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13Double integrals in polar coordinates and triple integrals
A past final opens with $\iint_D e^{-(x^{2}+y^{2})}\,dA$, where $D$ is the disk of radius $2$. You set it up the only way you know: $x$ runs across, $y$ runs from $-\sqrt{4-x^{2}}$ to $\sqrt{4-x^{2}}$, and the inner integral is $\int e^{-y^{2}}\,dy$. That antiderivative does not exist in elementary form, and the whole paper stalls on the first line. The region is a plain disk and the function only depends on distance from the origin, yet the grid of little squares cuts straight across every circle it should be following.
You will be able to take an integral whose region is a disk, wedge, or ring, rewrite it so the region becomes a rectangle in the new variables, finish it in two or three lines, and check the number against a known area or volume.
In 60 seconds
Match the coordinates to the region: circles and wedges want $r$ and $\theta$ with an extra factor of $r$ in the area, and a solid wants three nested integrals whose outer limits are always plain numbers.
Area element in polar
$$dA = r\,dr\,d\theta$$
the region is bounded by circles or rays, or the integrand depends on $x^{2}+y^{2}$
the solid is capped below and above by $z=u_1$ and $z=u_2$ over a flat region $D$
Volume
$$V=\iiint_E 1\,dV$$
you want the size of a solid and have its limits set up
Three most common mistakes
Writing $dA=dr\,d\theta$ and dropping the $r$. The area of a small polar patch grows with its distance from the origin, so the $r$ is not decoration; leaving it out changes the answer.
Letting an outer limit of a triple (or double) integral depend on the variable that is integrated later. The outermost limits are always constants, or the iteration is not defined.
Forgetting to turn $x^{2}+y^{2}$ into $r^{2}$ inside the integrand after switching to polar, so a clean $r^{2}$ is left disguised as a sum of two squares.
The syllabus lists a weekly quiz (10%) and homework (10%), and both draw straight from the setup skills on this page: converting a circular region to polar and reading limits off a solid. No larger exam weight is published for this week, so treat every setup here as a quiz rehearsal, done to a finished number.
How much time do you have?
10 minutes
You leave able to convert one circular region to polar with the r attached, and to set up a triple integral over a box in the right order.
In 60 seconds card, Where the extra r comes from, Reading the limits off a region, Stacking a triple integral over a box, Formula card
45 minutes
Add the cases that lose whole questions: an r that depends on the angle, the improper polar integral behind the bell curve, and a solid that is not a box.
everything in the 10 minute path, Volumes, areas, and the integral Cartesian cannot do, From a box to a general solid, Choosing the order and the average value, Full exam-style question, Practice B, Practice C
full read
The parts that transfer to the next block on cylindrical and spherical coordinates: how the region, not the formula, chooses the coordinates, and the ladder where the scaffolding comes off one rung at a time.
all blocks in order, Contrast pairs, Scaffolding comes off, Practice A to D, Mistake ledger, Self audit
By the end of this section
Explain why the area element in carries a factor of r, and produce that factor from the area of a small polar patch rather than memorising it.
Set up a double integral over a region bounded by rays and by curves of the form r equals g of theta, reading the theta limits and the r limits off a sketch.
Compute volumes and areas with a polar double integral, including one improper integral that has no elementary Cartesian antiderivative.
Evaluate a triple integral over a rectangular box, and state why the six possible orders of integration all give the same number.
Describe a solid as a flat region capped by two surfaces, and turn that description into ordered triple-integral limits.
Choose an order of integration that makes a solid tractable, reverse an order when the first choice stalls, and compute an average value over a solid.
Syllabus coverage
15.3
Double integrals in polar coordinates
The area element r dr dtheta and where its r comes from, the change-of-variable rule for a and for a region bounded by r = g(theta), and the applications: area, volume under a surface, and the improper integral behind the normal distribution.
covered
15.6
Triple integrals
The triple integral over a box and the six-order Fubini statement, the description of a general solid by a flat region and two capping surfaces, volume as the integral of 1, the average value of a function over a solid, and reversing the order of integration when the first order stalls.
covered
Recall first
Iterated double integral over a region
$\displaystyle\iint_D f\,dA=\int_{a}^{b}\!\int_{g_1(x)}^{g_2(x)} f(x,y)\,dy\,dx$ for a region $D$ described by $a\le x\le b$, $g_1(x)\le y\le g_2(x)$.
Both the polar rule and the triple-integral rule reduce, in their last steps, to an ordinary double integral over a plane region. If setting one up is shaky, the new material has nothing to stand on.
Polar coordinates
$x=r\cos\theta,\quad y=r\sin\theta,\quad x^{2}+y^{2}=r^{2}$, with $r$ the distance from the origin and $\theta$ the angle from the positive $x$-axis.
This is the substitution the whole first half rests on. In particular $x^{2}+y^{2}=r^{2}$ is what collapses a messy integrand into a clean power of $r$.
Every inner integral on this page is one of these. The multiple-integral machinery only arranges the limits; the actual integration is single-variable calculus you have already met.
Area of a circular sector
A sector of radius $r$ and central angle $\theta$ (in radians) has area $\tfrac12 r^{2}\theta$.
This elementary-geometry fact is what the factor $r$ in $dA=r\,dr\,d\theta$ is built from, in the proof inside the first concept.
Try it yourself first (3 questions)
1§13.0 — swapping the order of integration●●○○○
Here is the reflex this week has to sharpen. A student integrates over the triangle $D=\{0\le x\le 1,\ 0\le y\le x\}$ and, wanting the other order, simply swaps the limit stacks.
Now integrate the result $2x+2$ over $x$ from $0$ to $1$.
Hint 4/4
$\int_{0}^{1}(2x+2)\,dx=1+2=3$.
Show solutionInner integral in y
$$\int_{0}^{2}(x+y)\,dy=2x+2$$
x is constant; integrate y
Outer integral in x
$$\int_{0}^{1}(2x+2)\,dx=[x^{2}+2x]_{0}^{1}=3$$
now integrate x
Answer $$3$$
Check
By symmetry the average of $x+y$ on $[0,1]\times[0,2]$ is $\tfrac12+1=\tfrac32$, times the area $2$, gives $3$.
3§13.0 — a circle in polar●○○○○
One line of the polar dictionary, needed before every conversion on this page.
Given
the circle $x^{2}+y^{2}=9$
Find
Written in polar coordinates, this circle is $r=$ what number?
Hint 1/4
Recall the bridge $x^{2}+y^{2}=r^{2}$.
Hint 2/4
Replace $x^{2}+y^{2}$ by $r^{2}$, so $r^{2}=9$.
Hint 3/4
Take the nonnegative square root, since $r\ge 0$.
Hint 4/4
$r=3$.
Show solutionSubstitute
$$r^{2}=9\ \Rightarrow\ r=3$$
the bridge x squared plus y squared equals r squared, then the nonnegative root
Answer $$r=3$$
Check
A radius-3 circle should give $r=3$; reading $r=9$ off the equation confuses $r$ with $r^{2}$.
Notation
symbol
reads as
means
watch out
$dA=r\,dr\,d\theta$
dee A equals r dee r dee theta
The area of an infinitesimal polar patch. Its two sides are the radial step $dr$ and the arc $r\,d\theta$, and their product carries the $r$.
The $r$ is part of the area, not part of the function. Even when $f=1$ the integrand is $r$, not $1$.
$\iint_D,\ \iiint_E$
the double integral over D, the triple integral over E
A sum over a flat region D in the plane, or over a solid region E in space, of the integrand times an element of area or volume.
The subscript names the region, not a variable. It tells you the limits, which live in the iterated form, not in this compact symbol.
