5 concepts15 worked examples28 exercises5 exam-level4 figures
How often this comes up in exams: 31 questions, 26/39 of the years
all years (1988-2026)Final 19 · Midterm 2 11 · Midterm 1 1
last ten yearsFinal 6 · Midterm 2 3
also touched by33 questions that are filed under another section but still use this one
The questions in this section name themselves clearly in the archive.
Counted from the department's own exam archive: 106 exams, 599 question items. On an independent sample of 60 questions read by hand, the topic labels were 83% right, so treat these counts as a direction, not a certainty.
What are you here for?
14 Triple integrals in cylindrical and spherical coordinates, and change of variables
Start with this
One question before you read anything. Getting it wrong is the point: it shows you what this section is for.
§14.1 — is the r really necessary?
Before the new material, test the instinct you are carrying in from polar coordinates. A quick true/false.
Find(a) True or false, with one sentence of reason.
Givenclaim: in $dV=dz\,dr\,d\theta$
Hint 1/4
Think about whether relabelling coordinates can leave areas unchanged when the grid is now made of fans.
Hint 2/4
The polar area element is $dA=r\,dr\,d\theta$, and cylindrical just stacks that in $z$.
Hint 3/4
So the correct element carries a factor $r$: $dV=r\,dz\,dr\,d\theta$; the claim omits it.
Hint 4/4
The statement is false.
Show solution
Compare with the polar element
$$dA_{\text{polar}}=r\,dr\,d\theta$$
established for area integrals
$$dV=r\,dz\,dr\,d\theta$$
stacking area elements in z keeps the r
Answer $$\text{False}$$
Check
Any nonconstant integrand would come out wrong without the r; even plain volume would be off.
A grain silo is a cylinder with a rounded cap, and the weight it holds depends on packing that thickens with depth. Slice it into little Cartesian boxes and the round wall fights you at every limit, turning clean bounds into square roots.
By the end you can set up and evaluate the volume or total of any solid built from cylinders, cones, spheres and their caps, and say which of three coordinate systems makes the limits constant.
In 60 seconds
Match the coordinate system to the solid's surfaces, attach the right volume element, and let constant limits factor the integral.
a substitution flattens the region or the integrand
Three most common mistakes
Dropping the $r$ in cylindrical or the $\sin\varphi$ in spherical.
Letting $\varphi$ run to $2\pi$ instead of stopping at $\pi$.
Using $J$ with its sign instead of $\lvert J\rvert$.
Quizzes and homework are $10\%$ each; this material shows up most often as a single set-up-and-evaluate volume, so the marks are in choosing the system and getting the element and limits right, not in heavy arithmetic.
How much time do you have?
10 minutes
You leave with the two volume elements, the three costliest mistakes, and a worked cylindrical volume — enough to attempt a straight find the volume question.
The 60-second card · Formula card · Triple integrals in cylindrical coordinates · Mistake ledger
45 minutes
You add spherical set-ups and the Jacobian that explains where both volume elements come from, plus enough drilled computation to be fast.
The 60-second card · Triple integrals in cylindrical coordinates · Triple integrals in spherical coordinates · Change of variables and the Jacobian · The region first, and the Jacobian you can actually compute · Fading ladder · Practice B (computation)
Full reading
Everything in built order: the failure that forces new coordinates, the two systems, the Jacobian that unifies them, and interleaved practice where you choose the system yourself.
The opening pages · Recall first · Try it yourself first · All five concept blocks · Method boxes · Contrast pair · Fading ladder · Full exam-style question · Practice A to D · Mistake ledger · Check yourself
By the end of this section
Set up and evaluate triple integrals in cylindrical coordinates, including the factor r.
Set up and evaluate triple integrals in spherical coordinates, including the factor ρ² sinφ.
Compute a Jacobian and use it to change variables in a double integral.
Derive the cylindrical and spherical volume elements from the 3D Jacobian and apply it to stretched solids.
Choose the coordinate system that makes a given solid's bounding surfaces into constant limits.
Transform a region by sending its bounding curves to constant lines, and produce the area factor from a substitution given as $u(x,y),v(x,y)$ by taking the reciprocal of the determinant you can compute.
Syllabus coverage
covered
15.7 — covered
Triple integrals in cylindrical coordinates
covered
15.8 — covered
Triple integrals in spherical coordinates
Spherical coordinates and the volume element in c-spherical, with both radial cases: a sphere centred at the origin, where the limit is a constant, and a sphere through the origin, where it becomes $\rho=2a\cos\varphi$ and the three integrals stop separating.
covered
15.9 — covered
Change of variables in multiple integrals
Three concepts: the 2D Jacobian in c-change-2d, the region and the in c-region-map for the usual case where the substitution is given as $u(x,y),v(x,y)$, and the 3D case in c-change-3d.
Recall first
Polar area element
In polar coordinates $x=r\cos\theta,\ y=r\sin\theta$ and $dA=r\,dr\,d\theta$.
cylindrical coordinates are polar in the floor with z left alone, so the r reappears; spherical and the Jacobian generalise the same factor
Iterated triple integral
$\iiint_E f\,dV$ is evaluated inside-out, holding the outer variables fixed while the inner integral runs.
every set-up here is an iterated integral; only the coordinates and the volume element change
2x2 and 3x3 determinants
$\det\begin{pmatrix}a&b\\c&d\end{pmatrix}=ad-bc$, and a $3\times3$ determinant expands along any row or column with alternating signs.
the Jacobian is a determinant of partial derivatives
Try it yourself first (2 questions)
1§14.2 — which factor for the ball?
This one is a deliberate trap: the tempting answer is the one that mimics cylindrical coordinates. Pick the spherical volume element.
Find(a) Which is the correct volume element $dV$?
Givenspherical coordinates $\rho,\varphi,\theta$ with $\varphi$ measured from the $+z$-axis
Hint 1/4
Ask which element makes the unit ball integrate to $\tfrac43\pi$, not which looks like the cylindrical one.
Hint 2/4
The spherical element is $\rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta$.
Hint 3/4
Only one option has both the $\rho^2$ and the $\sin\varphi$; the others drop one of them.
Hint 4/4
The correct choice is $\rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta$.
You only need the $2\times2$ rule; no calculus is involved yet.
Hint 2/4
For a $2\times2$ matrix $\det=ad-bc$.
Hint 3/4
Here $a=3,\ b=1,\ c=2,\ d=4$, so $\det=3\cdot4-1\cdot2$.
Hint 4/4
$\det M=10$.
Show solution
Apply ad−bc
$$\det M=3\cdot4-1\cdot2=12-2=10$$
the standard 2x2 rule
Answer $$\det M=\boxed{10}$$
Check
Positive, so the map preserves orientation; the Jacobian will use this same computation.
Notation
symbol
reads as
means
watch out
$r,\theta,z$
r, theta, z
cylindrical coordinates: $r$ and $\theta$ are polar in the $xy$-plane, $z$ is the usual height.
$r\ge0$; the same $\theta$ as polar. Do not confuse this $r$ with the spherical $\rho$.
$\rho,\varphi,\theta$
rho, phi, theta
spherical coordinates: $\rho$ is the distance from the origin, $\varphi$ the angle from the $+z$-axis, $\theta$ the azimuth.
$\varphi$ runs only over $[0,\pi]$, while $\theta$ runs over $[0,2\pi]$.
$dV$
the volume element
the little piece of volume: $dz\,dr\,d\theta$ scaled by $r$, or the spherical scaling $\rho^2\sin\varphi$.
the scaling factor is not decoration; it carries the geometry of the coordinate grid.
