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12Double integrals over general regions, and polar coordinates
Two double integrals sit next to each other on a past final. The first runs over the square $0\le x\le 1$, $0\le y\le 1$, and last week's method finishes it in a line. The second has the very same integrand, $x+y$, but the region is the triangle below the line $y=x$ inside that square, and it is worth more marks. Feed the triangle into last week's rectangle machinery and you get $1$; the honest answer is exactly half of that.
You will be able to take any region pinned between two curves, read its picture into a pair of integration limits with the inner limits allowed to move, and evaluate the double integral to a number you can check independently.
In 60 seconds
A double integral over a region that is not a rectangle is an ordinary iterated integral whose inner limits are the two curves that bound the region; and polar coordinates rename each point by a distance and an angle so that circles become straight limits.
you meet a bare cosine or sine equation and need to know it is a circle, not centred at the origin
Three most common mistakes
Putting a variable in the outer limits: the outer pair is always two plain numbers, and only the inner pair may carry the other variable.
Reversing the order by swapping $dx\,dy$ and copying the same limits over, instead of redrawing the region and reading new limits off the picture.
Reading $r=2\cos\theta$ as a circle about the origin; it is a circle of radius $1$ centred at $(1,0)$, and the is only one point of it.
The syllabus line gives quizzes $10$ percent and homework $10$ percent of the course, so this week is live for both. No exam weight is stated on that line, so treat any confident claim that this is or is not on a midterm as unfounded and go by what your own instructor announces.
How much time do you have?
10 minutes
You leave able to set up and evaluate a double integral over a region bounded by two curves, in whichever of the two slicing directions the picture hands you. That is the single most-tested skill of the week.
In 60 seconds card, From rectangles to regions that bend, Slicing the other way, Formula card
45 minutes
Add the two moves that separate a full-mark answer from a stuck one: choosing the cheaper order when both are legal, and reversing the order when one order is impossible. Then pick up the polar dictionary you will need next week.
everything in the 10 minute path, Choosing the order, and splitting a region, When one order is impossible, A distance and an angle, Practice B and C
full read
The parts that transfer: the shapes that recur in polar form, the ladder where the scaffolding comes off one rung at a time, and the interleaved set where nothing tells you which coordinate system or which order the question wants.
all blocks in order, The curves that keep coming back, Contrast pairs, Scaffolding comes off, Practice A to D, Mistake ledger, Self audit
By the end of this section
Set up and evaluate a double integral over a , with the inner limits written as the two functions of x that bound it above and below.
Set up and evaluate a double integral over a Type II region, slicing horizontally so the inner limits are functions of y.
Choose the cheaper order for a region that is both types, and split a region that is neither into simple pieces.
Reverse the order of integration by redrawing the region, and use the reversal to evaluate an integral whose inner antiderivative does not exist in elementary form.
Convert points and equations between rectangular and polar coordinates, handling the quadrant of the angle and the non-uniqueness of the representation.
Identify and sketch the standard polar curves, convert their equations, and use symmetry to halve the plotting work.
Syllabus coverage
15.2
Double integrals over general regions
Type I and Type II regions and the iterated integrals they give, choosing the order, splitting a region that is neither type, , and the properties of the double integral including area and the bounding estimate.
covered
10.3
Polar coordinates
The polar coordinate system, the non-uniqueness of a point's representation and negative $r$, conversion of points and equations in both directions, the standard polar curves, and the three symmetry tests. Polar area and the polar double integral itself are left to next week, where they are the whole subject.
covered
Recall first
Fubini's theorem on a rectangle
On $R=[a,b]\times[c,d]$ with $f$ continuous, $\iint_R f\,dA=\int_a^b\int_c^d f\,dy\,dx=\int_c^d\int_a^b f\,dx\,dy$.
Everything this week is the same idea with the constant inner limits $c,d$ replaced by curves. If the rectangle case is not automatic, the general case has nothing to stand on.
Area between two curves
If $g_2(x)\ge g_1(x)$ on $[a,b]$, the area between them is $\int_a^b\big(g_2(x)-g_1(x)\big)\,dx$.
It is exactly $\iint_D 1\,dA$ over a Type I region, so it is both a check on your limits and the simplest double integral there is.
Two antiderivatives by substitution
$\int y\,e^{y^2}\,dy=\tfrac12 e^{y^2}+C$ and $\int x\sin(x^2)\,dx=-\tfrac12\cos(x^2)+C$.
These are the antiderivatives that appear after a reversal of order turns an impossible inner integral into a possible one. The reversal is useless if the resulting substitution is not fluent.
Unit-circle values and the range of arctangent
$\cos\tfrac{\pi}{3}=\tfrac12$, $\sin\tfrac{\pi}{3}=\tfrac{\sqrt3}{2}$, $\tan\tfrac{\pi}{4}=1$; and $\arctan$ returns an angle in $(-\tfrac{\pi}{2},\tfrac{\pi}{2})$.
Converting points to polar form leans on these constantly, and the restricted range of $\arctan$ is the reason a point in the left half-plane needs an extra $\pi$.
Try it yourself first (3 questions)
1§12.0 — a rectangle, before the region bends●○○○○
One line of last week, to make sure the rectangle case is automatic before we let the limits move.
Given
$$\iint_R xy\,dA,\qquad R=[0,1]\times[0,2]$$
Find
Evaluate the double integral over the rectangle $R$.
Hint 1/4
This is a rectangle, so both limits are constants and either order is legal. Do not overthink it.
Hint 2/4
On a rectangle, $\iint_R f\,dA=\int_a^b\int_c^d f\,dy\,dx$, and the inner integral treats the outer variable as a constant.
Hint 3/4
Integrate in $y$ first: $\int_0^2 xy\,dy=x\cdot\frac{y^2}{2}\big|_0^2=2x$. Then integrate $2x$ from $0$ to $1$.
Hint 4/4
The value is $1$.
Show solution
We integrate in $y$ first only because the region is a rectangle and the choice is free; nothing about $xy$ prefers one order.
$x$ is a constant while $y$ runs, so it comes out front of the $y$-integral
$$\int_0^1 2x\,dx=x^2\Big|_0^1=1$$
the inner integral has left a plain function of $x$, integrated normally
Answer $$1$$
Check
Reverse the order as a check: $\int_0^1 xy\,dx=\tfrac{y}{2}$, then $\int_0^2 \tfrac{y}{2}\,dy=1$. Same number, which is Fubini for a rectangle.
2§12.0 — the trap that costs setups●●○○○
A claim about iterated limits that sounds harmless. Decide before reading on; the reason matters more than the verdict.
Given
Claim: in $\int_a^b\int_{g_1(x)}^{g_2(x)}f\,dy\,dx$ the outer numbers $a$ and $b$ are allowed to depend on $x$.
Find
Is the claim true or false?
Hint 1/4
Ask what kind of object the whole double integral has to be when you are finished.
Hint 2/4
The outer integration is the last one; after it, no variable may remain.
Hint 3/4
If $a$ or $b$ contained an $x$, then after integrating out $y$ and then $x$ a stray $x$ would survive. Restate the claim: outer limits with an $x$ in them.
Hint 4/4
False: the outer limits must be constants.
Show solution
We test the claim against what the final object must be, a number, rather than against a memorised rule.
Ask what the outer integral produces
$$\int_{g_1(x)}^{g_2(x)}f(x,y)\,dy=A(x)$$
the inner integral in $y$ leaves a function of $x$ alone
$$\int_a^b A(x)\,dx=\text{a number}$$
the outer integral must remove the last variable; if $a$ or $b$ still held an $x$, an $x$ would survive into the answer
Answer $$\text{False}$$
Check
Independent check: the answer of a definite double integral is a pure number, and a surviving $x$ in the outer limit is dimensionally a function. The two cannot both be true.