$r^{2}=x^{2}+y^{2}$
r squared equals x squared plus y squared
The bridge between the two coordinate systems, together with $x=r\cos\theta$ and $y=r\sin\theta$.
After switching to polar, any leftover $x^{2}+y^{2}$ in the integrand should already be $r^{2}$. A sum of squares surviving the switch means the substitution reached the limits but not the function.
$\bar f=\dfrac{1}{V(E)}\iiint_E f\,dV$
f bar equals one over the volume of E times the triple integral of f
The average value of f over a solid E: its integral divided by the volume of the solid.
The denominator is the volume of the same solid E, computed with the same limits; it is not the length of any one edge.
Conventions used here
Every angle is in radians
Every $\theta$ limit on this page is a radian measure, so a full turn is $2\pi$ and a right angle is $\tfrac{\pi}{2}$. The factor $r$ in $dA=r\,dr\,d\theta$ is only correct when $d\theta$ is measured in radians.
The arc that gives the patch its width is $r\,d\theta$ only in radians; in degrees that arc length carries an extra $\tfrac{\pi}{180}$ and every area would be off by that factor.
The radius r is taken to be nonnegative
On this page $r\ge 0$. A point is reached by choosing the angle $\theta$ first and then walking outward a distance $r$. When a polar curve such as $r=2\sin\theta$ would ask for a negative $r$, we restrict $\theta$ to the range where $r$ stays nonnegative.
Allowing negative r lets the same point carry two different sets of limits, and a region would then be swept twice or missed entirely.
Order of an iterated integral is read inside out
In $\int_a^b\!\int_{c}^{d}\!\int_{p}^{q} f\,dz\,dy\,dx$ the differential nearest the integrand is done first. The innermost limits may depend on the outer variables, the middle limits on the outermost variable only, and the outermost limits must be constants.
Each integration removes one variable, so a limit may only mention variables not yet integrated away; an outer limit that mentions an inner variable leaves a symbol in the final number.
Volume and area are read off the same integral as 1
The area of a plane region is $\iint_D 1\,dA$ and the volume of a solid is $\iiint_E 1\,dV$; a positive integrand $f$ turns the double integral into the volume under the surface $z=f(x,y)$.
Keeping area, volume, and volume-under-a-surface as one machine with different integrands stops them from being memorised as three unrelated formulas.
13.1Where the extra r comes from
Circular regions turn into rectangles in polar, at the price of one factor of r in the area element.
The hook stalled because a square grid was chasing a circular region. We switch to coordinates that follow circles, $x=r\cos\theta$ and $y=r\sin\theta$, and the only surprise is what happens to the area element.
Solvable with what we have
$\iint$ over a rectangle $[a,b]\times[c,d]$ — the limits are four numbers
$\iint$ over the triangle under $y=x$ from the previous block — a type-I region
$\iint_D xy\,dA$ over a region with polynomial edges — ordinary iterated integration
Not solvable yet
$\iint_D e^{-(x^{2}+y^{2})}\,dA$ over a disk
the area between the circles $x^{2}+y^{2}=1$ and $x^{2}+y^{2}=4$
$\iint_D \sqrt{x^{2}+y^{2}}\,dA$ over a disk
Set the disk of radius $2$ up in Cartesian: $\int_{-2}^{2}\!\int_{-\sqrt{4-x^{2}}}^{\sqrt{4-x^{2}}} e^{-(x^{2}+y^{2})}\,dy\,dx$. The inner integral is $e^{-x^{2}}\!\int e^{-y^{2}}\,dy$, and $\int e^{-y^{2}}\,dy$ has no antiderivative in elementary terms. The limits are square roots, and the integrand fights the grid at every step.
Why it fails
Both the region and the integrand are organised around distance from the origin, and a grid of little squares is organised around two straight directions. The fix is not a cleverer antiderivative; it is to change the grid so its cells are wedges of rings that follow the circles the problem already lives on.
RuleChange to polar coordinates (polar rectangle)
Conditions
$f$ is continuous on the polar rectangle $R=\{\,a\le r\le b,\ \alpha\le\theta\le\beta\,\}$
Put r cosine theta wherever x stood and r sine theta wherever y stood, replace the region by the ranges of r and theta that trace it out, and replace the area element by r times dee r dee theta. The lone r at the end is the whole content of the rule: it is the factor that turns a step in r and a step in theta into an actual area.
Why the area element is r dr dtheta and not dr dtheta
A circular sector of radius $r$ and angle $\Delta\theta$ has area $\tfrac12 r^{2}\,\Delta\theta$. A polar patch is the sector out to $r_2$ with the sector out to $r_1$ removed, so its area is $\tfrac12 r_2^{2}\Delta\theta-\tfrac12 r_1^{2}\Delta\theta=\tfrac12(r_2+r_1)(r_2-r_1)\,\Delta\theta$. Write $r_2-r_1=\Delta r$ and call the average radius $r^{*}=\tfrac12(r_2+r_1)$; the area is $r^{*}\,\Delta r\,\Delta\theta$. As the patch shrinks, $r^{*}\to r$ and the element is $r\,dr\,d\theta$.
One polar patch. Its radial side is $dr$ and its curved side is the arc $r\,d\theta$, drawn in blue; the small angle $d\theta$ is in orange. The two blue sides multiply to the area $dA=r\,dr\,d\theta$, and the point of the figure is that the curved side, and so the area, grows with the distance $r$.
Looks like this, but is not
It is tempting to write $dA=dr\,d\theta$ by analogy with $dA=dx\,dy$, since $r$ and $\theta$ are just the new variables.
That would be the area of a patch only if the arc $r\,d\theta$ always had length $d\theta$, which happens only at $r=1$. At $r=3$ the same angular step spans an arc three times as long, so the patch is three times larger; the $r$ records exactly that stretch.
r
patch area r times delta r times delta theta
1
0.010
2
0.020
3
0.030
Tripling the distance from the origin triples the area of the patch, even though the steps in r and theta never change. That growing factor is the r.
The area element from the sector area
We build the factor r rather than quote it, because seeing it come from an area makes it impossible to forget.
Given
$$\text{A patch between radii } r_1\le r_2 \text{ and rays } \Delta\theta \text{ apart}$$
Find
the area of the patch, and its limiting form
Solution
We factor the difference of squares instead of expanding, because the factored form hands us the width and the average radius directly, and those are the two things that survive the limit.
Sweep the angle for the outer limits; for each angle, run the radius from the inner to the outer boundary.
The polar-rectangle rule wants four constants. Most regions give you only two — the angles $\alpha$ and $\beta$ — and the radius runs between curves $r=h_1(\theta)$ and $r=h_2(\theta)$ that depend on the direction you are looking.
RuleChange to polar over a general region
Conditions
$D=\{\,(r,\theta):\ \alpha\le\theta\le\beta,\ h_1(\theta)\le r\le h_2(\theta)\,\}$ with $0\le h_1\le h_2$
$h_1,h_2$ continuous on $[\alpha,\beta]$ and $0\le\beta-\alpha\le 2\pi$
The outer integral sweeps the angle across the whole region, from the smallest direction to the largest. The inner integral, for one frozen direction, walks the radius from the near boundary to the far boundary. The angle limits are numbers; the radius limits are allowed to be functions of the angle, never the other way round.
A region read for polar limits. The angle $\theta$ sweeps from $\alpha$ to $\beta$; for each fixed $\theta$ the dashed ray shows $r$ running from the origin out to the boundary $r=g(\theta)$. Those two sentences are exactly the two pairs of limits, outer in $\theta$ and inner in $r$.