$\dfrac{\partial(x,y)}{\partial(u,v)}$
the Jacobian of x, y with respect to u, v
the determinant $x_u y_v-x_v y_u$ of first partial derivatives.
the integral uses its absolute value $\lvert J\rvert$.
Conventions used here
Spherical angle convention
We follow the textbook's order $(\rho,\varphi,\theta)$ with $\varphi$ the angle measured down from the $+z$-axis, $0\le\varphi\le\pi$, and $\theta$ the azimuth $0\le\theta\le2\pi$. Some physics texts swap the names of $\varphi$ and $\theta$; whenever you read another source, check which letter is the angle from the axis before trusting a formula.
Radii are non-negative and angles in radians
Both $r$ and $\rho$ are distances, so $r\ge0$ and $\rho\ge0$ always, and every angle is in radians. A limit that would make a radius negative is a sign that the region was set up wrong.
The area and volume factor is never negative
The change-of-variables factor is $\lvert J\rvert$, the absolute value of the Jacobian. A negative $J$ only records that the map reverses orientation; it never makes a computed area or volume negative.
14.1Triple integrals in cylindrical coordinates
Trade x and y for polar r and θ while keeping z, and pay for it with one extra factor of r.
We already integrate over solids by stacking little boxes $dx\,dy\,dz$; the trouble starts the moment the solid has a round wall.
Solvable with what we have
Integrate over a box $[a,b]\times[c,d]\times[p,q]$ in Cartesian coordinates.
Integrate over a solid whose shadow on the floor is a rectangle.
Not solvable yet
Cleanly integrate over a solid inside $x^2+y^2=4$, because the floor limits become $-\sqrt{4-x^2}\le y\le\sqrt{4-x^2}$.
Handle a paraboloid or cone without square-root limits fighting back at every step.
For the disk of radius $2$ the Cartesian set-up forces $\int_{-2}^{2}\!\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}(\cdots)\,dy\,dx$, and every antiderivative drags a $\sqrt{4-x^2}$ along.
Why it fails
The geometry is round but the grid is square. Polar coordinates already fixed this for area integrals; cylindrical coordinates are just polar in the floor with the height $z$ left untouched.
DefinitionCylindrical coordinates and their volume element
Conditions
$r\ge 0$ and angles are in radians
$x=r\cos\theta,\ y=r\sin\theta,\ z=z$, so $x^2+y^2=r^2$
Replace $x$ and $y$ by polar $r,\theta$, leave $z$ alone, and multiply the integrand by $r$: the volume element is $dV=r\,dz\,dr\,d\theta$.
The patch at radius $\textcolor{#1f6feb}{r}$ has sides $dr$ and $r\,d\theta$, so its area is $\textcolor{#1f6feb}{r}\,dr\,d\theta$ — the factor $r$ is geometry, not a rule to memorise.
Looks like this, but is not
Writing $dV=dz\,dr\,d\theta$, as if switching to $r,\theta$ were a pure relabelling.
Dropping the $r$ makes the answer wrong. The little patch at radius $r$ has area $r\,dr\,d\theta$, not $dr\,d\theta$: an arc at radius $2$ is twice as long as one at radius $1$ for the same $d\theta$. The $r$ is the width of the fan, and it must be there.
Volume between $z=x^2+y^2$ and the plane $z=4$
Find the volume of the solid $E$ that sits above the paraboloid $z=x^2+y^2$ and below the plane $z=4$.
Findthe volume of $E$
Given
$E$ lies above $z=x^2+y^2$
$E$ lies below $z=4$
Solution
Find where the two surfaces meet
$$x^2+y^2 = 4$$
set the paraboloid equal to the plane; this is the shadow of the solid on the floor
$$r = 2$$
in cylindrical $x^2+y^2=r^2$, so the disk has radius $2$ — that fixes the $r$ and $\theta$ limits
Write the limits, floor to ceiling
$$r^2 \le z \le 4$$
for a fixed point in the disk, z runs from the paraboloid up to the flat lid
$\int_0^2(4r-r^3)dr$; the $\theta$-integral just gives $2\pi$
Answer $$V=\boxed{8\pi}$$
Check
Independent check by subtraction: a cylinder of radius $2$ and height $4$ has volume $16\pi$; the part scooped out under the paraboloid is $\iint_{r\le2} r^2\,dA=\int_0^{2\pi}\!\int_0^2 r^3\,dr\,d\theta=2\pi\cdot 4=8\pi$. So $16\pi-8\pi=8\pi$ — same answer, different route.
When z is trapped between two surfaces and the shadow is a disk, cylindrical coordinates make the z-integral a one-liner and turn the floor into constant r-limits.
Volume of the cone under $z=2$
Find the volume of the solid bounded by the cone $z=\sqrt{x^2+y^2}$ below and the plane $z=2$ above.
Findthe volume of the solid
Given
lower surface $z=\sqrt{x^2+y^2}$
upper surface $z=2$
Solution
Read the cone in cylindrical
$$z=\sqrt{x^2+y^2}=r$$
the cone is simply $z=r$: a $45^\circ$ wall rising from the origin
$$r \le z \le 2,\quad 0\le r\le 2$$
z runs from the cone up to the lid; the lid meets the cone at r=2
The solid is a right circular cone with base radius $2$ and height $2$, so $V=\tfrac13\pi R^2 h=\tfrac13\pi(4)(2)=\tfrac{8\pi}{3}$ — the elementary-geometry formula agrees.
A cone through the origin is $z=r$ in cylindrical, one of the cleanest surfaces the system has; hold this next to spherical, where the same cone becomes a constant angle.
Checkpoint
§14.1 — cylinder volume as a warm-up
You want a fast sanity check that the cylindrical set-up reproduces a volume you already know. Take the plainest solid there is.
Find
(a) Write $\iiint_E dV$ as an iterated integral in cylindrical coordinates.
(b) Evaluate it.
Given
$E$ is inside the cylinder $x^2+y^2=9$
$E$ lies between $z=0$ and $z=5$
Hint 1/4
The solid is a plain cylinder. What are the three constant limits for $r$, $\theta$ and $z$?
Hint 2/4
In cylindrical coordinates the volume element is $dV=r\,dz\,dr\,d\theta$ — the $r$ is not optional.
Hint 3/4
With $x^2+y^2=9$ the radius runs $0\le r\le 3$, the height $0\le z\le5$, and $0\le\theta\le2\pi$; put these on the three integral signs.
Hint 4/4
All limits are constant, so the triple integral factors into $2\pi\cdot\int_0^3 5r\,dr$.
For each direction set by the angles $\varphi,\theta$, march outward in $\rho$; the volume element is $dV=\rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta$, big far out and pinched near the poles.
A cross-section through the axis: $\textcolor{#1f6feb}{\rho}$ is the distance from the origin, $\textcolor{#d1690a}{\varphi}$ the angle down from the $z$-axis, and $\textcolor{#128a5a}{r=\rho\sin\varphi}$ the distance from the axis.
Looks like this, but is not
Guessing $dV=\rho^2\,d\rho\,d\varphi\,d\theta$ by analogy with the $r$ of cylindrical coordinates.
The $\sin\varphi$ is missing, and without it the ball comes out as $\tfrac{2\pi^2 a^3}{3}$ instead of $\tfrac43\pi a^3$. Near the poles ($\varphi\approx0$ or $\pi$) a step in $\theta$ barely moves you, so those slabs are thin — $\sin\varphi$ is the width of that slab.
Volume of a ball of radius $a$
Confirm the volume of a solid ball of radius $a$ directly from the spherical volume element.