3§12.0 — the substitution a reversal will hand back●○○○○
A one-step substitution you will meet again in disguise once we reverse an order of integration.
Given
$$\int_0^1 2y\,e^{y^2}\,dy$$
Find
Evaluate the integral.
Hint 1/4
Look for a factor that is the derivative of something else already in the integrand.
Hint 2/4
If the integrand is $g'(y)\,e^{g(y)}$, then $u=g(y)$ turns it into $\int e^{u}\,du$.
Hint 3/4
Here $g(y)=y^2$ and $g'(y)=2y$, which is present. So $u=y^2$, $du=2y\,dy$, and the integral becomes $\int_0^1 e^{u}\,du$.
Hint 4/4
The value is $e-1$.
Show solution
Substitution rather than anything else, because the derivative of the exponent $y^2$ is $2y$, which is standing in front.
Name the inside and its derivative
$$u=y^2,\qquad du=2y\,dy$$
the $2y\,dy$ is exactly $du$, so the substitution is complete with nothing left over
$$\int_{0}^{1}e^{u}\,du=e^{u}\Big|_0^1=e-1$$
the limits change with the variable: $y=0\Rightarrow u=0$ and $y=1\Rightarrow u=1$
Answer $$e-1$$
Check
Differentiate $e^{y^2}$: the chain rule returns $2y\,e^{y^2}$, the integrand, so $e^{y^2}$ is the right antiderivative and the value is $e^1-e^0=e-1$.
Notation
symbol
reads as
means
watch out
$\iint_D f\,dA$
the double integral of f over dee, dee ay
The total of $f(x,y)$ over the two-dimensional region $D$, built from a rectangle by letting the region be any shape. $dA$ is the area element.
$D$ is a region in the plane, not an interval. Writing a single pair of constant limits for it only works when $D$ is a rectangle.
$\int_a^b\!\int_{g_1(x)}^{g_2(x)}\!f\,dy\,dx$
integral from a to b, integral from g one of x to g two of x
The Type I iterated form: for each fixed $x$ integrate in $y$ from the lower curve to the upper curve, then integrate the result in $x$.
The inner differential $dy$ must match the inner limits, which are functions of $x$. Mismatching them is a guaranteed wrong answer.
$\int_c^d\!\int_{h_1(y)}^{h_2(y)}\!f\,dx\,dy$
integral from c to d, integral from h one of y to h two of y
The Type II iterated form: for each fixed $y$ integrate in $x$ from the left curve to the right curve, then integrate in $y$.
Here the inner differential is $dx$ and the inner limits are functions of $y$. The two forms are not obtained by relabelling; the region has to be re-read.
$(r,\theta)$
the ordered pair r, theta
Polar coordinates of a point: $r$ is the signed distance from the pole and $\theta$ is the angle the ray to the point makes with the .
The order matters and the pair is not unique. $(2,\tfrac{\pi}{3})$ and $(2,\tfrac{\pi}{3}+2\pi)$ are the same point.
$r$
r
The signed distance from the pole. For $r>0$ you walk out along the ray at angle $\theta$; for $r<0$ you walk backwards, through the pole.
$r$ is allowed to be negative, unlike an ordinary distance. $r^2=x^2+y^2$ only fixes $|r|$; the sign is a separate choice.
$dA=dy\,dx=dx\,dy$
dee ay equals dee y dee x equals dee x dee y
The area element in rectangular coordinates, written either way depending on the slicing direction chosen.
In polar coordinates the area element is not $dr\,d\theta$; it carries an extra factor, which is exactly the subject of next week and not assumed here.
Conventions used here
The outer limits are always numbers
In every iterated integral on this page the outer pair of limits is two constants, and only the inner pair may contain the other variable. So $\int_a^b\int_{g_1(x)}^{g_2(x)}f\,dy\,dx$ is legal and $\int_{g_1(x)}^{g_2(x)}\int_a^b f\,dy\,dx$ is not, because the final answer must be a number and a leftover $x$ in the outer limit would leave a function.
A variable surviving into the outer limit is the single most common way a setup produces a nonsensical answer, and it never shows up as an arithmetic slip; it shows up as an $x$ that cannot be there.
Area is reported as a positive number
The double integral $\iint_D 1\,dA$ equals the area of $D$, always a positive number. When a computation of an area comes out negative, the limits have been written with the lower boundary on top, and we fix the order rather than dropping the sign.
A negative area is a signal, not an answer; discarding the sign hides the swapped-boundary mistake that produced it.
Every angle is in radians
Every angle $\theta$ on this page is a radian measure, and every trigonometric value such as $\cos\tfrac{\pi}{3}=\tfrac12$ is read that way. A degree symbol never appears.
The calculus that follows next week differentiates and integrates these functions, and those operations are false in degrees while nothing on the page would reveal it.
How a polar point is reported
Given a choice we report a point with $r\ge 0$ and $\theta$ in $[0,2\pi)$. We allow negative $r$, where $(-r,\theta)$ means the point $(r,\theta+\pi)$ directly opposite through the pole, but we only use it when a curve's equation forces it.
Every point has infinitely many polar names, so without a stated preference two correct answers look like a disagreement.
The angle is fixed by the point, not the calculator
When we recover $\theta$ from $x$ and $y$ we start from $\tan\theta=\tfrac{y}{x}$ but then place $\theta$ in the quadrant where the point actually sits, since $\tan$ cannot tell the first quadrant from the third. The bare $\arctan$ value is only a starting angle.
$\arctan(y/x)$ always returns an angle in the right half-plane, so for any point with $x<0$ it is off by exactly $\pi$, and the error is invisible until you plot it.
12.1From rectangles to regions that bend
Over a region with a single floor curve and a single ceiling curve, the double integral is an iterated integral whose inner limits are those two curves.
Last week every region was a rectangle, so all four limits were numbers. The hook broke that: a triangle has no single number for the top edge, because the top moves as $x$ moves.
Solvable with what we have
$\iint_R xy\,dA$ over $[0,1]\times[0,2]$ — constant limits, done last week
$\iint_R (x+y)\,dA$ over the unit square
the average value of $f$ over a rectangle
Not solvable yet
$\iint_D (x+y)\,dA$ over the triangle below $y=x$
$\iint_D y\,dA$ over the region under a parabola
$\iint_D 1\,dA$ over a disk
Force the triangle below $y=x$ into last week's machinery by integrating over the smallest rectangle that contains it, the unit square. That gives $\iint_{[0,1]^2}(x+y)\,dA=1$. But the square is twice the triangle, and $x+y$ is symmetric across the diagonal, so the honest triangle answer is $\tfrac12$, not $1$.
Why it fails
A rectangle has a top edge at one fixed height $y=d$. The triangle's top edge is the line $y=x$, whose height changes with $x$. Constant inner limits cannot describe a boundary that moves, so the rectangle silently integrates over extra area.
TheoremDouble integral over a Type I region
Conditions
$f$ is continuous on $D$
$D=\{(x,y): a\le x\le b,\ g_1(x)\le y\le g_2(x)\}$ with $g_1,g_2$ continuous
the outer limits $a,b$ are constants; the inner limits $g_1(x),g_2(x)$ are functions of $x$ only
For each fixed $x$ between $a$ and $b$, walk up the vertical line from the floor $g_1(x)$ to the ceiling $g_2(x)$, adding $f$ along the way; that inner integral leaves a function of $x$ alone. Then sweep $x$ from $a$ to $b$, adding those slice-totals. The inner limits are curves because the top and bottom of the slice move; the outer limits are the two numbers where the region starts and stops.
The Type I region between $\textcolor{#1f6feb}{y=x}$ above and $\textcolor{#d1690a}{y=x^2}$ below. The $\textcolor{#6f42c1}{\text{vertical slice}}$ at a fixed $x$ shows the inner integral: $y$ runs from the lower curve to the upper curve, and then $x$ sweeps the slice across.