Looks like this, but is not
A student writes $\displaystyle\int_{0}^{2}\!\int_{0}^{\theta}(\dots)\,d\theta\,dr$, putting the angle limit $\theta$ on the outer integral and a number on the inner.
The outer integral is the last one done, so its limits must be pure numbers; a $\theta$ sitting in the outer limits would still be there after everything is integrated, and the answer would contain a leftover variable. Angle-dependent limits belong on the inner $r$-integral, and constant limits on the outer $\theta$-integral.
A quarter of a ring: ∬(x²+y²) over 1 ≤ r ≤ 2 in the first quadrant
The region is bounded by two circles and two axes, so every limit is a constant: a genuine polar rectangle.
We integrate r first because both r-limits are constants here, which keeps the inner integral a pure power; sweeping theta first would gain nothing and cost a step.
the inner value is constant, so multiply by the length of the theta range
Answer $$\dfrac{15\pi}{8}$$
Check
Order-of-magnitude check: the region has area $\tfrac14\pi(2^{2}-1^{2})=\tfrac{3\pi}{4}\approx 2.36$, and $r^{2}$ ranges from 1 to 4 on it, so the integral should sit between $2.36$ and $9.4$. Indeed $\tfrac{15\pi}{8}\approx 5.9$.
When both boundaries are circles the two integrals separate into a number times a number, and the whole problem is two one-line integrals.
A curve r = g(θ): the region inside r = 2 sin θ
Here the outer boundary is not a fixed radius but a circle written in polar, so the inner limit is a function of the angle.
Given
$$D:\ \ \text{the region enclosed by } r=2\sin\theta$$
Find
the angle range and the area of the region
Solution
We solve r = 2 sin theta ≥ 0 for the angle range before setting up, because guessing 0 to 2 pi would trace the circle twice and double the area.
13.3Volumes, areas, and the integral Cartesian cannot do
A height that depends only on distance becomes one polar integral, and the r is what makes the bell curve integrable.
With the limits under control, the payoff arrives. A surface whose height depends only on $r$ gives a volume that is a single easy integral, and the same $r$ rescues the integral the hook could not do.
The volume under a surface is the double integral of its height; in polar the height is rewritten in r and theta and the element carries its r. When the height depends only on the distance r, the theta integral is just a constant multiplier and the whole volume collapses to one integral in r.
The paraboloid $z=4-x^{2}-y^{2}$ seen in cross-section as $z=4-r^{2}$. The shaded area under this curve, spun around the $z$-axis, is the solid whose volume the worked example finds; because the height depends only on $r$, the volume is one clean polar integral.
Looks like this, but is not
For the bell curve one might write $\displaystyle\int_{0}^{2\pi}\!\int_{0}^{\infty} e^{-r^{2}}\,dr\,d\theta$, dropping the $r$ as a harmless constant.
Without the $r$ the inner integral is $\int_{0}^{\infty} e^{-r^{2}}\,dr$, which is again non-elementary — the very wall the hook hit. With the $r$ it is $\int_{0}^{\infty} e^{-r^{2}}\,r\,dr$, which the substitution $u=r^{2}$ finishes in one line. The $r$ is not a constant and it is exactly what makes the problem solvable.
Volume under the paraboloid z = 4 - x² - y² over the disk r ≤ 2
The surface sits above the disk where it is positive, and its height depends only on r. This is the picture in the figure.
the volume between the surface and the plane z = 0
Solution
We integrate r before theta because the height has no theta in it; the theta integral then contributes only its length 2 pi, turning a double integral into a single one.
distribute the r first, then a plain power-rule integral
$$V=\int_{0}^{2\pi} 4\,d\theta=4\cdot 2\pi=8\pi$$
the inner value is a constant, so multiply by the full turn
Answer $$8\pi$$
Check
Bounds: the solid fits inside the cylinder of radius $2$ and height $4$, volume $\pi(2)^{2}(4)=16\pi$, and it fills exactly half of that cylinder since a paraboloid cuts its bounding cylinder in half. Half of $16\pi$ is $8\pi$.
Whenever the height depends only on r, expect the answer to be 2 pi times a one-variable integral; that structure is the signal that polar was the right move.
The integral Cartesian cannot do: the bell curve
This is the hook. The Cartesian version stalls; the polar version, with the r, finishes in two lines and pins down the famous constant.
We square the one-dimensional integral first, because a product of two identical single integrals in x and y is secretly the plane integral, and only in the plane can we switch to polar and gain the r.
Square the target and read it as a double integral
Sign and size: $e^{-x^{2}}>0$, so $I>0$ and the positive root is correct; numerically $\sqrt{\pi}\approx 1.77$, and a rough box under the bump (height 1, effective width near 2) is of that order.
One substitution, one trivial angular integral. The Cartesian route has no first step at all.
This is why square roots of pi run through probability: the normalising constant of the normal distribution is this very integral, made finite by the polar r.
Checkpoint
§13.3 — what the r rescues●●○○○
Thirty seconds. In the plane integral of the bell curve you have reached $\int_{0}^{\infty} e^{-r^{2}}\,r\,dr$.
Given
the inner integral $\displaystyle\int_{0}^{\infty} e^{-r^{2}}\,r\,dr$
Find
Why is this integral elementary while $\int_{0}^{\infty} e^{-r^{2}}\,dr$ is not?
Hint 1/4
Compare the two integrands: one has an extra factor of $r$. Ask what that factor is good for.
Hint 2/4
A substitution $u=r^{2}$ needs $du=2r\,dr$. Look for the $r$ that supplies the $du$.
Hint 3/4
With the $r$ present, $\int e^{-r^{2}}r\,dr=-\tfrac12 e^{-r^{2}}$; without it there is no elementary antiderivative.
Hint 4/4
The extra $r$ is exactly the $du$ that the substitution $u=r^{2}$ requires.
Over a box the three integrals are independent; iterate inside out, and all six orders give the same number.
Leave the plane. A solid needs three integrations, one per direction. The simplest solid is a box, where every limit is a constant and the only new idea is that there are now three of them, nested.
TheoremFubini's theorem for a box
Conditions
$f$ is continuous on the box $B=[a,b]\times[c,d]\times[p,q]$
Integrate the innermost variable first with the other two held fixed, then the middle, then the outer, each over its own constant limits. Because the limits are all numbers, any of the six orders of dx, dy, dz gives the same value; you pick the order that makes the arithmetic shortest.
A triple integral over a box, read inside out. Fix a point of the base $D$; the inner integral runs $z$ from the bottom face to the top face (the blue arrow). Then the outer two integrals sweep that point across the base in $x$ and $y$. For a box all six sweep orders enclose the same solid, so they give the same number.
Looks like this, but is not
Changing the order over a box seems to demand new limits, so a student rewrites $\int_0^1\!\int_0^2\!\int_0^3 f\,dz\,dy\,dx$ as $\int_0^3\!\int_0^2\!\int_0^1 f\,dz\,dy\,dx$ when switching to $dx$ first.
Over a box each variable owns a fixed interval no matter when it is integrated, so the limits travel with their differential, not with their position. Reordering to do $x$ first gives $\int_0^2\!\int_0^3\!\int_0^1 f\,dx\,dz\,dy$: the $x$-limits $0$ to $1$ stay glued to $dx$. Only for non-box solids do the numbers themselves change.
A triple integral over a box: ∭ (x + 2y + z) dV
Every limit is a constant, so the only work is three ordinary integrals done in sequence.