Findthe volume $\tfrac43\pi a^3$
Giventhe ball is $x^2+y^2+z^2\le a^2$
Solution
Read the ball in spherical
$$0\le \rho\le a$$
the ball is every point within distance a of the origin
$$0\le\varphi\le\pi,\quad 0\le\theta\le2\pi$$
to sweep the whole ball the two angles cover their full ranges
This is the volume every geometry course quotes for a sphere, recovered here with no special tricks — a strong sign the $\rho^2\sin\varphi$ factor is right.
The three-way split is the reason spherical coordinates feel easy on balls: constant limits mean the answer is a product of three separate integrals.
An ice-cream cone: inside $\rho=2$, above $\varphi=\pi/3$
Find the volume of the region inside the sphere $\rho=2$ and above the cone $\varphi=\pi/3$ (the cone opening upward from the origin).
Findthe volume of the region
Given
inside the sphere $\rho=2$
above the cone $\varphi=\tfrac{\pi}{3}$
Solution
Turn the two surfaces into limits
$$0\le\rho\le 2$$
inside the sphere means the distance from the origin is at most 2
Solid-angle check: the cone $\varphi\le\tfrac{\pi}{3}$ captures a fraction $\tfrac{1-\cos(\pi/3)}{2}=\tfrac{1-1/2}{2}=\tfrac14$ of every direction, and $\tfrac14$ of the full ball $\tfrac43\pi(2)^3=\tfrac{32\pi}{3}$ is exactly $\tfrac{8\pi}{3}$.
Two of the three integrals were single lines; only the ϕ-limit carried the geometry.
A cone measured from the origin is a constant ϕ; a sphere is a constant ρ. When both bounding surfaces are constants in a system, that system is the right one.
A sphere through the origin: $x^{2}+y^{2}+z^{2}=2z$
Convert the sphere $x^{2}+y^{2}+z^{2}=2z$ to spherical coordinates and find the volume it encloses. Note that this sphere is not centred at the origin, so its radial limit will not be a constant.
Findthe surface in the form $\rho=\rho(\varphi)$, and the enclosed volume
Given
boundary surface $x^{2}+y^{2}+z^{2}=2z$
$\rho^{2}=x^{2}+y^{2}+z^{2}$ and $z=\rho\cos\varphi$
Solution
Convert the surface
$$\rho^{2}=2\rho\cos\varphi$$
the left side is $\rho^{2}$ by definition and $z=\rho\cos\varphi$
$$\rho=2\cos\varphi$$
divide by $\rho$; the discarded root $\rho=0$ is the origin, which the surface passes through anyway
Read the limits, and notice which one moves
$$0\le\rho\le 2\cos\varphi$$
for a fixed direction the ray starts at the origin and stops on the surface, and where it stops depends on $\varphi$
Complete the square in Cartesian form: $x^{2}+y^{2}+(z-1)^{2}=1$ is the sphere of radius $1$ centred at $(0,0,1)$, whose volume is $\tfrac43\pi(1)^{3}=\tfrac{4\pi}{3}$ by the elementary formula. The integral with the moving limit reproduces it exactly.
A sphere centred at the origin gives a constant $\rho$; a sphere through the origin gives $\rho=2a\cos\varphi$. Once the radial limit moves, the three integrals no longer factor, so integrate in $\rho$ first and let the result become the next integrand.
Checkpoint
§14.2 — a spherical shell
You need to be sure you can turn two spheres into ρ-limits and keep the angle factors straight. A shell between two radii is the cleanest test.
Find
(a) Set up the volume in spherical coordinates.
(b) Evaluate it and check against $\tfrac43\pi(R_2^3-R_1^3)$.
Giventhe region lies between the spheres $\rho=1$ and $\rho=2$
Hint 1/4
The region is everything whose distance from the origin is between $1$ and $2$, in every direction. What does that make the three limits?
Hint 2/4
The volume element is $dV=\rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta$; the angles sweep their full ranges.
Hint 3/4
With $1\le\rho\le2$, $0\le\varphi\le\pi$, $0\le\theta\le2\pi$, the integral splits into $2\pi\cdot 2\cdot\int_1^2\rho^2\,d\rho$.
Hint 4/4
$\int_1^2\rho^2 d\rho=\tfrac{8-1}{3}=\tfrac73$, so $V=4\pi\cdot\tfrac73$.
Any substitution multiplies areas by the absolute value of a determinant, the Jacobian.
Both new systems bought their simplicity with a mystery factor — the $r$ and the $\rho^2\sin\varphi$; the single idea that produces every such factor is the Jacobian.
TheoremChange of variables in a double integral
Conditions
$T:(u,v)\mapsto(x,y)$ is one-to-one with continuous partials on the region $S$
the Jacobian $J=\dfrac{\partial(x,y)}{\partial(u,v)}$ is nonzero on the interior of $S$
Rewrite the integrand in the new variables, swap the region $R$ for its preimage $S$, and multiply by the absolute Jacobian, which measures how much the map stretches area at each point.
Proof
A tiny rectangle $du\times dv$ at $(u,v)$ maps to a tiny parallelogram spanned by $T_u\,du$ and $T_v\,dv$.
The area of that parallelogram is $\lvert T_u\times T_v\rvert\,du\,dv$, and $\lvert T_u\times T_v\rvert=\left|x_u y_v-x_v y_u\right|=\lvert J\rvert$.
Summing $f$ times these little areas over $S$ gives the right-hand side.
The map $T$ sends the $\textcolor{#1f6feb}{uv}$ unit square to an $\textcolor{#d1690a}{xy}$ parallelogram; its area is $\textcolor{#128a5a}{\lvert J\rvert}$ times $du\,dv$.
Looks like this, but is not
Writing $\iint_R f\,dA=\iint_S f\,du\,dv$, treating the substitution as a pure renaming with no factor.
That is only right when $|J|=1$ (a rigid move or a shear). A stretch changes area, and the ellipse example makes it plain: $x=au,\ y=bv$ scales every area by $ab$, so dropping $|J|$ underestimates the ellipse by exactly that factor.
Where the polar $r$ comes from
Compute the Jacobian of the polar map $x=r\cos\theta,\ y=r\sin\theta$ and show it equals $r$.
This is exactly the factor we were told to insert for polar area integrals; the Jacobian derives it instead of asking us to memorise it, and $|r|=r$ since $r\ge0$.
Every coordinate change hides its scaling factor in a Jacobian — polar, cylindrical and spherical are all one formula with different maps plugged in.
Area of an ellipse by a linear change of variables
Find the area enclosed by the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ using the substitution $x=au,\ y=bv$.
the ellipse becomes the unit disk in the uv-plane — a region we can integrate over blindfolded
Compute the Jacobian
$$J=\begin{vmatrix} a & 0\\ 0 & b\end{vmatrix}=ab$$
$x_u=a,\ x_v=0,\ y_u=0,\ y_v=b$
Transfer the integral
$$A=\iint_{u^2+v^2\le1} ab\,du\,dv$$
$dA=|J|\,du\,dv=ab\,du\,dv$ since $ab>0$
$$=ab\cdot(\text{area of unit disk})=ab\cdot\pi$$
the leftover integral is just the area of the unit disk
Answer $$A=\boxed{\pi ab}$$
Check
When $a=b$ the ellipse is a circle of radius $a$ and the formula gives $\pi a^2$, the familiar circle area — a clean special-case check.
A linear substitution stretches every area by the constant $|J|$, so a hard region can be traded for an easy one at the cost of one determinant.