Looks like this, but is not
A ring, the region between two circles $x^2+y^2=1$ and $x^2+y^2=4$, looks like something we should slice vertically in one go.
A vertical line through the ring meets its boundary in four places, not two, so there is no single floor $g_1(x)$ and single ceiling $g_2(x)$. The Type I form needs exactly one lower and one upper curve, so the ring must first be cut into pieces that each have that shape.
The triangle from the top of the page: ∬ (x+y) over y ≤ x
The integrand is the same $x+y$ from the hook. The region is the lower triangle of the unit square, cut by the diagonal $y=x$.
We slice vertically, in $y$ first, because the region is described most simply that way: for a fixed $x$ the height runs from the floor $y=0$ up to the diagonal $y=x$.
Read the inner limits off the picture
$$a=0,\ b=1;\qquad y:\ 0\ \to\ x$$
$x$ sweeps the whole base from $0$ to $1$; at each $x$ the segment rises from the x-axis to the line $y=x$
Slice the other way as an independent check: $\int_0^1\int_y^1(x+y)\,dx\,dy$ also gives $\tfrac12$. And the naive rectangle $[0,1]^2$ gives $1$, exactly twice this, because the triangle is half of it and $x+y$ is symmetric across the diagonal.
One vertical slice, no splitting; the region has a single floor and a single ceiling.
The move that mattered was turning the upper boundary curve into the inner upper limit. Everything after that was one-variable calculus.
Between a line and a parabola: ∬ y over x² ≤ y ≤ x
A lens-shaped region: the line $y=x$ on top, the parabola $y=x^2$ underneath, meeting at $(0,0)$ and $(1,1)$.
Given
$$\iint_D y\,dA,\quad D\ \text{bounded above by } y=x,\ \text{below by } y=x^2$$
Find
the value of the double integral
Solution
Vertical slices again: on $0\le x\le 1$ we have $x\ge x^2$, so the ceiling is $y=x$ and the floor is $y=x^2$, and the inner integral in $y$ is immediate.
Find where the boundaries meet, to get the outer limits
Independent check by slicing horizontally: at height $y$ the region runs from $x=y$ to $x=\sqrt{y}$, and $\int_0^1\!\int_y^{\sqrt y} y\,dx\,dy=\int_0^1\big(y^{3/2}-y^2\big)dy=\tfrac25-\tfrac13=\tfrac{1}{15}$. The two orders agree.
Whenever the region is bounded by two curves that cross, the crossing points are the outer limits, and finding them is step one, not an afterthought.
Area as a double integral: ∬ 1 over the same lens
Set the integrand to $1$ and the double integral returns the plain area of the region. It is the cheapest check you can run on a set of limits.
Given
$$\iint_D 1\,dA,\quad D\ \text{between } y=x \text{ and } y=x^2$$
Find
the area of D
Solution
We keep the vertical slices from the previous example, because only the integrand has changed; the region and its limits are identical.
Integrate the constant 1 in y
$$\int_{x^2}^{x} 1\,dy=x-x^2$$
integrating $1$ in $y$ just returns the height of the slice, ceiling minus floor
this is exactly the area-between-curves integral from single-variable calculus
Answer $$\dfrac{1}{6}$$
Check
Cross-check against the recalled formula: the area between $y=x$ and $y=x^2$ is $\int_0^1(x-x^2)\,dx=\tfrac16$, the same object viewed as $\iint_D 1\,dA$. A positive number, as an area must be.
Because $\iint_D 1\,dA$ is the area, a quick mental estimate of the area is a free sanity check on any limits you write down.
Checkpoint
§12.1 — reading a triangle into limits●●○○○
Thirty seconds, no computing. The triangle has vertices $(0,0)$, $(2,0)$ and $(2,4)$, and you only have to choose the correct setup.
Given
Triangle with vertices $(0,0)$, $(2,0)$, $(2,4)$.
The slanted edge runs from $(0,0)$ to $(2,4)$, so it is the line $y=2x$.
Find
Which iterated integral equals $\iint_D f\,dA$?
Hint 1/4
Draw it. For a fixed $x$ between the two vertical extremes, where does the segment start and end in $y$?
Hint 2/4
For a Type I region the outer limits are the constant range of $x$, and the inner limits are the floor and ceiling curves in $y$.
Hint 3/4
The base runs $0\le x\le 2$; at each $x$ the segment goes from the x-axis $y=0$ up to the slanted edge $y=2x$. Restated: the corners are $(0,0),(2,0),(2,4)$ and the top edge is $y=2x$.
Hint 4/4
The setup is $\int_0^2\int_0^{2x} f\,dy\,dx$.
Show solution
Vertical slices, because the ceiling is a single line and the floor is the single segment of x-axis.
Fix the range of x and the two boundaries
$$0\le x\le 2$$
the leftmost point is $x=0$, the two right vertices are at $x=2$
$$y:\ 0\ \to\ 2x$$
floor is the x-axis, ceiling is the slanted edge $y=2x$
A Type II region is described by a left curve and a right curve as $y$ runs over an interval, and the inner integral is taken in $x$.
Vertical slices are not the only choice. Turn the picture ninety degrees: fix a height $y$ and let $x$ run from a left curve to a right curve. Some regions are ugly one way and clean the other.
TheoremDouble integral over a Type II region
Conditions
$f$ is continuous on $D$
$D=\{(x,y): c\le y\le d,\ h_1(y)\le x\le h_2(y)\}$ with $h_1,h_2$ continuous
the outer limits $c,d$ are constants; the inner limits $h_1(y),h_2(y)$ are functions of $y$ only
For each fixed height $y$ between $c$ and $d$, walk across the horizontal line from the left curve $h_1(y)$ to the right curve $h_2(y)$, adding $f$ as you go; that inner integral leaves a function of $y$ alone. Then sweep $y$ from $c$ to $d$. The differential is now $dx$ first, and the inner limits are curves written as $x$ in terms of $y$.
A Type II region between $\textcolor{#1f6feb}{x=y^2}$ on the left and $\textcolor{#d1690a}{x=2-y^2}$ on the right. The $\textcolor{#6f42c1}{\text{horizontal slice}}$ at a fixed $y$ runs from the left curve to the right curve; then $y$ sweeps from $-1$ to $1$.
Looks like this, but is not
A region shaped like a sideways letter C, opening to the right, looks horizontally simple: pick a height and slide across.
At a height that passes through the mouth of the C, a horizontal line enters the region, leaves it through the gap, and enters again, meeting the boundary four times. There is no single left curve and single right curve at that height, so the region must be split into pieces that each have exactly one of each.
One horizontal slice where a vertical one would need two: area between two parabolas
The region is caught between $x=y^2$ on the left and $x=2-y^2$ on the right. Read vertically it needs two integrals; read horizontally it needs one.
Horizontal slices, because at every height $y$ the region has one clean left curve $x=y^2$ and one clean right curve $x=2-y^2$; slicing vertically instead, the ceiling would switch from one parabola to the other at $x=1$.
Independent check by symmetry: the region is symmetric about the x-axis, so its area is $2\int_0^1(2-2y^2)\,dy=2\big(2-\tfrac23\big)=\tfrac{8}{3}$. Same value, reached without the negative limit.
One Type II integral. As Type I it would have split at x = 1 into two integrals with different ceilings.
When one slicing direction makes the boundary switch curves partway across and the other does not, take the direction that does not.