Given
$$B=[0,1]\times[0,2]\times[0,3],\qquad f=x+2y+z$$
Find
the value of the triple integral
Solution
We integrate z first only because it appears most simply; any order works over a box, and the linearity check below uses a different decomposition to confirm the number.
the last integration removes x and leaves a number
Answer $$24$$
Check
Independent check by linearity: $\iiint x\,dV=\tfrac12\cdot2\cdot3=3$, $\iiint 2y\,dV=2(1)(2)(3)=12$, $\iiint z\,dV=(1)(2)\tfrac92=9$, and $3+12+9=24$. Each piece is a product of three single integrals because the box separates.
Over a box a sum integrand splits into a sum of products of one-variable integrals; that shortcut is the practical content of Fubini.
When the box separates completely: ∭ xyz dV over the unit cube
A product integrand over a box factors into three independent integrals, one per variable.
Given
$$B=[0,1]^{3},\qquad f=xyz$$
Find
the value of the triple integral
Solution
We factor rather than iterate line by line, because a product integrand over a box is exactly the case where the three integrals are independent and no order matters at all.
each variable ranges over a constant interval and the integrand is a product, so the triple integral is a product of three single integrals
Evaluate the three equal factors
$$=\tfrac12\cdot\tfrac12\cdot\tfrac12=\tfrac18$$
each factor is one half by the power rule
Answer $$\tfrac18$$
Check
Reasonableness: $xyz$ averages $\tfrac18$ on the unit cube (its mean value), and the cube has volume $1$, so the integral is $\tfrac18\cdot 1=\tfrac18$.
Separation only works when both the region is a box and the integrand is a product; lose either and you must iterate in full.
Checkpoint
§13.4 — why the six orders agree●●○○○
Thirty seconds. You computed a triple integral over a box in the order $dz\,dy\,dx$ and a classmate used $dx\,dy\,dz$; you got the same number.
Given
$f$ continuous on a box $B$
Find
What guarantees the two orders give the same value?
Hint 1/4
This is a named theorem about interchanging the order of integration. What is its one hypothesis here?
Hint 2/4
Fubini's theorem: if the integrand is continuous on the box, every order of integration yields the same value.
Hint 3/4
The box has constant limits and $f$ is continuous, so Fubini applies and the order is free to choose.
Hint 4/4
Continuity of $f$ on the box is what licenses swapping the order (Fubini's theorem).
Show solutionState it
$$f\in C(B)\ \Rightarrow\ \text{orders agree}$$
continuity on the box is Fubini's hypothesis
Answer $$\text{Fubini's theorem}$$
Check
Without continuity one can build integrands whose iterated integrals in two orders disagree; continuity rules that out.
⚠ Changing the box limits when changing the order
reordering feels like it should renumber the limits, as it does for non-box solids
Cap the solid below and above by two surfaces; integrate z between them, then handle the shadow region D as a double integral.
Real solids are not boxes. The move is to split the three integrations: the inner one climbs from the floor surface to the ceiling surface, and the outer two are the double integral over the shadow the solid casts on a coordinate plane.
Freeze a point of the shadow region D. Integrate z straight up from the bottom surface to the top surface; that inner integral leaves a function of x and y. Then integrate that function over D exactly as a double integral, using whatever description of D — rectangular or polar — is convenient.
A general solid, read as a flat region $D$ with a lid. For each point of $D$ the inner integral runs $z$ from the bottom surface $z=u_1(x,y)$ up to the top surface $z=u_2(x,y)$; what is left is an ordinary double integral over $D$, which you already know how to set up.
Looks like this, but is not
For the tetrahedron under $x+y+z=1$ a student writes the inner limits $0\le z\le 1$, treating the ceiling as the constant plane $z=1$.
The ceiling is the slanted plane $z=1-x-y$, not $z=1$. Using $z=1$ integrates over the whole unit cube's worth of height and overcounts everything above the slanted face. The upper $z$-limit is the surface the solid actually reaches, which here falls as $x$ and $y$ grow.
Volume of the tetrahedron under x + y + z = 1
The classic first solid: bounded by the plane x + y + z = 1 and the three coordinate planes. Every limit is found by asking how far the solid reaches in one direction with the others fixed.
Given
$$E:\ \ x\ge 0,\ y\ge 0,\ z\ge 0,\ \ x+y+z\le 1$$
Find
the volume of the tetrahedron
Solution
We integrate z first because the solid is described by a floor and a ceiling in z; that inner integral collapses the problem to a double integral over the triangle, which we already know how to do.
Inner: z from the floor to the slanted ceiling
$$0\le z\le 1-x-y$$
the floor is z = 0 and the ceiling is the plane solved for z; the ceiling depends on x and y
$$\int_{0}^{1-x-y} 1\,dz=1-x-y$$
the inner integral of 1 is just the height of the column at (x,y)
Project to the shadow triangle D and integrate
$$D:\ 0\le x\le 1,\ \ 0\le y\le 1-x$$
setting z = 0 in x + y + z ≤ 1 gives the triangle x + y ≤ 1 in the first quadrant
the outer integral of (1 - x) squared over two from 0 to 1
Answer $$\tfrac16$$
Check
Formula check: a corner tetrahedron with legs of length $1$ has volume $\tfrac16\,(\text{leg})^{3}=\tfrac16$. Equivalently, base triangle area $\tfrac12$ times height $1$ times $\tfrac13$ is $\tfrac16$.
The pattern is permanent: inner z between two surfaces, then a familiar double integral over the shadow. The only new skill is finding the shadow, done by setting the caps equal or setting z = 0.
A sloping lid: volume under z = y over the rectangle [0,1] × [0,2]
Here the shadow is a rectangle but the ceiling z = y tilts, so the inner limit is not a constant.
Given
$$E:\ 0\le x\le 1,\ 0\le y\le 2,\ 0\le z\le y$$
Find
the volume of the solid
Solution
We keep the rectangle in x and y and integrate z first, because the only complication is the sloping lid, which the inner z-integral turns into the plain integrand y.
Inner integral in z
$$\int_{0}^{y} 1\,dz=y$$
the column at (x,y) has height y, the value of the sloping lid there
the y-integral gives a constant 2, and the x-integral multiplies by the length 1
Answer $$2$$
Check
Geometric check: the solid is a triangular prism, cross-section the triangle $0\le z\le y\le 2$ of area $\tfrac12(2)(2)=2$, extruded a length $1$ in $x$, giving volume $2\cdot 1=2$.
A non-constant lid is no harder than a constant one; the inner integral simply returns the lid's height as the new integrand.
Checkpoint
§13.5 — the inner z-limit of the tetrahedron●●○○○
Thirty seconds. You are setting up $\iiint_E f\,dV$ over the tetrahedron $x,y,z\ge 0$, $x+y+z\le 1$, integrating $z$ first.
Given
$E:\ x,y,z\ge 0,\ \ x+y+z\le 1$
Find
For a fixed $(x,y)$ in the base, what are the $z$-limits?
Hint 1/4
Ask: at a point $(x,y)$ of the base, where does the solid start and stop in the $z$-direction?
Hint 2/4
The floor is the plane $z=0$; the ceiling is the slanted face, found by solving $x+y+z=1$ for $z$.
Hint 3/4
Solving $x+y+z=1$ gives $z=1-x-y$, so $z$ runs from $0$ up to $1-x-y$.
Hint 4/4
The limits are $0\le z\le 1-x-y$.
Show solutionSolve the cap for z
$$z=1-x-y$$
the ceiling face solved for z
$$0\le z\le 1-x-y$$
floor zero, ceiling the slanted plane
Answer $$0\le z\le 1-x-y$$
Check
At the origin corner the height is 1; along the edge x + y = 1 the height falls to 0, as a tetrahedron should.