Checkpoint
§14.3 — the Jacobian of a linear map
You want to confirm you can turn a substitution into an area-scaling number. A plain linear map is the place to test it, because you can check the answer by hand.
The unit square spanned by $(1,0)$ and $(0,1)$ maps to the parallelogram spanned by $(2,1)$ and $(1,3)$, whose area is $|2\cdot3-1\cdot1|=5$ — the determinant read geometrically.
⚠ Forgetting the absolute value
the determinant is written without bars, and a negative orientation is easy to keep by accident
wrong$$dA=J\,du\,dv$$
right$$dA=\lvert J\rvert\,du\,dv$$
⚠ Inverting the Jacobian
there are two Jacobians and only one of them gets flipped; the one already in the theorem's own direction does not
14.4The region first, and the Jacobian you can actually compute
Turn the bounding curves into constant lines to get the new region, and get the Jacobian you need by flipping the one you can compute.
The theorem asks for $x$ and $y$ as functions of $u$ and $v$, but a question almost never hands you the map that way round: it hands you the region and a substitution written $u=u(x,y)$, $v=v(x,y)$.
RuleThe region rule and the reciprocal rule
Conditions
$T$ is one-to-one with continuous partials and a nonzero Jacobian on the interior
the boundary curves of $R$ are exactly the curves on which $u$ or $v$ is constant
Boundaries map to boundaries, so each curve bounding the region becomes a line on which one new variable is constant; and the Jacobian you want is one over the Jacobian you can compute directly from the substitution as given.
Proof
Going from $(u,v)$ to $(x,y)$ and back again is the identity map, whose Jacobian matrix is the identity matrix.
By the chain rule the matrix of partials of a composition is the product of the two matrices of partials, so those two matrices are inverses of each other.
Determinants multiply, so the two determinants multiply to $1$ — which is the reciprocal rule, and it also explains why a zero Jacobian is forbidden: nothing has a reciprocal there.
The four bounding lines of $\textcolor{#d1690a}{R}$ are exactly $u=0,\ u=3,\ v=0,\ v=3$, so the awkward parallelogram becomes the square $\textcolor{#1f6feb}{S}$ — and no limit in the new integral depends on the other variable.
Looks like this, but is not
Computing $\dfrac{\partial(u,v)}{\partial(x,y)}$ from the substitution as given, and putting that number straight into the integral as the area factor.
It is the reciprocal of what the theorem asks for. For $u=y-x,\ v=y+2x$ the direct determinant is $-3$, so a small patch in the $uv$-plane corresponds to a patch three times smaller in the $xy$-plane and the correct factor is $\tfrac13$, not $3$ — an answer wrong by a factor of nine.
A parallelogram flattened into a square
Evaluate $\displaystyle\iint_R (y-x)\,dA$, where $R$ is the parallelogram with vertices $(0,0),(1,1),(0,3),(-1,2)$, using $u=y-x$ and $v=y+2x$.
Findthe value of the integral
Given
$R$ is bounded by $y=x$, $y=x+3$, $y=-2x$, $y=-2x+3$
constant limits let the double integral split into two single ones
Answer $$\iint_R(y-x)\,dA=\boxed{\dfrac92}$$
Check
Two independent checks. First the Jacobian: solving the substitution gives $x=\tfrac{v-u}{3}$, $y=\tfrac{2u+v}{3}$, whose determinant is $\left(-\tfrac13\right)\!\left(\tfrac13\right)-\left(\tfrac13\right)\!\left(\tfrac23\right)=-\tfrac13$, matching the reciprocal. Then the value: the area of $R$ is $3\cdot\tfrac13=1\cdot3=3$, which agrees with the cross product $|1\cdot2-1\cdot(-1)|=3$ of the two edge vectors $\langle1,1\rangle$ and $\langle-1,2\rangle$; and $y-x=u$ is spread evenly over $[0,3]$, so its mean is $\tfrac32$ and the integral must be $3\cdot\tfrac32=\tfrac92$.
Read the substitution backwards to see why it works: the four sides of $R$ were two families of parallel lines, and each family became a single new variable held constant.
Curved boundaries: $xy$ and $y/x$ as the new variables
Evaluate $\displaystyle\iint_R xy\,dA$, where $R$ is the region in the first quadrant bounded by the hyperbolas $xy=1$ and $xy=4$ and by the lines $y=x$ and $y=4x$, using $u=xy$ and $v=y/x$.
Findthe value of the integral
Given
$R$: $1\le xy\le4$ and $1\le y/x\le4$, with $x>0$
$u=xy,\quad v=y/x$
Solution
The region is already written in the new variables
$$S=\{(u,v):1\le u\le4,\ 1\le v\le4\}$$
each of the four boundary curves is one of $u=1,u=4,v=1,v=4$, which is exactly why these two combinations were chosen
Invert the map explicitly as a cross-check: $x=\sqrt{u/v}$ and $y=\sqrt{uv}$ give $\dfrac{\partial(x,y)}{\partial(u,v)}=\dfrac{1}{2v}$ by direct differentiation, the same factor the reciprocal rule produced in one line. As a size check, the same method gives area $\iint_S\tfrac{1}{2v}\,du\,dv=3\ln2\approx2.08$, and the integrand $xy$ runs only between $1$ and $4$ on $R$, so the integral must lie between $2.08$ and $8.32$ — and $5.20$ does.
Choose the new variables by looking at the boundary, not at the integrand: whatever combination is constant along a bounding curve is the variable worth naming.
Checkpoint
§14.6 — flip the Jacobian in thirty seconds
The substitution comes in the direction you cannot use directly. This is a thirty-second check that you can still produce the factor the theorem wants.
Solve directly: $x=\tfrac{u+v}{2}$, $y=\tfrac{u-v}{2}$, whose determinant is $\left(\tfrac12\right)\!\left(-\tfrac12\right)-\left(\tfrac12\right)\!\left(\tfrac12\right)=-\tfrac12$, matching in both size and sign.
⚠ Using the determinant you computed instead of its reciprocal
the direct determinant is the only one you actually calculated, so it is the number in front of you
Exactly the 2D rule with one more variable: rewrite the integrand, swap the solid for its preimage, and multiply by the absolute value of the $3\times3$ determinant of partial derivatives.
Proof
A small box $du\,dv\,dw$ maps to a parallelepiped spanned by $T_u\,du,\ T_v\,dv,\ T_w\,dw$.
Its volume is the absolute scalar triple product, which equals $\lvert J\rvert\,du\,dv\,dw$.
Looks like this, but is not
Believing you must memorise $r$ and $\rho^2\sin\varphi$ as separate, unrelated facts.
They are not separate: feeding the cylindrical map into the determinant gives $r$, and the spherical map gives $\rho^2\sin\varphi$. One law, two evaluations — the worked examples do both from scratch.
This is precisely the factor we inserted by hand for cylindrical integrals; the leftover $z$-row being $(0,0,1)$ is why cylindrical is just polar with an untouched height.
Cylindrical coordinates are 2D polar in disguise: the z-direction contributes a trivial 1 to the determinant, so the whole factor is the polar r.
The 3D Jacobian recovers the spherical $\rho^2\sin\varphi$
Using $x=\rho\sin\varphi\cos\theta,\ y=\rho\sin\varphi\sin\theta,\ z=\rho\cos\varphi$, compute $\dfrac{\partial(x,y,z)}{\partial(\rho,\varphi,\theta)}$.
Since $0\le\varphi\le\pi$ we have $\sin\varphi\ge0$, so $|J|=\rho^2\sin\varphi$ with no sign worry — and it matches the factor used throughout the spherical section.
One 3×3 determinant, expanded along the row with a zero to cut the work in half.