A clean Type II triangle: ∬ 2x over 0 ≤ x ≤ y
A triangle with the y-axis on the left, the line $x=y$ on the right, capped at $y=1$.
so $x=y^2$ is the left boundary and $x=y$ the right
$$\int_0^1\!\!\int_{y^2}^{y} f\,dx\,dy$$
inner limits left-to-right, outer the constant $y$-range
Answer $$\int_0^1\int_{y^2}^{y} f\,dx\,dy$$
Check
Test with $f=1$: $\int_0^1(y-y^2)\,dy=\tfrac12-\tfrac13=\tfrac16$, a positive area, so the left and right limits are in the right order.
⚠ Left and right boundaries written in the wrong order
the two curves are given without a picture, and it is easy to copy them top-to-bottom instead of left-to-right
wrong$$\int_0^1\!\!\int_{y}^{y^2} f\,dx\,dy$$
right$$\int_0^1\!\!\int_{y^2}^{y} f\,dx\,dy$$
⚠ Solving the boundary for the wrong variable
a Type II setup needs $x$ as a function of $y$, and a curve given as $y=\sqrt{x}$ gets copied instead of inverted
wrong$$x=\sqrt{y}$$
right$$x=y^2$$
12.3Choosing the order, and splitting a region
Pick the slicing direction whose slices stay in one piece and whose inner integral is doable; split into simple pieces only when neither direction is simple.
Now there are two legal setups for many regions, so the question becomes which to use. Two things decide it: whether one order avoids a split, and whether one order gives an inner integral you can actually do.
RuleProperties of the double integral
Conditions
$f,g$ continuous on $D$; $\alpha,\beta$ constants
$D_1,D_2$ overlap at most along a boundary curve (zero area)
Four facts. Constants and sums pull through the integral. Integrating the constant $1$ returns the plain area. A region cut into non-overlapping pieces has an integral equal to the sum over the pieces. And if the integrand never leaves the band between $m$ and $M$, the integral is trapped between $m$ times the area and $M$ times the area. The last one lets you bound an integral you cannot evaluate.
The tent triangle. A $\textcolor{#6f42c1}{\text{horizontal slice}}$ crosses in one piece from $\textcolor{#1f6feb}{x=y}$ to $\textcolor{#d1690a}{x=2-y}$, so it is one Type II integral; a vertical slice would have to split at the dashed line $x=1$, where the ceiling changes curve.
Looks like this, but is not
You might think that once a region is Type I you should always slice it vertically, since that is the first tool you learned.
The tent is Type I, but slicing it vertically forces two integrals because the ceiling changes at $x=1$. Being Type I means a vertical setup is possible, not that it is the cheapest. The cost is decided per region, by which direction keeps the slice in one piece.
One integral horizontally, two vertically: ∬ y over the tent
The triangle with corners $(0,0)$, $(1,1)$, $(2,0)$. Its ceiling is the line $y=x$ on the left half and $y=2-x$ on the right half, so vertical slicing changes ceiling at $x=1$.
Horizontal slices, to avoid the split: at height $y$ the left edge is $x=y$ (from $y=x$) and the right edge is $x=2-y$ (from $y=2-x$), both holding for the whole range $0\le y\le 1$.
Check by doing it the hard way, as Type I in two pieces: $\int_0^1\!\int_0^{x} y\,dy\,dx+\int_1^2\!\int_0^{2-x} y\,dy\,dx=\tfrac16+\tfrac16=\tfrac13$. The split answer matches, and it took twice the work.
One Type II integral versus two Type I integrals; the horizontal reading paid for itself.
A ceiling that is made of two different curves is the signal to try the other slicing direction before committing to a split.
Bounding an integral you cannot yet evaluate: ∬ e^{−(x²+y²)} over the unit disk
The disk $x^2+y^2\le 1$. We have no elementary way to evaluate this integral yet, so we trap it between two numbers instead.
Sanity check the bounds: $\tfrac{\pi}{e}\approx 1.16$ and $\pi\approx 3.14$, and the true value happens to be $\pi(1-e^{-1})\approx 1.99$, which sits between them. The estimate is honest.
No integration at all; two evaluations of the integrand and one known area.
When an integrand has no elementary antiderivative but is boxed between constants, the comparison property still returns something you can write down.
A region that is neither type: ∬ (x+y) over an L-shape
An L-shaped region: the rectangle $[0,2]\times[0,1]$ with the tall block $[0,1]\times[1,2]$ stacked on its left. No single pair of curves bounds it, so we split it.
Independent check by subtraction: over the full square $[0,2]^2$, $\iint(x+y)=8$; the missing corner $[1,2]\times[1,2]$ contributes $3$; and $8-3=5$. Building up and cutting down give the same number.
Additivity works in both directions: add up simple pieces, or start from a simple whole and subtract what is not there.
Checkpoint
§12.3 — which reading avoids a split●●●○○
The tent triangle with corners $(0,0)$, $(1,1)$, $(2,0)$. You are asked only how it should be set up, not to integrate it.
Given
Corners $(0,0),(1,1),(2,0)$; left edge $y=x$, right edge $y=2-x$.
Find
Which description is correct?
Hint 1/4
At a fixed height $y$, is the region one connected segment across, or does it come in two pieces? Now ask the same for a fixed $x$.
Hint 2/4
A region is Type II in one integral when a single left curve and a single right curve hold for the whole $y$-range; it is Type I in one integral when a single floor and single ceiling hold for the whole $x$-range.
Hint 3/4
Horizontally, every slice runs from $x=y$ to $x=2-y$ with no break. Vertically, the ceiling is $y=x$ for $x<1$ and $y=2-x$ for $x>1$. Restated: left edge $y=x$, right edge $y=2-x$.
Hint 4/4
It is one Type II integral, but Type I only after splitting at $x=1$.
Show solution
We check connectivity of a slice in each direction; that is what decides how many integrals a setup needs.
Horizontal slice
$$x:\ y\ \to\ 2-y\ \text{for all }0\le y\le 1$$
one unbroken segment at every height, so Type II needs one integral
a region cut into parts is easy to miscount, leaving one rectangle or triangle out of the sum
wrong$$\iint_{R_1} f$$
right$$\iint_{R_1} f+\iint_{R_2} f$$
12.4When one order is impossible
Reverse an order of integration by redrawing the region and re-reading its limits in the other direction; the region is fixed, only its description changes.
Choosing the cheaper order was a matter of effort. Now the stakes rise: sometimes one order cannot be finished at all, because the inner antiderivative does not exist in elementary form, and the only way through is to reverse.
MethodReversing the order of integration
Conditions
the region described by the given limits is sketched first
the reversal is read off the picture, not obtained by swapping $dx\,dy$ for $dy\,dx$
the value is unchanged; Fubini guarantees both orders give the same number
The two outer integrals are equal because both compute the very same double integral over the very same region $D$. To pass from one to the other you sketch $D$ from the first pair of limits, then describe that same $D$ with the roles of the slices swapped, reading new inner limits off the boundary curves. Reversing is a geometry step, not an algebra step.
The triangle above $\textcolor{#6f42c1}{y=x}$. Read as $\textcolor{#1f6feb}{dy\,dx}$, $y$ runs from the line up to $1$; read as $\textcolor{#d1690a}{dx\,dy}$, $x$ runs from the y-axis across to the line. Same region, two limit descriptions.
Looks like this, but is not
Reversing looks like it should be a purely mechanical swap: turn $dy\,dx$ into $dx\,dy$ and carry the same limits over.
Copying the limits keeps an $x$ inside a $dx\,dy$ integral, which is meaningless. The inner limits after reversal come from the boundary curves solved for the new inner variable, and finding them requires the picture. The differentials swap; the numbers almost never do.
The integral with no antiderivative in sight: reversing ∫∫ e^{y²}
Take the inner integral as written and you are asked for $\int e^{y^2}\,dy$, which has no elementary antiderivative. The region, not the formula, is what to look at.