⚠ Using a constant ceiling instead of the slanted surface
the number 1 in x + y + z ≤ 1 is mistaken for the top z-value
wrong$$0\le z\le 1$$
right$$0\le z\le 1-x-y$$
⚠ Forgetting the middle limit shrinks with x
the base looks like a square, so y is given the full range 0 to 1
wrong$$0\le x\le 1,\ 0\le y\le 1$$
right$$0\le x\le 1,\ 0\le y\le 1-x$$
13.6Choosing the order and the average value
Integrate the trivial direction first and project onto the plane where the shadow is simplest; average value divides the integral by the volume.
Two loose ends. First, when a solid has an easy direction, integrate it first and let the hard part become a plane region you can choose how to slice. Second, the average value of a quantity over a solid is its integral divided by the solid's volume.
DefinitionAverage value over a solid
Conditions
$E$ a solid of finite, nonzero volume $V(E)=\iiint_E 1\,dV$
The average value of f over a solid is the total, the triple integral of f, shared out over the size of the solid, its volume. It is the single number that, if f were constant at that value, would give the same total. The denominator is the volume of the very same solid, computed with the very same limits.
The shadow region for the last volume: everything above the parabola $y=x^{2}$ and below the line $y=4$. Whichever way you slice it — vertical strips in $x$ or horizontal strips in $y$ — this is the region the outer double integral runs over once the trivial $z$-integral is done.
Looks like this, but is not
To average $z$ over the tetrahedron a student divides the integral $\iiint_E z\,dV$ by an edge length, say $1$, instead of by the volume.
The average divides by the volume $V(E)=\tfrac16$, not by any single edge. Dividing $\tfrac{1}{24}$ by $1$ gives $\tfrac{1}{24}$, but dividing by $\tfrac16$ gives $\tfrac14$; only the second is a value that $z$ actually takes near the middle of the solid.
Average value of z over the tetrahedron
We already have the volume of this solid; now we need the total of z over it, then we divide.
We compute the numerator and denominator over the very same triangle D, because reusing the volume setup means only the integrand changes and no new limits are needed.
Total of z: inner z-integral first
$$\int_{0}^{1-x-y} z\,dz=\tfrac{(1-x-y)^{2}}{2}$$
with x and y fixed, integrate z from floor to slanted ceiling
Symmetry check: $x$, $y$, $z$ play identical roles in this tetrahedron, so $\bar x=\bar y=\bar z$. Their sum satisfies $\bar x+\bar y+\bar z=\overline{x+y+z}\le 1$, and by symmetry each is $\tfrac14$, matching the centroid at $(\tfrac14,\tfrac14,\tfrac14)$.
Average value is never a new integral technique; it is one integral you can already do, divided by a volume you can already find.
Integrate the easy direction first: a solid over a parabolic region
The solid is trapped between z = 0 and z = 3, so z is trivial. Everything interesting is the plane shadow, drawn in the figure.
Given
$$E:\ x^{2}\le y\le 4,\ \ 0\le z\le 3$$
Find
the volume of the solid
Solution
We integrate z first because it lies between two constants and returns a constant height 3; that turns the triple integral into three times a plane area, which is the whole computation.
Inner z-integral is a constant height
$$\int_{0}^{3} 1\,dz=3$$
z runs between two flat planes, so the column height is 3 everywhere over the shadow
Double integral over the parabolic region D
$$D:\ -2\le x\le 2,\ \ x^{2}\le y\le 4$$
the region above y = x squared and below y = 4; the curves meet where x squared = 4, at x = plus and minus 2
the even integrand gives twice the value on 0 to 2, namely two times sixteen over three times three
Answer $$32$$
Check
Check by prisms: the shadow has area $\int_{-2}^{2}(4-x^{2})\,dx=\tfrac{32}{3}$, and the solid is that shadow extruded to height $3$, so $V=3\cdot\tfrac{32}{3}=32$.
When a solid is capped by two flat parallel planes, the volume is just the height times the area of the shadow; the triple integral is only bookkeeping around a double integral.
Checkpoint
§13.6 — the denominator of an average●●○○○
Thirty seconds. You have $\iiint_E z\,dV=\tfrac{1}{24}$ over the tetrahedron of volume $\tfrac16$, and you want the average value of $z$.
a product over a rectangle separates; no circular limits exist to convert
Answer $$1$$
Check
Switching to polar here would create $\theta$-dependent limits for a plain rectangle and gain nothing.
Both are double integrals, but the first has a circular region and a distance integrand while the second has a rectangle and a product; the region and the integrand together decide the coordinates.
How to tell them apart
Reach for polar only when circles or rays bound the region or an x squared plus y squared sits in the integrand; a rectangle with a polynomial integrand is already in its best coordinates.
z gives the height 1 minus x squared, then a plain double integral
Answer $$\tfrac23$$
Check
The solid is the area under $1-x^2$ (which is $\tfrac23$) extruded a unit in $y$.
The box has three constant limits and lets you integrate in any order; the general solid has a surface for one limit, which forces that variable to be integrated first.
How to tell them apart
If every bound is a number the solid is a box and order is free; if any bound is a surface, integrate the variable that surface controls first, then project.
Scaffolding comes off
The common skeleton
Sketch the region D and confirm that circles and rays describe it more simply than x and y.
Read the angle range $\alpha\le\theta\le\beta$; these outer limits are constants.
For a fixed direction $\theta$, read the radius range from the near boundary to the far boundary.
Substitute $x=r\cos\theta$, $y=r\sin\theta$, and every $x^{2}+y^{2}=r^{2}$ in the integrand.
Attach the area element $dA=r\,dr\,d\theta$; the $r$ is not optional.
Integrate r first with theta fixed, then integrate theta.
Check the number against a known area or volume, or against its order of magnitude.
1 · fully worked
∬ x dA over the quarter disk r ≤ 2 in the first quadrant
We separate the double integral into a product of two single integrals because the integrand r squared cosine theta factors into a function of r times a function of theta over constant limits.
Region and integrand into polar
$$0\le\theta\le\tfrac{\pi}{2},\quad 0\le r\le 2$$
first quadrant is a right angle; the disk of radius 2 gives r up to 2, both constant
each one-variable integral is elementary; sine gives 1, the cube gives eight thirds
Answer $$\tfrac{8}{3}$$
Check
The region has area $\tfrac14\pi(2)^{2}=\pi$ and $x$ ranges $0$ to $2$ on it, so the integral should be a modest positive number; $\tfrac83\approx 2.7$ fits.
A product integrand over a polar rectangle splits into a theta-integral times an r-integral; watch for that separation, it saves a line.
2 · you write the reasoning
Same skeleton, easier: the plain area of the quarter disk of radius $2$, $\displaystyle\iint_D 1\,dA$. The steps are given; write your own reason for each line before opening the model reasons, because the reasons are what carry over to the next problem.
reasoning
The quarter disk is a right angle of sweep and a radius reaching 2, both constant limits, so this is a polar rectangle.
reasoning
The integrand is 1 because we want plain area, but the area element still carries its r; this is the line where the r is most often dropped.
reasoning
The inner r-integral of r is r squared over two, evaluated at 2 gives 2; theta was held fixed throughout.
reasoning
The inner value 2 is constant, so the outer integral just multiplies it by the length of the theta range, a right angle.
3 · find the buried error
Harder, and now you are the grader. Below is a student's solution to $\displaystyle\iint_D (x^{2}+y^{2})\,dA$ over the unit disk $x^{2}+y^{2}\le 1$. Exactly two of the five lines contain an error a tired person makes. Find them before looking at the choices; the correct value is $\tfrac{\pi}{2}$.
the two buried errors (2)
⚠ step 2
The area element lost its $r$. The integrand is $r^{2}$ but $dA=r\,dr\,d\theta$, so the line should read $\int_{0}^{2\pi}\!\int_{0}^{1} r^{2}\cdot r\,dr\,d\theta=\int_{0}^{2\pi}\!\int_{0}^{1} r^{3}\,dr\,d\theta$.