The two factors that looked like rules to memorise are both just this determinant; learn the map, turn the crank, and the volume element falls out.
Checkpoint
§14.4 — a diagonal 3D map
You want to see the 3D Jacobian scale a volume the way the 2D one scaled area. A pure stretch along the axes is the transparent case.
with 0≤φ≤π the answer must be non-negative, and a negative result flags an algebra error
wrong$$J=-\rho^2\sin\varphi$$
right$$J=\rho^2\sin\varphi$$
Setting up a cylindrical triple integral
the solid has a round wall, or is bounded by a cone or paraboloid, and its shadow on the xy-plane is a disk or polar region
Find the shadow
Project the solid onto the $xy$-plane and describe that region in polar $r,\theta$.
Set the z-limits
For a fixed $(r,\theta)$, let $z$ run from the lower surface to the upper surface, each written as a function of $r,\theta$.
Attach the element
Write $dV=r\,dz\,dr\,d\theta$ and convert the integrand using $x=r\cos\theta,\ y=r\sin\theta$.
Integrate inside-out
Do the $z$-integral first, then $r$, then $\theta$; if all limits are constant the three integrals separate.
Where it goes wrong
Dropping the factor $r$.
Using a nonconstant z-limit where the lid is flat, or vice versa.
Setting up a spherical triple integral
the solid is bounded by spheres (constant ρ) and/or cones from the origin (constant φ)
Convert the surfaces
Turn each bounding surface into $\rho=$const or $\varphi=$const; a cone $z=\sqrt{x^2+y^2}$ becomes $\varphi=\tfrac{\pi}{4}$.
Order the limits
Set $\theta$ and $\varphi$ (usually constant), then $\rho$ from the inner to the outer radial surface.
Attach the element
Write $dV=\rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta$ and convert the integrand.
Separate if you can
Constant limits let the integral factor into three single-variable integrals.
Where it goes wrong
Forgetting the $\sin\varphi$.
Letting $\varphi$ exceed $\pi$.
Changing variables with a Jacobian
a substitution turns the region into a simple shape or the integrand into something elementary
Map the region
Push the boundary of $R$ through the substitution to find the new region $S$ in $u,v$.
Compute the Jacobian
Form $\dfrac{\partial(x,y)}{\partial(u,v)}$ and take its absolute value.
Rewrite the integrand
Express $f$ in $u,v$.
Assemble
Write $\iint_S f\,\lvert J\rvert\,du\,dv$ and evaluate over the simple region $S$.
Where it goes wrong
Forgetting $\lvert\cdot\rvert$.
Using the reciprocal Jacobian by mistake.
When the substitution is handed to you as $u(x,y)$, $v(x,y)$
the question names the new variables in terms of the old ones, or the region is bounded by two families of curves such as $xy=c$ and $y/x=c$
Turn each bounding curve into a constant
Substitute each boundary equation and read off the value of $u$ or $v$ it forces. Four curves giving four constants means the new region is a rectangle.
Take the determinant you can take
Compute $\dfrac{\partial(u,v)}{\partial(x,y)}$ directly from the substitution — no inverting, no square roots.
Flip it
The factor in the theorem is $\left|\dfrac{\partial(u,v)}{\partial(x,y)}\right|^{-1}$. Do the reciprocal before the absolute value or after; the result is the same.
Rewrite the integrand and integrate
Express $f$ in $u$ and $v$, attach the factor, and integrate over the new region.
Where it goes wrong
using the determinant as computed instead of its reciprocal
changing the integrand but leaving the limits in $x$ and $y$
wasting time solving for $x(u,v)$ and $y(u,v)$ when only the factor is needed
Same answer, more work — a substitution was needed.
Cylindrical works but pays with a √ and a substitution the sphere avoids.
Both give $\tfrac{4\pi}{3}$, but spherical had three constant limits while cylindrical needed a square-root z-limit and a substitution.
How to tell them apart
If the solid's boundary is a sphere or a cone from the origin, spherical makes those surfaces constant; reach for cylindrical only when a straight round wall or a flat/paraboloid lid dominates.
Scaffolding comes off
The common skeleton
Identify the natural coordinate system from the bounding surfaces (round wall or paraboloid → cylindrical; sphere or cone from the origin → spherical).
Find the region's limits: the angles first, then the radial variable from the surfaces that bound it.
Attach the correct volume element (r dz dr dθ, or ρ² sinφ dρ dφ dθ).
If the limits are constant, factor the integral into single-variable pieces; otherwise integrate inside-out.
Check against a known volume or a solid-angle fraction.
1 · fully worked
Volume under the paraboloid $z=4-x^2-y^2$
Find the volume of the solid below $z=4-x^2-y^2$ and above the plane $z=0$.
Findthe volume
Given
top surface $z=4-x^2-y^2$
bottom $z=0$
Solution
Identify the system and the shadow
$$z=4-r^2$$
the paraboloid is round, so cylindrical; the floor is where z=0
Under a paraboloid capped at height 4 over a radius-2 disk, the average height works out to 2, and 2 times the disk area 4π is 8π.
This is the shared skeleton: name the system, find the shadow, set floor-to-ceiling limits, attach the element, evaluate.
2 · you write the reasoning
Set up and evaluate the volume of the plain cylinder $x^2+y^2\le1$, $0\le z\le 2$ in cylindrical coordinates. The steps are given; supply the reason for each, then check yourself against the model reasons.
reasoning
The radius is bounded by the cylinder wall $r=1$, the height by the two flat lids, and $\theta$ sweeps the full circle — all constant.
reasoning
Attach the cylindrical element $r\,dz\,dr\,d\theta$; the $r$ is the part that is easy to drop.
reasoning
With constant limits the integral factors: $\int_0^2 dz=2$ and $\int_0^1 r\,dr=\tfrac12$, so $2\pi\cdot2\cdot\tfrac12=2\pi$.
reasoning
Geometry agrees: $\pi R^2 h=\pi(1)(2)=2\pi$.
3 · find the buried error
A student computes the volume inside the sphere $\rho=1$ and above the cone $\varphi=\pi/4$. Two of the steps are wrong. Find them.
the two buried errors (2)
⚠ step 2
the cone caps the region, so $\varphi$ must stop at $\pi/4$, not run to $\pi$. Writing $0\le\varphi\le\pi$ integrates over the entire ball and ignores the cone.
the phrase 'inside the sphere' is read as the whole ball, and the second bounding surface — the cone — is quietly forgotten
right
Use $0\le\varphi\le\tfrac{\pi}{4}$; the region is the part of the ball inside a $45^\circ$ cone from the origin.
⚠ step 3
the volume element is missing its $\sin\varphi$: it must be $\rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta$.
the ρ² is remembered because it is prominent, while the sinφ — the factor that pinches the poles — is the one students most often drop
right
Write $dV=\rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta$; with both fixes $V=\tfrac{2\pi}{3}\big(1-\tfrac{\sqrt2}{2}\big)$.
4 · the bare problem
§14.2 — hemisphere, on your own
No scaffolding this time. A hemisphere is the reward for keeping the angle limits and the volume element straight.
Find
(a) Set up the volume in spherical coordinates.
(b) Evaluate it.
Giventhe solid is the upper half of the ball $\rho\le3$, i.e. $z\ge0$
Hint 1/4
'Upper half' restricts one of the two angles — which one, and to what range?
Hint 2/4
The volume element is $\rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta$; $z\ge0$ means $\varphi$ stops at the equator.