Given
$$\int_0^1\!\!\int_x^1 e^{y^2}\,dy\,dx$$
Find
the value of the integral
Solution
Reverse the order, because the inner integral in $y$ is impossible while the inner integral in $x$ will be trivial; the integrand $e^{y^2}$ does not involve $x$ at all.
Read the region from the given limits
$$0\le x\le 1,\qquad x\le y\le 1$$
the outer says $x$ from $0$ to $1$; the inner says $y$ from the line $y=x$ up to $y=1$
$$D=\{(x,y): 0\le y\le 1,\ 0\le x\le y\}$$
the same triangle, re-read horizontally: at height $y$, $x$ runs from the y-axis to the line $x=y$
the recalled substitution $u=y^2$; the stray $y$ is exactly the $\tfrac12\,du$ it needs
Answer $$\dfrac{e-1}{2}$$
Check
Bound check: on the triangle $e^{y^2}$ ranges over $[1,e]$ and the area is $\tfrac12$, so the value must lie in $[\tfrac12,\tfrac{e}{2}]\approx[0.5,1.36]$. Our $\tfrac{e-1}{2}\approx 0.86$ sits inside.
One reversal turned an impossible integral into a one-line substitution.
When the integrand is a function of $y$ alone but the inner variable is $y$, reversing so the inner variable is $x$ makes the inner integral a bare multiplication.
The same rescue with a trig integrand: reversing ∫∫ sin(x²)
Here the impossible inner integral is $\int \sin(x^2)\,dx$. The integrand depends only on $x$, so we want $x$ to be the outer variable, not the inner one.
Given
$$\int_0^1\!\!\int_y^1 \sin(x^2)\,dx\,dy$$
Find
the value of the integral
Solution
Reverse the order, for the mirror-image reason of the last example: $\sin(x^2)$ has no elementary antiderivative in $x$, but integrating it in $y$ is free.
Read and re-describe the region
$$0\le y\le 1,\qquad y\le x\le 1$$
the inner runs $x$ from the line $x=y$ to $x=1$; the region is the triangle below $x=1$ and above $y=x$
$$D=\{(x,y): 0\le x\le 1,\ 0\le y\le x\}$$
re-read with vertical slices: at a fixed $x$, $y$ runs from $0$ up to the line $y=x$
the recalled substitution $u=x^2$; the leading $x$ is the $\tfrac12\,du$
Answer $$\dfrac{1-\cos 1}{2}$$
Check
Bound check: $\sin(x^2)\in[0,\sin 1]$ on the triangle since $x^2\in[0,1]$, area $\tfrac12$, so the value is in $[0,\tfrac{\sin 1}{2}]\approx[0,0.42]$. Ours is $\tfrac{1-\cos 1}{2}\approx 0.23$, inside.
The pattern is identical to the previous one; recognising it is what makes a reversal a reflex rather than a discovery.
Checkpoint
§12.4 — reversing without re-reading●●●○○
You have decided to reverse $\int_0^1\int_x^1 e^{y^2}\,dy\,dx$. Only the setup is asked for.
Given
$\int_0^1\int_x^1 e^{y^2}\,dy\,dx$, region the triangle $0\le x\le 1$, $x\le y\le 1$.
Find
Which is the correctly reversed integral?
Hint 1/4
Draw the region first. Reversing means describing that same triangle with $y$ outside and $x$ inside, not swapping the two $d$-symbols.
Hint 2/4
After reversing, the outer limits are the constant range of $y$, and the inner limits give $x$ as functions of $y$.
Hint 3/4
The triangle is $0\le y\le 1$ with $x$ from the y-axis to the line $x=y$. Restated: at height $y$, $x$ runs from $0$ to $y$.
Hint 4/4
The reversed form is $\int_0^1\int_0^{y} e^{y^2}\,dx\,dy$.
Show solution
We re-read the region horizontally rather than manipulate symbols.
Fix the y-range and read x across
$$0\le y\le 1$$
the triangle spans heights $0$ to $1$
$$x:\ 0\ \to\ y$$
at height $y$, the slice runs from the y-axis to the diagonal $x=y$
Assemble
$$\int_0^1\!\!\int_0^{y} e^{y^2}\,dx\,dy$$
outer constant $y$-range, inner $x$ from $0$ to $y$
Answer $$\int_0^1\int_0^{y} e^{y^2}\,dx\,dy$$
Check
Check it evaluates: the inner gives $y\,e^{y^2}$, and $\int_0^1 y e^{y^2}\,dy=\tfrac{e-1}{2}$, a finite number, whereas the original inner integral had no elementary form. The reversal did real work.
⚠ Swapping the differentials but keeping the limits
reversal is misremembered as a symbol move, so the old limits get carried straight over
wrong$$\int_0^1\!\!\int_x^1 e^{y^2}\,dx\,dy$$
right$$\int_0^1\!\!\int_0^{y} e^{y^2}\,dx\,dy$$
⚠ Reversing onto the wrong side of the boundary
the diagonal $y=x$ splits the square into two triangles, and it is easy to describe the other one
wrong$$\int_0^1\!\!\int_{y}^1(\cdots)\,dx\,dy$$
right$$\int_0^1\!\!\int_0^{y}(\cdots)\,dx\,dy$$
12.5A distance and an angle
Polar coordinates locate a point by a signed distance $r$ and an angle $\theta$; convert with $x=r\cos\theta$, $y=r\sin\theta$ and $r^2=x^2+y^2$.
A change of gear. Next week the regions worth integrating over are disks and sectors, which are miserable in $x$ and $y$. Polar coordinates name a point by how far out it is and which way it faces, and in that language a disk becomes a rectangle.
DefinitionPolar coordinates and the conversion formulas
Conditions
$r$ is the signed distance from the pole; $\theta$ is measured from the polar axis, counterclockwise positive
the same point has many names: $(r,\theta)=(r,\theta+2\pi)=(-r,\theta+\pi)$
$\tan\theta=\tfrac{y}{x}$ fixes only a reference angle; the quadrant of $(x,y)$ fixes $\theta$
To go from polar to rectangular, drop a perpendicular from the point: its shadow on the horizontal axis is $r\cos\theta$ and on the vertical axis is $r\sin\theta$. To come back, the distance is the ordinary $\sqrt{x^2+y^2}$, and the angle starts from $\tan\theta=y/x$ but must then be pushed into the quadrant where the point actually sits.
A point at $\textcolor{#6f42c1}{(r,\theta)}$. Its horizontal shadow is $\textcolor{#1f6feb}{x=r\cos\theta}$ and its vertical shadow is $\textcolor{#d1690a}{y=r\sin\theta}$; the angle $\theta$ opens from the polar axis.
Looks like this, but is not
It is tempting to treat $(r,\theta)$ like $(x,y)$: one point, one name, and $r$ an ordinary distance that cannot be negative.
Neither holds. A single point has infinitely many polar names, differing by full turns in $\theta$, and $r$ is allowed to be negative, meaning the point directly opposite through the pole. Polar coordinates trade the uniqueness of rectangular coordinates for the convenience of describing circles.
Polar to rectangular: where is (2, π/3)?
A distance of $2$ from the pole, along the ray at angle $\tfrac{\pi}{3}$. The conversion formulas turn that into an $(x,y)$ pair directly.
Given
$$(r,\theta)=\Big(2,\ \tfrac{\pi}{3}\Big)$$
Find
the rectangular coordinates of the point
Solution
We use $x=r\cos\theta$ and $y=r\sin\theta$ straight off, since polar to rectangular is the easy direction with no quadrant ambiguity.
Check the distance back out: $\sqrt{x^2+y^2}=\sqrt{1+3}=2=r$, and the point $(1,\sqrt3)$ sits in the first quadrant, matching an angle of $\tfrac{\pi}{3}$.