The converted integrand r squared looks complete on its own, and the r from the element is a separate factor that is easy to leave off.
right
Attach the $r$: the inner integrand is $r^{3}$, and $\int_{0}^{1} r^{3}\,dr=\tfrac14$.
⚠ step 4
The full turn is $2\pi$, not $\pi$. Even with the wrong integrand, $\int_{0}^{2\pi}\tfrac13\,d\theta=\tfrac13\cdot 2\pi=\tfrac{2\pi}{3}$; the student halved the angle.
A half-turn $\pi$ and a full-turn $2\pi$ are swapped under time pressure, especially right after a fraction has appeared.
right
Multiply the constant inner value by the full $2\pi$. With the correct inner value $\tfrac14$, the answer is $\tfrac14\cdot 2\pi=\tfrac{\pi}{2}$.
By symmetry $\bar y$ over the half-disk is $\tfrac{4\cdot 3}{3\pi}=\tfrac{4}{\pi}$; times area $\tfrac{9\pi}{2}$ gives $18$.
Full exam-style question
Volume between two paraboloids: z = x² + y² and z = 8 - x² - y²exam format
A whole exam question in one move: two paraboloids trap a solid, and the region under it is a disk, so the volume is a polar double integral of the gap between the two surfaces.
Given
$$\text{solid between } z=x^{2}+y^{2}\ (\text{below}) \text{ and } z=8-x^{2}-y^{2}\ (\text{above})$$
Find
the volume of the enclosed solid
Solution
We equate the two surfaces first because their intersection is the boundary of the shadow disk; without that circle we would not know how far r runs.
Find where the surfaces meet, to get the shadow disk
the inner value is constant, so multiply by the full turn
Answer $$V=16\pi$$
Check
Midpoint check: at the axis $r=0$ the gap is $8-0=8$; averaged over the disk (area $4\pi$) the mean gap works out to $4$, and $4\times 4\pi=16\pi$, matching. The two surfaces are mirror images about $z=4$, which is why the mean height is exactly half the maximum.
One intersection, one polar double integral of a difference of surfaces. The whole question is the single setup line for the height.
Volume between two surfaces over a disk is always the double integral of (top minus bottom); the only work is writing both heights in r and remembering the area-element r.
Practice
A · concept 4 questions
1§13.1 — area of a disk in polar●●○○○
A concept trap, not a computation. You want the area of the disk of radius $3$ set up as a polar integral.
Given
disk $x^{2}+y^{2}\le 9$
Find
Which iterated integral gives the area?
Hint 1/4
Area is $\iint_D 1\,dA$. The only question is what $dA$ becomes and how far $r$ reaches.
Hint 2/4
The disk of radius $3$ is $0\le r\le 3$, $0\le\theta\le 2\pi$, and $dA=r\,dr\,d\theta$.
Hint 3/4
Attach the $r$ and use $r$ up to $3$, not $9$: $\int_0^{2\pi}\!\int_0^3 r\,dr\,d\theta$.
Hint 4/4
The area is $\int_0^{2\pi}\!\int_0^3 r\,dr\,d\theta=9\pi$.
sine is nonnegative on the first half-turn, which traces the circle once
Answer $$0\le\theta\le\pi$$
Check
Integrating to $2\pi$ gives area $2\pi$, twice the true area $\pi$ of a radius-1 circle.
B · computation 6 questions
1§13.2 — area of a ring●●○○○
A full annulus between two circles: constant limits, plain area.
Given
$$D:\ 1\le x^{2}+y^{2}\le 4$$
Find
Find the area of the ring.
Hint 1/4
The region is between two circles; find the radius range and the angle range.
Hint 2/4
$1\le r\le 2$ and $0\le\theta\le 2\pi$; area is $\iint_D 1\,dA=\int\!\int r\,dr\,d\theta$.
Hint 3/4
Inner integral $\int_1^2 r\,dr=\tfrac{4-1}{2}=\tfrac32$; then multiply by the full turn.
Hint 4/4
Area $=\tfrac32\cdot 2\pi=3\pi$.
Show solution
We integrate the constant $1$ over a full polar rectangle because both boundaries are circles, so the two integrals separate into a plain number times $2\pi$; a Cartesian setup would drag in square-root limits for nothing.
Both the region and the integrand are circular; convert everything to $r$ and $\theta$.
Hint 2/4
$x^{2}+y^{2}=r^{2}$, $0\le r\le 3$, $0\le\theta\le 2\pi$, and attach $dA=r\,dr\,d\theta$.
Hint 3/4
The inner integrand is $r^{2}\cdot r=r^{3}$: $\int_0^3 r^3\,dr=\tfrac{81}{4}$.
Hint 4/4
Multiply by the full turn: $\tfrac{81}{4}\cdot 2\pi=\tfrac{81\pi}{2}$.
Show solution
We convert before integrating because the integrand $x^{2}+y^{2}$ is exactly $r^{2}$ and the region is a disk; leaving it in Cartesian would force the non-separating limits $\pm\sqrt{9-x^{2}}$.
Find the volume between the cone and the plane $z=0$.
Hint 1/4
The height depends only on distance from the axis; write it in $r$.
Hint 2/4
$z=\sqrt{r^{2}}=r$, over $0\le r\le 2$, $0\le\theta\le 2\pi$; volume is $\iint_D z\,dA$.
Hint 3/4
The integrand $r$ times the element $r$ is $r^{2}$: $\int_0^2 r^2\,dr=\tfrac{8}{3}$.
Hint 4/4
Multiply by the full turn: $\tfrac{8}{3}\cdot 2\pi=\tfrac{16\pi}{3}$.
Show solution
We write the height in $r$ first because $z=\sqrt{x^{2}+y^{2}}=r$ has no $\theta$ in it, so the $\theta$-integral contributes only its length $2\pi$ and the work collapses to a single $r$-integral.
the cone height r times the area-element r gives r squared
Answer $$\tfrac{16\pi}{3}$$
Check
Cylinder of radius $2$ and height $2$ has volume $8\pi$; the part above the cone $z=r$ is itself a cone of volume $\tfrac13\pi(2)^2(2)=\tfrac{8\pi}{3}$; the solid below $z=r$ is the rest, $8\pi-\tfrac{8\pi}{3}=\tfrac{16\pi}{3}$.
When the height depends on $r$ alone the volume is $2\pi$ times a one-variable integral, and the answer cross-checks against the cylinder-minus-cone picture.
4§13.4 — a triple integral over a box●●○○○
A box and a simple linear integrand; iterate inside out.
Integrate the innermost variable first, holding the others fixed.
Hint 2/4
Inner $\int_0^1(2x+z)\,dz=2x+\tfrac12$; the box limits are all constants.
Hint 3/4
Middle $\int_0^2(2x+\tfrac12)\,dy=4x+1$; then integrate over $x$.
Hint 4/4
Outer $\int_0^1(4x+1)\,dx=2+1=3$.
Show solution
We iterate $z$ then $y$ then $x$ because the box has constant limits and no order is cheaper; taking the innermost variable that appears linearly keeps every step a one-line power rule.
Inner in z, middle in y
$$\int_0^1(2x+z)\,dz=2x+\tfrac12$$
z moves, x and y fixed
$$\int_0^2(2x+\tfrac12)\,dy=4x+1$$
multiply by the y-length 2
Outer in x
$$\int_0^1(4x+1)\,dx=[2x^2+x]_0^1=3$$
last integration removes x
Answer $$3$$
Check
By linearity $\iiint 2x\,dV=2\cdot\tfrac12\cdot2\cdot1=2$ and $\iiint z\,dV=1\cdot2\cdot\tfrac12=1$, sum $3$.