Hint 3/4
Use $0\le\rho\le3$, $0\le\varphi\le\tfrac{\pi}{2}$, $0\le\theta\le2\pi$; then $V=2\pi\cdot\big[-\cos\varphi\big]_0^{\pi/2}\cdot\tfrac{27}{3}$.
Hint 4/4
$\big[-\cos\varphi\big]_0^{\pi/2}=1$ and $\int_0^3\rho^2 d\rho=9$, so $V=2\pi\cdot1\cdot9$.
A hemisphere is half a ball: $\tfrac12\cdot\tfrac43\pi(3)^3=\tfrac12\cdot36\pi=18\pi$.
Full exam-style question
Sphere above a cone: $x^2+y^2+z^2=4$ above $z=\sqrt{x^2+y^2}$exam format
Find the volume of the solid bounded above by the sphere $x^2+y^2+z^2=4$ and below by the cone $z=\sqrt{x^2+y^2}$.
Findthe volume of the solid
Given
upper surface: sphere $x^2+y^2+z^2=4$
lower surface: cone $z=\sqrt{x^2+y^2}$
Solution
On an exam the marks are in the set-up: recognising that a sphere and a cone-from-the-origin both become constants in spherical coordinates is worth more than the arithmetic that follows.
Solid-angle check: the cone $\varphi\le\tfrac{\pi}{4}$ takes a fraction $\tfrac{1-\cos(\pi/4)}{2}=\tfrac{1-\sqrt2/2}{2}$ of the ball $\tfrac{32\pi}{3}$, which is $\tfrac{16\pi}{3}\big(1-\tfrac{\sqrt2}{2}\big)\approx4.91$ — matching the integral.
One determinant-free set-up: both bounding surfaces became constant limits, so the whole integral factored.
Practice
A · concept 4 questions
1§14.1 — the r is not optional
A classmate says the factor $r$ can be dropped whenever the solid is symmetric about the axis, since symmetry 'evens things out'. Decide.
Find(a) True or false, with a one-line reason.
Givenclaim: for axially symmetric solids, $dV=dz\,dr\,d\theta$ is fine
Hint 1/4
Symmetry moves the θ-integral out front, but ask whether it touches the radial weighting at all.
Hint 2/4
The $r$ comes from the area of the polar patch, independent of any symmetry.
Hint 3/4
Even for the symmetric cylinder $r\le R$, $0\le z\le h$, the correct volume $\pi R^2 h$ needs the $r$.
Swapping u and v flips the sign of J but cannot change a volume.
3§14.5 — which system fits the solid?
A solid is bounded below by the cone $z=\sqrt{x^2+y^2}$ and above by the sphere $x^2+y^2+z^2=9$. You must pick the coordinate system before writing a single limit.
Find(a) Which system makes both bounding surfaces into constant limits?
Given
lower: cone $z=\sqrt{x^2+y^2}$
upper: sphere $x^2+y^2+z^2=9$
Hint 1/4
Look at what each surface becomes in each system: which system turns both into single equations like 'variable = constant'?
Hint 2/4
A cone from the origin is $\varphi=$ const and a sphere is $\rho=$ const — both constants in spherical.
Hint 3/4
In cylindrical the sphere becomes $z=\sqrt{9-r^2}$, a nonconstant limit; spherical keeps both clean.
Hint 4/4
Spherical coordinates are the fit.
Show solution
Convert each surface
$$\text{sphere}:\ \rho=3$$
constant ρ
$$\text{cone}:\ \varphi=\tfrac{\pi}{4}$$
constant φ; both constants ⇒ spherical
Answer $$\text{spherical}$$
Check
Both limits constant means the triple integral factors — the hallmark of the right system.
4§14.2 — does φ ever reach π?
Weigh this statement about the angle limits: 'for a full ball, $\varphi$ runs from $0$ to $2\pi$, just like $\theta$.' Decide.
Find(a) True or false, with a reason.
Givenclaim: full ball needs $0\le\varphi\le2\pi$
Hint 1/4
Picture sweeping φ from the north pole downward; when have you already covered every point once?
Hint 2/4
$\varphi$ is the angle from the $+z$-axis and runs only over $[0,\pi]$.
Hint 3/4
By $\varphi=\pi$ you are at the south pole; $\theta\in[0,2\pi]$ already spins you around, so $\varphi$ to $2\pi$ would double-count.
Hint 4/4
The claim is false.
Show solution
Range of each angle
$$0\le\varphi\le\pi$$
colatitude, pole to pole
$$0\le\theta\le2\pi$$
azimuth, full spin; φ to 2π would cover the ball twice
Answer $$\text{False}$$
Check
Doubling φ's range would double the computed volume — a fast way to catch the slip.
B · computation 7 questions
1§14.1 — mass inside a cylinder
A cylinder $x^2+y^2\le4$ between $z=0$ and $z=3$ carries density $\delta=x^2+y^2$ (heavier toward the wall). You want its total mass.
Find
(a) Set up $\iiint_E (x^2+y^2)\,dV$ in cylindrical coordinates.
(b) Evaluate it.
Given
region: $x^2+y^2\le4$, $0\le z\le3$
density $\delta=x^2+y^2$
Hint 1/4
First turn the density into the new variables; then attach the element and read off the constant limits.
Hint 2/4
In cylindrical $x^2+y^2=r^2$ and $dV=r\,dz\,dr\,d\theta$, so the integrand times element is $r^2\cdot r=r^3$.
Hint 3/4
With $0\le r\le2$, $0\le z\le3$, $0\le\theta\le2\pi$: $M=\int_0^{2\pi}\!\int_0^2\!\int_0^3 r^3\,dz\,dr\,d\theta$.
Hint 4/4
$\int_0^3 dz=3$, $\int_0^2 r^3 dr=4$, $\int_0^{2\pi}d\theta=2\pi$, so $M=2\pi\cdot3\cdot4$.
Both heights are positive on $0<r<1$ (paraboloid above cone there), so a positive volume is expected, and $\tfrac{5\pi}{6}\approx2.6$ is a sensible size for a solid inside radius 1.
3§14.2 — a spherical cap by angle
Find the volume of the region inside the ball $\rho\le1$ lying between the cones $\varphi=\tfrac{\pi}{6}$ and $\varphi=\tfrac{\pi}{3}$.
Find
(a) Set up the volume.
(b) Evaluate it, leaving $\sqrt3$ in the answer.
Given
$0\le\rho\le1$
$\tfrac{\pi}{6}\le\varphi\le\tfrac{\pi}{3}$
Hint 1/4
The two cones bound $\varphi$; the ball bounds $\rho$; nothing bounds $\theta$. Write the three ranges.
Hint 2/4
Use $dV=\rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta$ with $\tfrac{\pi}{6}\le\varphi\le\tfrac{\pi}{3}$.
Hint 3/4
$V=2\pi\big[-\cos\varphi\big]_{\pi/6}^{\pi/3}\cdot\int_0^1\rho^2 d\rho$, and $\int_0^1\rho^2 d\rho=\tfrac13$.
Check as an average: $\tfrac{4\pi/5}{4\pi/3}=\tfrac35$, so the mean of $\rho^2$ over the ball is $\tfrac35$, between 0 and 1 as it must be.
5§14.3 — a change of variables on a region
Evaluate $\iint_R (x+y)\,dA$ where $R$ is the square with vertices $(0,0),(1,1),(2,0),(1,-1)$, using $u=x+y,\ v=x-y$.
Find
(a) Find the region in the $uv$-plane and the Jacobian $\dfrac{\partial(x,y)}{\partial(u,v)}$.
(b) Evaluate the integral.