Polar to rectangular never needs a quadrant decision; the signs of $\cos\theta$ and $\sin\theta$ place the point automatically.
Rectangular to polar, the hard direction: convert (−1, 1)
A point in the second quadrant. Getting $r$ is easy; getting $\theta$ is where the quadrant trap lives.
Given
$$(x,y)=(-1,\ 1)$$
Find
a polar representation with $r>0$ and $\theta\in[0,2\pi)$
Solution
We take $r$ from $r^2=x^2+y^2$ and then fix $\theta$ by the point's actual quadrant, not by a bare arctangent.
Find r
$$r=\sqrt{(-1)^2+1^2}=\sqrt2$$
the positive root, since we agreed to report $r>0$
Find the reference angle, then place it
$$\tan\theta=\tfrac{y}{x}=\tfrac{1}{-1}=-1$$
this gives a reference angle of $\tfrac{\pi}{4}$, but $\tan$ cannot see the quadrant
$$(-1,1)\ \text{is in quadrant II}\ \Rightarrow\ \theta=\pi-\tfrac{\pi}{4}=\tfrac{3\pi}{4}$$
the point is up and to the left, so $\theta$ is between $\tfrac{\pi}{2}$ and $\pi$
Report
$$(r,\theta)=\Big(\sqrt2,\ \tfrac{3\pi}{4}\Big)$$
positive $r$, angle in the correct quadrant
Answer $$\Big(\sqrt2,\ \tfrac{3\pi}{4}\Big)$$
Check
Convert back as a check: $\sqrt2\cos\tfrac{3\pi}{4}=\sqrt2\cdot(-\tfrac{\sqrt2}{2})=-1$ and $\sqrt2\sin\tfrac{3\pi}{4}=\sqrt2\cdot\tfrac{\sqrt2}{2}=1$, returning $(-1,1)$. The naive $\arctan(-1)=-\tfrac{\pi}{4}$ would have placed the point in quadrant IV, at $(1,-1)$.
For any point with $x<0$, the calculator's arctangent is off by exactly $\pi$; add it and the point lands in the right quadrant.
One point, many names: representations of (2, π/3)
The same point from the first example, written several legal ways, to make the non-uniqueness concrete rather than a warning.
Given
$$\text{Point } (2,\tfrac{\pi}{3})\ \text{from the polar-to-rectangular example}$$
Find
three other polar names for the same point
Solution
We generate other names by the two moves that never change a point: adding a full turn to $\theta$, and flipping the sign of $r$ while adding half a turn.
Check the negative-$r$ name against the polar-to-rectangular conversion: $-2\cos\tfrac{4\pi}{3}=-2\cdot(-\tfrac12)=1$ and $-2\sin\tfrac{4\pi}{3}=-2\cdot(-\tfrac{\sqrt3}{2})=\sqrt3$, giving $(1,\sqrt3)$, the same point.
Because a point has infinitely many polar names, an answer key states a preferred form; without it, two correct answers can look like a contradiction.
Checkpoint
§12.5 — the quadrant trap●●○○○
You are converting the second-quadrant point $(-1,1)$ to polar form with $r>0$ and $\theta\in[0,2\pi)$.
Given
$(x,y)=(-1,1)$, which lies in quadrant II.
Find
Which is the correct polar form?
Hint 1/4
Sketch the point first. It is up and to the left, so its angle is somewhere between $\tfrac{\pi}{2}$ and $\pi$.
Hint 2/4
$\tan\theta=y/x$ gives only a reference angle; the quadrant of the point decides the actual $\theta$, and $r=\sqrt{x^2+y^2}$.
Hint 3/4
Here $r=\sqrt2$ and the reference angle is $\tfrac{\pi}{4}$; a quadrant-II point has $\theta=\pi-\tfrac{\pi}{4}$. Restated: the point $(-1,1)$ is up-left.
Hint 4/4
The form is $\big(\sqrt2,\ \tfrac{3\pi}{4}\big)$.
Show solution
We compute $r$ by the distance formula and $\theta$ by placing the reference angle in the right quadrant.
the formula $r^2=x^2+y^2$ is remembered as if the left side were $r$
wrong$$r=x^2+y^2=2$$
right$$r=\sqrt{x^2+y^2}=\sqrt2$$
12.6The curves that keep coming back
A short catalogue of polar curves plus three symmetry tests lets you name and sketch a curve from its equation in seconds.
With the dictionary in hand, the point is no longer single points but whole curves. A handful of polar equations describe the shapes you will integrate over next week, and a bare cosine or sine hides a circle that is not where you expect it.
RuleA catalogue of standard polar curves
Conditions
$a>0$ is a constant; convert any equation by multiplying or dividing by $r$ as needed
symmetry about the polar axis if $\theta\to-\theta$ leaves the equation unchanged
symmetry about the line $\theta=\tfrac{\pi}{2}$ if $\theta\to\pi-\theta$ leaves it unchanged; about the pole if $r\to-r$ does
$$\boxed{\begin{aligned}&r=a &&\text{circle, radius } a,\ \text{centre } O\\&\theta=\alpha &&\text{line through the pole}\\&r=2a\cos\theta &&\text{circle, radius } a,\ \text{centre } (a,0)\\&r=2a\sin\theta &&\text{circle, radius } a,\ \text{centre } (0,a)\\&r=a(1+\cos\theta) &&\text{cardioid, cusp at the pole}\\&r=a\cos(n\theta) &&\text{rose: } n\text{ petals if } n\text{ odd, } 2n\text{ if even}\end{aligned}}$$
The two circles centred at the origin and passing through it are the ones to keep straight: $r=a$ is centred at the origin, while $r=2a\cos\theta$ and $r=2a\sin\theta$ pass through the origin but are centred off it, along the x-axis and the y-axis respectively. Add a constant to a single trig term and the circle opens into a ; put a multiple of $\theta$ inside and it splits into the petals of a .
$\textcolor{#1f6feb}{r=2\cos\theta}$ drawn out: a circle of radius $1$ centred at $(1,0)$, through the pole and $(2,0)$. The $\textcolor{#d1690a}{\text{sample radius}}$ at $\theta=\tfrac{\pi}{4}$ has length $r=2\cos\tfrac{\pi}{4}=\sqrt2$.
Looks like this, but is not
$r=2\cos\theta$ looks like it should be a circle of radius $2$ about the origin, by analogy with $r=2$.
The $\cos\theta$ changes everything. It makes $r$ shrink to $0$ at $\theta=\tfrac{\pi}{2}$ and grow to $2$ at $\theta=0$, so the curve is a circle of radius $1$ that passes through the origin and is centred at $(1,0)$. Only $r=$ constant is centred at the origin.
$\theta$
$r=1+\cos\theta$
point reached
$0$
$2$
farthest right, $(2,0)$
$\tfrac{\pi}{3}$
$\tfrac{3}{2}$
upper right
$\tfrac{\pi}{2}$
$1$
top, $(0,1)$
$\tfrac{2\pi}{3}$
$\tfrac{1}{2}$
upper left
$\pi$
$0$
the pole, the cusp
As $\theta$ turns from $0$ to $\pi$, $r$ falls steadily from $2$ to $0$, so the curve spirals inward to the cusp at the pole; reflecting across the polar axis completes the heart.
The equation that fooled everyone: what is r = 2 cos θ?
A bare cosine equation. The instinct is to read it as a circle about the origin; converting to rectangular shows it is not.