Over a box a linear integrand splits term by term through linearity, and that split is the fastest independent check on the iterated answer.
5§13.5 — volume under a parabolic sheet●●●○○
A solid trapped under $z=4-y^{2}$ over a rectangle in $x$ and $y$.
The lid $z=4-y^{2}$ depends only on $y$; integrate $z$ first.
Hint 2/4
Inner $\int_0^{4-y^{2}}dz=4-y^{2}$; the shadow is the rectangle $0\le x\le 2$, $-2\le y\le 2$.
Hint 3/4
Integrate over the rectangle: $\int_0^2\!\int_{-2}^2(4-y^{2})\,dy\,dx$; the $y$-integral is $\tfrac{32}{3}$.
Hint 4/4
Multiply by the $x$-length $2$: $\tfrac{32}{3}\cdot 2=\tfrac{64}{3}$.
Show solution
We integrate $z$ first because the lid $z=4-y^{2}$ is the column height directly; the remaining double integral is over a plain rectangle, and the $x$ that never appears only contributes its length.
Write the distance power in $r$: $(x^{2}+y^{2})^{3/2}=(r^{2})^{3/2}$.
Hint 2/4
$(r^{2})^{3/2}=r^{3}$, over $0\le r\le 1$, $0\le\theta\le\tfrac{\pi}{2}$, with $dA=r\,dr\,d\theta$.
Hint 3/4
Inner integrand $r^{3}\cdot r=r^{4}$: $\int_0^1 r^4\,dr=\tfrac15$; the angle range is a right angle.
Hint 4/4
Value $=\tfrac15\cdot\tfrac{\pi}{2}=\tfrac{\pi}{10}$.
Show solution
We rewrite the distance power as $r^{3}$ before anything else, because $(x^{2}+y^{2})^{3/2}=(r^{2})^{3/2}=r^{3}$ turns a fractional-power integrand into a monomial that the $r$ from $dA$ only raises by one.
Convert and attach r
$$(r^{2})^{3/2}=r^{3},\quad r^{3}\cdot r=r^{4}$$
the three-halves power of r squared is r cubed; times the element r gives r to the fourth
constant separation: a right angle times one fifth
Answer $$\tfrac{\pi}{10}$$
Check
On the quarter disk (area $\tfrac{\pi}{4}$) the integrand ranges $0$ to $1$, so the value is below $\tfrac{\pi}{4}\approx 0.79$; $\tfrac{\pi}{10}\approx 0.31$ fits.
Any integer or half-integer power of $x^{2}+y^{2}$ becomes a single power of $r$, so once the region is a polar rectangle these integrals are one monomial integration.
C · exam level 3 questions
1§13.3 — volume between a paraboloid and a plane●●●●○
An exam-style solid: the bowl $z=x^{2}+y^{2}$ closed off by the flat lid $z=4$.
Given
$$\text{solid: } x^{2}+y^{2}\le z\le 4$$
Find
What is the volume of the solid?
Hint 1/4
The solid sits between a floor surface and a ceiling; the shadow is where the floor is below the ceiling.
Hint 2/4
Floor $z=r^{2}$, ceiling $z=4$; they meet at $r^{2}=4$, so the disk is $0\le r\le 2$, and the height is $4-r^{2}$.
Hint 3/4
Set up $\int_0^{2\pi}\!\int_0^2 (4-r^{2})\,r\,dr\,d\theta$; do not forget the area-element $r$.
Hint 4/4
The inner integral is $\int_0^2(4r-r^{3})\,dr=8-4=4$, and the full turn gives $8\pi$.
Show solution
We find where the two surfaces meet before integrating, because $r^{2}=4$ fixes the shadow disk $r\le 2$; without that radius the height $4-r^{2}$ has nowhere to run.
the r-integral of 4r minus r cubed is 4, times the full turn
Answer $$8\pi$$
Check
The solid fills half the cylinder of radius 2 and height 4 (volume $16\pi$), as a paraboloid always does; half of $16\pi$ is $8\pi$.
A solid trapped between a paraboloid and a flat lid always fills exactly half its bounding cylinder, a one-line check on any such volume.
2§13.5 — volume of a tetrahedron●●●●○
A full exam setup: describe the solid, read all three limits, and integrate.
Given
$$E:\ x,y,z\ge 0,\ \ x+y+z\le 2$$
Find
(a) State the $z$-limits for a fixed $(x,y)$ in the base.
(b) Compute the volume of the tetrahedron.
Hint 1/4
For a fixed point $(x,y)$ of the base, where does $z$ start and stop?
Hint 2/4
Floor $z=0$, ceiling the plane $z=2-x-y$; the base is the triangle $x+y\le 2$ in the first quadrant.
Hint 3/4
Volume $=\int_0^2\!\int_0^{2-x}\!\int_0^{2-x-y} dz\,dy\,dx$; the inner integral is the height $2-x-y$.
Hint 4/4
The $y$-integral gives $\tfrac{(2-x)^{2}}{2}$ and the $x$-integral gives $\tfrac{4}{3}$.
Show solution
We integrate z first because the solid is described by a floor and a slanted ceiling in z; the remaining double integral is over the triangular shadow.
Corner-tetra formula $\tfrac16(\text{leg})^{3}=\tfrac16(2)^{3}=\tfrac{8}{6}=\tfrac{4}{3}$; also this is $2^{3}=8$ times the volume $\tfrac16$ of the unit tetra, since scaling all three legs by $2$ scales volume by $8$.
Scaling a corner tetrahedron's legs by a factor multiplies its volume by the cube of that factor, a quick check on any such answer.
3§13.6 — average value over a tetrahedron●●●●○
An exam favourite: an average value, which is one triple integral divided by a volume.
Symmetry: $x$, $y$, $z$ are interchangeable in this tetra, so $\bar x=\bar y=\bar z=\tfrac14$, placing the centroid at $(\tfrac14,\tfrac14,\tfrac14)$, the known centroid of a corner tetrahedron.
The centroid of a corner tetrahedron with legs on the axes sits one quarter of the way along each leg; average value and centroid are the same computation.
D · interleaved 3 questions
1§13.2 — reverse the order to make it doable●●●●○
Interleaved with the general-region work of the previous blocks: the given order cannot be integrated, so the region has to be re-read the other way.
Given
$$\int_0^1\!\int_x^1 e^{y^{2}}\,dy\,dx$$
Find
Evaluate by reversing the order of integration.
Hint 1/4
Try the inner integral as written. $\int e^{y^{2}}\,dy$ has no elementary form, so the given order is a dead end; the fix is to swap the order.
Hint 2/4
Sketch the region $0\le x\le 1$, $x\le y\le 1$: a triangle. Re-read it with $y$ outer: $0\le y\le 1$ and, for each $y$, $0\le x\le y$.
Hint 3/4
The integral becomes $\int_0^1\!\int_0^{y} e^{y^{2}}\,dx\,dy=\int_0^1 y\,e^{y^{2}}\,dy$, now with the extra $y$ that the substitution needs.
Hint 4/4
With $u=y^{2}$, $\int_0^1 y e^{y^{2}}\,dy=\tfrac12(e-1)$.
Show solution
We swap the order because integrating x first turns the impossible inner integral into a trivial strip width y, which then supplies the du for u = y squared.
the freed y is exactly the du the substitution needs
Answer $$\tfrac{e-1}{2}$$
Check
Positivity: $e^{y^{2}}\ge 1$ on the triangle of area $\tfrac12$, so the value is at least $\tfrac12$; $\tfrac{e-1}{2}\approx 0.86$ fits.