Given
$R$ has vertices $(0,0),(1,1),(2,0),(1,-1)$
$u=x+y,\ v=x-y$
Hint 1/4
Push the four vertices through the map to see the new region; it should be a rectangle in $u,v$.
Hint 2/4
Invert to $x=\tfrac{u+v}{2},\ y=\tfrac{u-v}{2}$ and compute $\dfrac{\partial(x,y)}{\partial(u,v)}$.
Hint 3/4
The vertices map to $u\in[0,2]$, $v\in[0,2]$; $J=\begin{vmatrix}1/2 & 1/2\\ 1/2 & -1/2\end{vmatrix}=-\tfrac12$, so $|J|=\tfrac12$. The integrand $x+y=u$.
Setting $a=b=c=R$ recovers the sphere $\tfrac43\pi R^3$, the right special case.
7§14.2 — a moving radial limit meets a cone
The two spherical surfaces that produce non-constant limits meet in one solid: a sphere sitting on the origin, cut by a cone from the origin. Exams like this pairing because a student who only ever saw constant limits stalls at the first line.
Find(a) Find the volume of the solid that lies inside the sphere $\rho=2\cos\varphi$ and above the cone $\varphi=\tfrac{\pi}{4}$.
Given
inside the sphere $\rho=2\cos\varphi$
above the cone $\varphi=\tfrac{\pi}{4}$
Hint 1/4
Only one limit changes compared with the full sphere through the origin: the cone stops $\varphi$ early. Everything else is the same set-up.
Hint 2/4
Integrate in $\rho$ first, from $0$ to $2\cos\varphi$, then in $\varphi$; $\theta$ still factors out to $2\pi$.
Hint 3/4
The inner integral leaves $\dfrac{8\cos^{3}\varphi}{3}\sin\varphi$, and $\displaystyle\int_0^{\pi/4}\cos^{3}\varphi\sin\varphi\,d\varphi=\left[-\tfrac{\cos^{4}\varphi}{4}\right]_0^{\pi/4}$.
Hint 4/4
With $\cos^{4}\tfrac{\pi}{4}=\tfrac14$ the bracket is $\tfrac{3}{16}$, so $V=\tfrac{16\pi}{3}\cdot\tfrac{3}{16}=\pi$.
multiply through by the constant collected in front
Answer $$V=\boxed{\pi}$$
Check
Keep the cone angle general: the same computation gives $V(\alpha)=\tfrac{4\pi}{3}\left(1-\cos^{4}\alpha\right)$. At $\alpha=\tfrac{\pi}{2}$ the cone has opened into the whole half-space and $V=\tfrac{4\pi}{3}$, the volume of the whole unit ball — the independent check. At $\alpha=\tfrac{\pi}{4}$ it returns $\tfrac{4\pi}{3}\cdot\tfrac34=\pi$.
Whenever the radial limit is a function of $\varphi$, the $\rho$-integral has to run first; the result then becomes an ordinary one-variable integral in $\varphi$.
C · exam level 5 questions
1§14.2 — value of a spherical integral
On a timed exam you meet $\displaystyle\iiint_B z\,dV$ over the upper half-ball $\rho\le1,\ z\ge0$. Rather than grind, set it up in spherical and pick the value.
Find(a) Which is the value of $\iiint_B z\,dV$?
Given
$B:\ \rho\le1,\ z\ge0$
integrand $z=\rho\cos\varphi$
Hint 1/4
Write $z$ and $dV$ in spherical, then note which single integral carries all the φ-dependence.
Hint 2/4
$z\,dV=\rho\cos\varphi\cdot\rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta$, which simplifies to $\rho^3\sin\varphi\cos\varphi\,d\rho\,d\varphi\,d\theta$, with $0\le\varphi\le\tfrac{\pi}{2}$.
$x_u=2u,\ x_v=-2v,\ y_u=2v,\ y_v=2u$; substitute into $x_u y_v-x_v y_u$.
Hint 4/4
$2u\cdot2u-(-2v)(2v)=4u^2+4v^2$.
Show solution
Partials and determinant
$$x_u=2u,\ x_v=-2v,\ y_u=2v,\ y_v=2u$$
four first partials
$$J=2u\cdot2u-(-2v)(2v)=4u^2+4v^2$$
ad−bc
Answer $$4(u^2+v^2)$$
Check
It is non-negative and vanishes only at the origin, where the squaring map is not locally invertible — a meaningful check.
4§14.2 — average distance from the centre of a ball
Find the average value of the distance to the origin over the solid ball $\rho\le a$. (The average of $g$ over a solid is $\tfrac{1}{\text{vol}}\iiint g\,dV$.)
Find
(a) Compute $\iiint_B \rho\,dV$.
(b) Divide by the ball's volume to get the average.
Given
ball $\rho\le a$
function $g=\rho$
Hint 1/4
Distance to the origin is just $\rho$; set up $\iiint\rho\,dV$ and remember what to divide by at the end.
Hint 2/4
$\rho\,dV=\rho\cdot\rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta=\rho^3\sin\varphi\,\dots$, and divide by $\tfrac43\pi a^3$.
The result is between 0 and a and closer to the surface — sensible, since more of a ball's volume sits at larger radius.
5§14.6 — a trapezoid with a non-rectangular image
An integrand that no ordering of $x$ and $y$ can handle, over a region whose four sides are two pairs of level curves. The new region is not a rectangle this time, so the inner limits carry the outer variable.
Find
(a) Find the region $S$ in the $uv$-plane and the area factor.
$R$ is the trapezoid with vertices $(1,0),(2,0),(0,-2),(0,-1)$
$u=x+y,\quad v=x-y$
integrand $e^{(x+y)/(x-y)}$
Hint 1/4
The integrand is $e^{u/v}$ the moment you substitute, so the whole question is really about the region and the factor.
Hint 2/4
Map the four sides. Two of them are the lines $x-y=1$ and $x-y=2$; the other two are the axes $y=0$ and $x=0$, which become $u=v$ and $u=-v$.
Hint 3/4
So $S$ is the triangle-like strip $1\le v\le2$, $-v\le u\le v$, and $\dfrac{\partial(u,v)}{\partial(x,y)}=\begin{vmatrix}1&1\\1&-1\end{vmatrix}=-2$, giving the factor $\tfrac12$.
Hint 4/4
Integrate in $u$ first: $\displaystyle\int_{-v}^{v}e^{u/v}du=v\left(e-e^{-1}\right)$, then in $v$ over $[1,2]$ to reach $\tfrac34\left(e-e^{-1}\right)$.
Bound it independently. The area of $R$ is $\tfrac12\cdot\text{area}(S)=\tfrac12\cdot3=\tfrac32$, and on $R$ the exponent $\tfrac{x+y}{x-y}$ runs over $[-1,1]$, so the integrand lies between $e^{-1}$ and $e$ and the integral must lie between $\tfrac32 e^{-1}\approx0.55$ and $\tfrac32 e\approx4.08$. The value $1.76$ sits inside, and it is nearer the middle than either end, as an exponential averaged over a symmetric exponent range should be.
A non-rectangular image is not a failure of the substitution: it just means one of the new limits carries the other variable, exactly as in an ordinary double integral.
D · interleaved 3 questions
1§14.5 — pick the tool, then use it
No section heading is attached to this one. Decide the cleanest method yourself, then find the volume of the solid inside both the cylinder $x^2+y^2=1$ and the sphere $x^2+y^2+z^2=4$.
Find
(a) Choose cylindrical or spherical, with a one-line reason.
(b) Compute the volume.