Given
$$r=2\cos\theta$$
Find
the rectangular equation and the shape
Solution
Multiply through by $r$, because that manufactures the combinations $r^2=x^2+y^2$ and $r\cos\theta=x$ that the dictionary translates directly.
multiplying by $r$ is legal and creates $r^2$ and $r\cos\theta$, both of which convert
$$x^2+y^2=2x$$
using $r^2=x^2+y^2$ and $r\cos\theta=x$
Complete the square to name the shape
$$x^2-2x+y^2=0\ \Rightarrow\ (x-1)^2+y^2=1$$
grouping the $x$ terms and completing the square
Answer $$(x-1)^2+y^2=1$$
Check
Check two points: $\theta=0$ gives $r=2$, the point $(2,0)$, which is on the circle; $\theta=\tfrac{\pi}{2}$ gives $r=0$, the pole, also on it. A circle of radius $1$ centred at $(1,0)$, not at the origin.
One multiplication by r and one completed square.
A bare $\cos\theta$ or $\sin\theta$ in a polar equation is the signature of a circle that has been pushed off the origin, never a circle centred on it.
The other direction: put x² + y² = 2y into polar form
A rectangular circle, this one offset upward. We translate it into polar, where its equation collapses to one clean term.
Given
$$x^2+y^2=2y$$
Find
the polar equation, and a description
Solution
Substitute the dictionary entries directly: $x^2+y^2$ is $r^2$ and $y$ is $r\sin\theta$, which turns the whole equation into a relation in $r$ and $\theta$.
Substitute the rectangular pieces
$$r^2=2\,r\sin\theta$$
$x^2+y^2=r^2$ on the left, $y=r\sin\theta$ on the right
Cancel one r
$$r=2\sin\theta$$
dividing by $r$; the solution $r=0$ we drop is just the pole, which $r=2\sin\theta$ already includes at $\theta=0$
Answer $$r=2\sin\theta$$
Check
Recognise it against the catalogue: $r=2a\sin\theta$ with $a=1$ is a circle of radius $1$ centred at $(0,1)$, which is exactly $x^2+(y-1)^2=1$, the same as $x^2+y^2=2y$.
Dividing by $r$ is the mirror of multiplying by $r$: one sets a polar equation up for the dictionary, the other tidies it afterwards.
Sketching a cardioid from symmetry: r = 1 + cos θ
The classic heart-shaped curve. Rather than plot dozens of points, we use one symmetry to halve the work and a short table for the rest.
Given
$$r=1+\cos\theta$$
Find
a sketch, justified by symmetry and a few points
Solution
We test for symmetry first, because a confirmed symmetry means we only plot half the angles and reflect the rest.
the curve reaches the pole at $\theta=\pi$, forming the cusp
Answer $$\text{cardioid, cusp at the pole, pointing along the positive }x\text{-axis}$$
Check
Check the cusp: $r=0$ only at $\theta=\pi$, so the curve touches the pole exactly once, from the left, which is the inward dimple of a cardioid. Maximum reach $r=2$ at $\theta=0$ matches the point $(2,0)$.
Confirming a symmetry before plotting is the difference between six points and twelve; on an exam it is the difference between finishing and not.
Checkpoint
§12.6 — naming a bare-sine curve●●○○○
Identify the curve from its polar equation, without a full plot.
Given
$r=4\sin\theta$
Find
What curve is this?
Hint 1/4
Compare it with the catalogue entry $r=2a\sin\theta$. What is $a$, and what does a bare sine mean for the centre?
Hint 2/4
$r=2a\sin\theta$ is a circle of radius $|a|$ centred at $(0,a)$; multiply by $r$ if you want to confirm it rectangularly.
Hint 3/4
Matching $4\sin\theta=2a\sin\theta$ gives $a=2$. Restated: the equation is $r=4\sin\theta$, a bare sine.
Hint 4/4
It is a circle of radius $2$ centred at $(0,2)$.
Show solution
We match the equation against the standard and read off the centre.
Same number as the Type I reading, as Fubini promises.
The same region and integrand give the same value either way; here neither order is harder, so the choice is free.
How to tell them apart
When both orders are clean, they must agree; when they disagree, one setup has a swapped or variable limit. Use the second order as a check on the first.
A bare cosine: r = 2 cos θ
Coefficient times a single cosine, nothing added.
Given
$$r=2\cos\theta$$
Find
the shape
Solution
A bare trig term with only a coefficient is a circle off the origin.
$r=0$ at $\theta=\tfrac{\pi}{2}$ only, so the curve touches the pole once and closes into a circle.
A constant plus a cosine: r = 1 + cos θ
The same cosine, but with a constant $1$ added.
Given
$$r=1+\cos\theta$$
Find
the shape
Solution
A constant added to a single trig term opens the circle into a cardioid.
Read the anchors
$$\theta=0:r=2;\ \theta=\pi:r=0$$
the reach shrinks to zero at $\theta=\pi$
$$\text{one cusp at the pole}$$
$r=0$ once, but approached so the curve dimples inward
Answer $$\text{cardioid, cusp at the pole}$$
Check
Unlike the circle, $r$ ranges over $[0,2]$ and returns to the pole only at $\theta=\pi$, forming the heart's dimple.
Both curves carry a $\cos\theta$ and both pass through the pole, which is why they are mistaken for each other.
How to tell them apart
A coefficient alone in front of a single $\cos\theta$ or $\sin\theta$ is a circle; a constant added to that trig term is a cardioid. Look for the added constant.
Scaffolding comes off
The common skeleton
Sketch the region and pick the slicing direction that keeps every slice in one piece.
Read the outer limits as the constant range of the outer variable, and the inner limits as the boundary curves.
Do the inner integral, holding the outer variable constant.
Do the outer integral down to a number.
Check: redo it with the integrand $1$ for the area, or evaluate the other order and compare.
1 · fully worked
Full worked rung: ∬ 6xy over the triangle below y = x
Every step and reason shown. Read it once with the reasons covered, then again with them showing.
Reverse as a check: $\int_0^1\int_y^1 6xy\,dx\,dy=\int_0^1 3y(1-y^2)\,dy=3\big(\tfrac12-\tfrac14\big)=\tfrac34$. Both orders agree.
The ceiling curve became the inner upper limit; everything after that was one-variable calculus.
2 · you write the reasoning
Same region, easier integrand: $\iint_D 2x\,dA$ over the triangle $0\le x\le 1$, $0\le y\le x$. The steps are given; write your own reason for each before opening the model answers.
reasoning
The triangle sits under the diagonal $y=x$; vertical slices put the inner integral in $y$, with $y$ running from $0$ to $x$.
reasoning
The factor $2x$ is constant in $y$, so integrating it over a slice of height $x$ just multiplies it by $x$, giving $2x^2$.
reasoning
Power rule: $2x^2$ integrates to $\tfrac{2x^3}{3}$, which is $\tfrac{2}{3}$ at $x=1$ and $0$ at $x=0$.
3 · find the buried error
Harder: a region bounded by two curves. Below is a student's full solution to $\iint_D x\,dA$ over the region between $y=x^2$ and $y=2x$. Exactly two of the five lines contain an error, each a slip a careful person makes under time pressure. Find them before looking at the choices.
the two buried errors (2)
⚠ step 2
The floor and ceiling are swapped. On $(0,2)$, at $x=1$ the line gives $2x=2$ and the parabola gives $x^2=1$, so the line $y=2x$ is on top. The inner limits should be $\int_{x^2}^{2x}$, not $\int_{2x}^{x^2}$.
the parabola is the curved boundary, so it is instinctively called the top one, without testing a sample point
right
Test $x=1$: $2x=2>1=x^2$, so the ceiling is $2x$. The setup is $\int_0^2\int_{x^2}^{2x} x\,dy\,dx$.
⚠ step 3
The constant factor $x$ was dropped. $\int_{2x}^{x^2} x\,dy = x\big[(x^2)-(2x)\big]=x^3-2x^2$, not $x^2-2x$. Integrating $x$ in $y$ multiplies $x$ by the slice width.
with $x$ held constant it reads like part of the background and only the width $x^2-2x$ gets written down
right
Keep the factor: $\int x\,dy = x\cdot(\text{width})$, so the inner integral is $x(2x-x^2)=2x^2-x^3$ once the limits are also fixed.