When an inner integral has no elementary antiderivative, reversing the order is the first thing to try; it often produces the very factor a substitution was missing.
2§13.3 — a Cartesian integral that is secretly polar●●●●○
Interleaved: a double integral handed to you in $x$ and $y$, whose square-root limit is the tell that its region is a disk.
Read the region: $0\le x\le 1$ and $0\le y\le\sqrt{1-x^{2}}$. What curve is $y=\sqrt{1-x^{2}}$?
Hint 2/4
It is the quarter of the unit disk in the first quadrant: $0\le r\le 1$, $0\le\theta\le\tfrac{\pi}{2}$; and $x^{2}+y^{2}=r^{2}$.
Hint 3/4
Convert: $\int_0^{\pi/2}\!\int_0^1 r^{2}\cdot r\,dr\,d\theta$, remembering the area-element $r$.
Hint 4/4
The value is $\tfrac{\pi}{2}\cdot\tfrac14=\tfrac{\pi}{8}$.
Show solution
We convert because the limit y = sqrt(1 minus x squared) is a circle and the integrand is x squared plus y squared; both signal polar, which turns a hard Cartesian inner integral into r cubed.
Identify the region and convert
$$y=\sqrt{1-x^{2}}\Rightarrow x^{2}+y^{2}=1$$
the upper limit is the unit circle; with x from 0 to 1 this is the first-quadrant quarter disk
r cubed inner integral is one quarter, times the right angle
Answer $$\tfrac{\pi}{8}$$
Check
The Cartesian inner integral $\int_0^{\sqrt{1-x^2}}(x^2+y^2)\,dy$ is messy but doable; converting avoids it entirely, and $\tfrac{\pi}{8}\approx 0.39$ is a plausible value over a quarter disk of area $\tfrac{\pi}{4}\approx 0.79$.
A square-root limit of the form sqrt(a squared minus x squared) is almost always a circle in disguise; converting to polar is the reflex it should trigger.
3§13.3 — the volume of a sphere from a double integral●●●●●
Interleaved with the volume-under-a-surface idea: the top half of a sphere is a surface over a disk, so its volume is a polar double integral, and doubling it gives the whole sphere.
Given
$$\text{sphere } x^{2}+y^{2}+z^{2}=9$$
Find
Find the volume of the sphere of radius $3$ using a polar double integral.
Hint 1/4
The upper hemisphere is the surface $z=\sqrt{9-x^{2}-y^{2}}$ over the disk of radius $3$; volume of the sphere is twice the volume under it.
Hint 2/4
In polar $z=\sqrt{9-r^{2}}$, so $V=2\int_0^{2\pi}\!\int_0^3 \sqrt{9-r^{2}}\,r\,dr\,d\theta$.
Hint 3/4
For the inner integral use $u=9-r^{2}$, $du=-2r\,dr$; the $r$ from the area element is exactly what it needs.
Hint 4/4
$\int_0^3\sqrt{9-r^{2}}\,r\,dr=9$, so $V=2\cdot 2\pi\cdot 9=36\pi$.
Show solution
We double the hemisphere rather than integrating the full sphere as a solid, because the hemisphere is a single surface over a disk, which is precisely a polar double integral.
u = 9 minus r squared; the r is the du, and the bracket runs from 9 down to 0
$$V=2\cdot 2\pi\cdot 9=36\pi$$
twice the full turn times the inner value 9
Answer $$36\pi$$
Check
Sphere-volume formula $\tfrac43\pi R^{3}=\tfrac43\pi(3)^{3}=\tfrac43\pi\cdot 27=36\pi$, matching exactly.
The classic $\tfrac43\pi R^{3}$ is not an axiom; it falls out of one polar double integral, with the area-element r doing the crucial work in the substitution.
Mistake ledger (12 entries)
⚠ Dropping the r in the polar area element
The area of a polar patch grows with distance from the origin; the r records that growth and is never optional.
wrong$$dA=dr\,d\theta$$
right$$dA=r\,dr\,d\theta$$
⚠ Measuring the angle in degrees
The factor r in the area element is only valid when the angular step is in radians.
wrong$$\int_0^{360}\!\int_0^1 r\,dr\,d\theta$$
right$$\int_0^{2\pi}\!\int_0^1 r\,dr\,d\theta$$
⚠ Putting an angle-dependent limit on the outer integral
Outer limits must be constants; a variable left in an outer limit survives into the final answer.
Close the page and write from memory: the polar area element with its factor, the one sentence that says which coordinate the outer limits belong to and why they are constants, the setup that turned the bell curve into an elementary integral, Fubini's one hypothesis for a box, the two-surface description of a general solid with the order it forces, and the definition of average value with the correct denominator.
Produce the factor $r$ in $dA=r\,dr\,d\theta$ from the area of a polar patch, not from memory.
c-polar-why
Read the $\theta$-limits and the $r$-limits off a sketch, and say which of them may depend on the other.
c-polar-regions
Compute a volume with a polar integral and explain why the r makes the bell-curve integral elementary.
c-polar-apps
Evaluate a triple integral over a box and state the hypothesis that makes the six orders agree.
c-triple-box
Turn a floor-and-ceiling description of a solid into ordered triple-integral limits.
c-triple-general
Compute an average value over a solid, dividing by the volume rather than an edge.
c-triple-order
Glossary (8 terms)
polar coordinateskutupsal koordinatlar
A point located by its distance $r$ from the origin and its angle $\theta$ from the positive $x$-axis, with $x=r\cos\theta$, $y=r\sin\theta$.
area elementalan elemanı
The area of an infinitesimal patch of the region. In polar coordinates it is $r\,dr\,d\theta$, not $dr\,d\theta$.
polar rectanglekutupsal dikdörtgen
A region $\{a\le r\le b,\ \alpha\le\theta\le\beta\}$ with all four limits constant, bounded by two circular arcs and two rays.
triple integralüçlü integral
An integral $\iiint_E f\,dV$ of a function over a solid region in space, summing $f$ times a volume element.
iterated integralardışık integral
A multiple integral written as nested single integrals, evaluated one variable at a time from the inside out.
Fubini's theoremFubini teoremi
The statement that for a continuous integrand on a box, the multiple integral equals the iterated integral in any order of the variables.
type-1 solidbirinci tip cisim
A solid described by a flat region $D$ and two capping surfaces $u_1(x,y)\le z\le u_2(x,y)$, so the inner integration is in $z$.
average valueortalama değer
The integral of a function over a region divided by the size of the region; over a solid, $\bar f=\tfrac{1}{V(E)}\iiint_E f\,dV$.
What comes next
§14 · Triple integrals in cylindrical and spherical coordinates, and change of variables
The next block builds coordinate systems for the solids that resisted a plain triple integral here: cylindrical coordinates wrap the polar trick around a vertical axis, spherical coordinates suit anything built from distance to a point, and the change-of-variables rule turns the area-element r and its three-dimensional cousins into one general Jacobian factor.
Sources
James Stewart, Calculus, Ninth Edition — sections 15.3 and 15.6 The change-to-polar rule, the triple-integral definition and Fubini's theorem are the ones this book states; every function integrated on this page is a different one, worked from scratch.
Course syllabus, week 13: Multiple Integrals 15.3, 15.6 The two section numbers on this page are the two the syllabus line names; the quiz and homework weights on the card come from the same syllabus.
Every numerical value on this page Each area, volume and average was recomputed by an independent route — a geometry formula, a symmetry argument, or an order-of-magnitude bound — before publication.