_0^2 x\,dx)(\int_0^3 dy)=6$ and $\iint_R y\,dA=(\int_0^2 dx)(\int_0^3 y\,dy)=9$, sum $15$; the other order also gives $15$.
2§11.3 — carry the frozen variable●●○○○
Here the inner integrand keeps an $x$ that must ride through.
Given
$\displaystyle\int_0^1\!\int_0^2 (x+2y)\,dy\,dx$
Find
Evaluate the integral.
Hint 1/4
Inner integral in $y$, with $x$ a constant.
Hint 2/4
$\int_0^2 (x+2y)\,dy=\big[xy+y^{2}\big]_0^2$.
Hint 3/4
That is $2x+4$; integrate it in $x$ from $0$ to $1$.
Independent check by the average: the plane's value at the centre $(\tfrac12,\tfrac12)$ is $3$, and for a plane the centre value equals the average, so volume $=3\cdot 1=3$.
3§11.6 — bound without evaluating●●●○○
The integrand has no elementary antiderivative. Use the comparison property to pin its integral.
$R=[a,b]\times[c,d]$ is a rectangle; $[a,b]$ is cut into $m$ equal pieces and $[c,d]$ into $n$ equal pieces, giving $mn$ cells of area $\Delta A=\Delta x\,\Delta y$; $(x_{ij}^{*},y_{ij}^{*})$ is any sample point in cell $(i,j)$
$R=[a,b]\times[c,d]$ split into $mn$ equal cells of area $\Delta A$; $\bar x_i$ is the midpoint of the $i$-th $x$-interval, $\bar y_j$ the midpoint of the $j$-th $y$-interval
$R=[a,b]\times[c,d]$ with area $A(R)=(b-a)(d-c)>0$
Check yourself
Close the page and write, from memory: the one definition the whole section is built on, the recipe for turning it into two single integrals, and the one theorem that says the two orders agree.
Say what product a double integral sums over a grid, and what the integral means when the function is nonnegative.
c-double-integral
Produce a Midpoint Rule estimate on a small grid and say whether it is likely high or low.
c-midpoint
Evaluate an iterated integral over a rectangle, keeping the frozen variable through the inner integral.
c-iterated
State Fubini's hypothesis and swap the order of integration to reach an easier inner integral.
c-fubini
Recognise a product integrand and split it into two single integrals.
c-separable
Compute an average value, and bound an integral using the largest and smallest values of the integrand.
c-average
Glossary (11 terms)
double integraliki katlı integral
The limit of double Riemann sums of a function over a two dimensional region; over a rectangle it is the volume under the surface when the function is nonnegative.
double Riemann sumiki katlı Riemann toplamı
A sum over a grid of cells in which each cell area is multiplied by a sampled function value, approximating the double integral.
iterated integralardışık integral
A double integral written as one single integral inside another, evaluated inner first with the outer variable held constant.
Fubini teoremi
The result that a continuous function's double integral over a rectangle equals either iterated integral, so the order of integration may be swapped.
Midpoint Ruleorta nokta kuralı
A numerical estimate of a double integral that samples the function at the centre of each cell; it is exact for a first degree function.
average valueortalama değer
The double integral of a function over a region divided by the area of the region; the constant height with the same enclosed volume.
kesit
The region exposed when a solid is cut by a plane; for a fixed value of one variable it gives the slice area used in slicing.
sample pointörnek nokta
The point chosen inside a cell at which the function is evaluated in a Riemann sum; any point of the cell is allowed in the limit.
area elementalan elemanı
The small area of one cell of the grid, written dA and equal to the product of the two cell side lengths.
ayrılabilir çarpım
An integrand that factors into a function of one variable times a function of the other, letting the double integral split into a product of single integrals.
signed volumeişaretli hacim
The value of a double integral when the function can be negative, counting the part of the solid below the base plane as negative.
What comes next
§12 · Multiple integrals 15.2, 10.3
Every rectangle here had constant limits, and that is exactly what let the two orders be swapped freely. Next the region stops being a rectangle: the inner limits start to depend on the outer variable, and choosing the order becomes a real decision.
Sources
James Stewart, Calculus, Ninth Edition — section 15.1 The definition, the Midpoint Rule, the iterated integral, Fubini's Theorem and the average value follow this section; the rectangles and the numbers here are chosen fresh, not copied.
Course syllabus, week 11: Multiple Integrals 15.1 The single token 15.1 covered above comes from this line; the only stated assessment weights are quizzes 10 percent and homework 10 percent, and nothing finer is claimed.