4 · the bare problem
§12 — the scaffolding is off●●●○○
No steps, no reasons, no order suggested. The region is under a square-root curve.
Check as Type II: $0\le y\le 2$, $y^2\le x\le 4$, and $\int_0^2\int_{y^2}^{4} y\,dx\,dy=\int_0^2 y(4-y^2)\,dy=8-4=4$. Both orders give $4$.
Full exam-style question
Exam-style: reversing to evaluate ∫∫ e^{x²} over a triangleexam format
A double integral whose inner integral, as written, has no elementary antiderivative. Worth full marks only if you reverse the order.
Given
$$\int_0^1\!\!\int_{3y}^{3} e^{x^2}\,dx\,dy$$
Find
the exact value of the integral
Solution
Reverse the order: $\int e^{x^2}\,dx$ is impossible, but with $x$ outside, the inner integral in $y$ is a bare multiplication, and the leftover $x$ is exactly what the substitution $u=x^2$ needs.
Read the region from the given limits
$$0\le y\le 1,\qquad 3y\le x\le 3$$
the region is the triangle with corners $(0,0)$, $(3,0)$, $(3,1)$, bounded on the left by the line $x=3y$
$u=x^2$, $du=2x\,dx$; the $\tfrac{x}{3}$ supplies the $x$
$$=\tfrac{1}{6}\big(e^{9}-1\big)$$
evaluating $e^{x^2}$ from $0$ to $3$
Answer $$\dfrac{e^{9}-1}{6}$$
Check
Bound check: the triangle has area $\tfrac32$ and $e^{x^2}\in[1,e^9]$ on it, so the value lies in $[\tfrac32,\tfrac32 e^9]\approx[1.5,\,12155]$. Our $\tfrac{e^9-1}{6}\approx 1350$ sits inside, and is positive as it must be.
One reversal and one substitution; the un-reversed integral cannot be finished at all.
The exam signature is an $e^{x^2}$, $\sin(x^2)$ or similar trapped as the inner integrand; the reversal is the whole question.
Practice
A · concept 4 questions
1§12.4 — does reversing change the value?●●○○○
Decide before reading on; the reason is the point.
Given
Reversing the order of integration can change the value of a double integral.
Find
True or false?
Hint 1/4
State the claim in your own words and ask what would have to be true for it to hold.
Hint 2/4
Weigh it against a definition or a small example rather than intuition.
Hint 3/4
Restated: Reversing the order of integration can change the value of a double integral.
Hint 4/4
False. Both orders compute the same double integral over the same region, so by Fubini they are equal.
Show solution
The two orders are equal because they name the same double integral.
the completed square shows centre $(1,0)$, radius $1$
Answer $$\text{False}$$
Check
At $\theta=\tfrac{\pi}{2}$, $r=0$: the curve passes through the origin, which a circle centred there never does.
4§12.3 — which region forces a split●●●○○
Only one of these cannot be written as a single Type I or Type II integral without first being cut into pieces.
Given
Four regions: a ring, a triangle, a disk, a rectangle.
Find
Which one must be split first?
Hint 1/4
For each, ask whether a single line across it, in some direction, stays inside from one boundary to the other.
Hint 2/4
A region is simple in a direction when one slice meets the boundary exactly twice; it must be split when some slice meets it four times.
Hint 3/4
A vertical line through a ring crosses the outer circle, the inner circle, the inner circle again, and the outer circle: four crossings. Restated: a ring, a triangle, a disk, a rectangle.
Hint 4/4
The ring must be split.
Show solution
A single iterated integral needs every slice to meet the boundary exactly twice.
Close the page and write, from memory: the Type I and Type II iterated forms with which variable is inner in each; the rule that the outer limits are always constants; the four-step recipe for reversing an order; the sign that a reversal is needed; the four conversion formulas between rectangular and polar; the quadrant rule for recovering $\theta$; and which polar equations give a circle centred at the origin, a circle off it, and a cardioid.
Set up $\iint_D f\,dA$ for the triangle with vertices $(0,0),(2,0),(2,4)$ as a Type I integral, with correct limits.
c-general-regions
Turn a region given by a left curve $x=y^2$ and a right curve $x=y$ into a Type II integral without mixing up which is which.
c-type-2
Say, for the tent triangle $(0,0),(1,1),(2,0)$, which order needs a split and which does not, and why.
c-choose-order
Reverse $\int_0^1\int_x^1 e^{y^2}\,dy\,dx$ and finish it to $\tfrac{e-1}{2}$.
c-reverse-order
Convert $(-1,1)$ to polar with $r>0$ and $\theta\in[0,2\pi)$, and say why the calculator's arctangent is wrong here.
c-polar-coords
Name the curve $r=4\sin\theta$ and give its centre and radius from the catalogue.
c-polar-curves
Glossary (12 terms)
genel bölge üzerinde katlı integral
The integral $\iint_D f\,dA$ over a region $D$ of any shape, built from the rectangle case by extending $f$ to zero outside $D$.
Type I region
A region described by $a\le x\le b$, $g_1(x)\le y\le g_2(x)$: one lower curve and one upper curve as $x$ runs across. Integrated with $dy$ inside.
Type II region
A region described by $c\le y\le d$, $h_1(y)\le x\le h_2(y)$: one left curve and one right curve as $y$ runs across. Integrated with $dx$ inside.
iterated integralardışık integral
A double integral written as two nested single integrals, evaluated inside-out, the inner one treating the outer variable as constant.
reversing the order of integration
Rewriting an iterated integral with the inner and outer variables exchanged, by redrawing the region and reading new limits off it. The value is unchanged.
Fubini's theorem
The result that a double integral equals either iterated integral, so the two orders always give the same number for a continuous integrand.
polar coordinateskutupsal koordinatlar
A way of locating a point by a signed distance $r$ from the pole and an angle $\theta$ from the polar axis, related to $x,y$ by $x=r\cos\theta$, $y=r\sin\theta$.
polekutup
The origin of the polar system, the point $r=0$, whose angle $\theta$ is undefined since every ray meets it.
polar axiskutup ekseni
The ray from which $\theta$ is measured, coinciding with the positive $x$-axis.
cardioidkardioid
A heart-shaped curve $r=a(1\pm\cos\theta)$ or $r=a(1\pm\sin\theta)$: a constant added to a single trig term, with one cusp at the pole.
rose
A curve $r=a\cos(n\theta)$ or $r=a\sin(n\theta)$ with $n$ petals when $n$ is odd and $2n$ petals when $n$ is even.
off-origin circle
A circle whose polar equation is a bare $r=2a\cos\theta$ or $r=2a\sin\theta$, centred at $(a,0)$ or $(0,a)$ and passing through the pole.
What comes next
§13 · Double integrals in polar coordinates, and triple integrals
Next week the two halves of this page meet: the general-region double integral is set up over a disk or a sector, and the polar coordinates you just learned turn that region into a plain rectangle in $r$ and $\theta$. The one new ingredient is the area element, which in polar form is not $dr\,d\theta$ but carries an extra factor of $r$, and getting that factor right is the whole of the next section.
Sources
James Stewart, Calculus, Ninth Edition — sections 15.2 and 10.3 The definitions of Type I and Type II regions, the properties of the double integral, and the polar coordinate system are the ones this book gives; every worked region and every point on this page is a different one.
Course syllabus, week 12: Multiple Integrals 15.2, 10.3 The two section numbers used here are the two the syllabus line contains, and the only assessment weights stated there are quizzes at $10$ percent and homework at $10$ percent.
Every numerical value on this page Each double integral was evaluated in both orders where possible, and each polar conversion was checked by converting back, before publication.