Given
inside the cylinder $x^2+y^2=1$
inside the sphere $x^2+y^2+z^2=4$
Hint 1/4
One surface is a cylinder and the other a sphere; ask which system keeps the cylinder's wall as a constant limit.
Hint 2/4
The cylinder wall $r=1$ is a constant only in cylindrical; the sphere gives the z-limits $z=\pm\sqrt{4-r^2}$. Use $dV=r\,dz\,dr\,d\theta$.
Numerically $\tfrac{4\pi}{3}(8-5.196)\approx11.75$; the enclosing cylinder $r\le1$ up to $z=\pm\sqrt3$ has volume $2\pi\sqrt3\approx10.9$, wait — the sphere bulges above that near the axis, so a slightly larger value is right.
2§14.5 — a double integral in disguise
Type hidden. Evaluate $\displaystyle\int_0^{2}\!\int_0^{\sqrt{4-x^2}} e^{-(x^2+y^2)}\,dy\,dx$.
Find
(a) Identify the region and switch to polar coordinates.
(b) Evaluate the integral.
Given
inner $y:0\to\sqrt{4-x^2}$
outer $x:0\to2$
Hint 1/4
Read the limits as a region in the plane: what shape is $0\le x\le2,\ 0\le y\le\sqrt{4-x^2}$?
Hint 2/4
It is the quarter-disk of radius 2 in the first quadrant; in polar $e^{-(x^2+y^2)}=e^{-r^2}$ and $dA=r\,dr\,d\theta$.
Hint 3/4
$\int_0^{\pi/2}\!\int_0^2 e^{-r^2}r\,dr\,d\theta$, with $\theta$ over the first quadrant only.
Hint 4/4
$\int_0^2 e^{-r^2}r\,dr=\tfrac12(1-e^{-4})$, times $\tfrac{\pi}{2}$.
Show solution
Recognise the region
$$0\le x\le2,\ 0\le y\le\sqrt{4-x^2}$$
this is the first-quadrant quarter of the disk of radius 2
Without the polar switch the Cartesian integral has no elementary antiderivative; the factor $r$ from $dA$ is exactly what makes $e^{-r^2}r$ integrable. The value $\approx0.77$ is below the quarter-disk area $\pi\approx3.14$ times the max $1$, as it should be.
3§14.5 — order of integration meets coordinates
Type hidden. A solid $E$ is bounded by the paraboloid $z=x^2+y^2$ and the plane $z=9$. Find $\iiint_E z\,dV$.
Find
(a) Choose coordinates and set up $\iiint_E z\,dV$.
(b) Evaluate it.
Given
below $z=9$, above $z=x^2+y^2$
integrand $z$
Hint 1/4
The solid is the paraboloid problem again, but now with a weight $z$; cylindrical still fits.
Hint 2/4
$z$ runs from $r^2$ to $9$, $r$ from $0$ to $3$ (since the surfaces meet at $r^2=9$), and $dV=r\,dz\,dr\,d\theta$.
Hint 3/4
$\iiint z\,dV=\int_0^{2\pi}\!\int_0^3\!\int_{r^2}^9 z\,r\,dz\,dr\,d\theta$; the inner integral is $\tfrac12(81-r^4)\,r$.
The plain volume of this solid is $\tfrac{81\pi}{2}$ and its z-values lie in $[0,9]$, so a z-weighted total of $243\pi\approx763$ against volume $\approx127$ gives a mean height $\approx6$, sensibly biased toward the top.
Mistake ledger (12 entries)
⚠ The factor r goes missing
changing to r, θ feels like relabelling, but the polar patch genuinely has area r dr dθ
books disagree on the names, so the reflex is unreliable
wrong$$z=\rho\cos\theta$$
right$$z=\rho\cos\varphi$$
⚠ Letting φ run to 2π
θ runs to 2π, so φ is assumed to as well; but φ pole-to-pole is only π
wrong$$0\le\varphi\le2\pi$$
right$$0\le\varphi\le\pi$$
⚠ Forgetting the absolute value
the determinant is written without bars and a flipped orientation is easy to keep by accident
wrong$$dA=J\,du\,dv$$
right$$dA=\lvert J\rvert\,du\,dv$$
⚠ Inverting the Jacobian
there are two Jacobians; flipping $\partial(u,v)/\partial(x,y)$ is the rule, flipping $\partial(x,y)/\partial(u,v)$ after you already have it is the mistake
with 0≤φ≤π the Jacobian must be non-negative, so a negative result flags an algebra slip before it reaches the answer
wrong$$J=-\rho^2\sin\varphi$$
right$$J=\rho^2\sin\varphi$$
⚠ Freezing when the radial limit is not a constant
a sphere centred at the origin gives a constant limit and a sphere through the origin does not; integrate in $\rho$ first and the rest is a one-variable integral
Without looking, write the volume element in cylindrical and in spherical coordinates, state the range of each angle, and say in one line where the r and the ρ² sinφ come from.
Set up and evaluate the volume of the solid under $z=4-x^2-y^2$ over its disk, with the $r$ in place?
c-cylindrical
Compute the volume of an ice-cream cone (cone from the origin capped by a sphere) in spherical coordinates?
c-spherical
Compute a $2\times2$ Jacobian and use $\lvert J\rvert$ to change variables in a double integral?
c-change-2d
Derive the spherical $\rho^2\sin\varphi$ from the $3\times3$ Jacobian?
c-change-3d
Look at a solid's bounding surfaces and name the coordinate system that makes them constant?
c-spherical
Turn four bounding curves into the new region and get the area factor without ever solving for $x(u,v)$ and $y(u,v)$?
c-region-map
Glossary (9 terms)
cylindrical coordinatessilindirik koordinatlar
Coordinates $(r,\theta,z)$ using polar $r,\theta$ in the $xy$-plane and the usual height $z$; here $dV=r\,dz\,dr\,d\theta$.
spherical coordinatesküresel koordinatlar
Coordinates $(\rho,\varphi,\theta)$ where $\rho$ is distance from the origin, $\varphi$ the angle from the $+z$-axis, and $\theta$ the azimuth; here $dV=\rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta$.
volume elementhacim elemanı
The small piece of volume $dV$ in a given coordinate system, equal to the Jacobian times the product of the coordinate differentials.
Jacobian
The determinant of the matrix of first partial derivatives of a change of variables; its absolute value is the local area or volume scaling factor.
change of variablesdeğişken değiştirme
Replacing the integration variables by new ones through a map $T$, which rewrites the integrand and multiplies the element by the absolute Jacobian.
azimut açısı
The angle $\theta$ around the $z$-axis, the same in both cylindrical and spherical coordinates, running over $[0,2\pi]$.
kutup açısı
The spherical angle $\varphi$ measured down from the $+z$-axis, running over $[0,\pi]$; it controls the $\sin\varphi$ in the volume element.
bölgenin görüntüsü
The set of points a transformation sends a region onto; because boundaries map to boundaries, it is found by mapping the bounding curves one at a time.
reciprocal ruleters çevirme kuralı
The fact that the Jacobian of a map and the Jacobian of its inverse multiply to one, so either can be obtained from the other by a single division.
What comes next
That is the last block of Calculus II. The useful thing to do now is to shuffle solids from the whole chapter together and practise naming the coordinate system in a few seconds, before writing any limits — because on the exam the marks are in that first choice.
Sources
textbookJames Stewart, Calculus, Ninth Edition — sections 15.7, 15.8 and 15.9 Section numbers are quoted only as they appear on the week 14 line of the syllabus.
standard resultStandard conventions for cylindrical and spherical coordinates The angle from the axis is written φ and runs over [0, π]; other sources may name the angles differently.