Triple integrals in Cartesian, cylindrical, spherical coordinates: Worked examples

The scenario matrix

Before we compute anything, let us list every distinct situation a triple integral can put in front of you. Each worked example below is tagged with the cell it fills.

Cell	What makes it distinct	Example
C1 Cartesian, straight walls	limits are constants or simple lines/planes	Ex 1
C2 Cartesian, order matters	picking $d z d y d x$ vs another order changes difficulty	Ex 2
CYL Cylindrical, axis symmetry	region rotates about the $z$ -axis	Ex 3
SPH Spherical, full round	boundary is $ρ =$ const, integrand has $x^{2} + y^{2} + z^{2}$	Ex 4
DEG Degenerate / zero region	limits collapse — volume is $0$ ; a sanity trap	Ex 5
LIM Limiting behaviour	let a parameter $\to 0$ or $\to \infty$ and check it makes sense	Ex 6
SIGN Negative coordinates / sign-changing integrand	region crosses an axis; $f$ goes $+$ then $-$	Ex 7
WORD Real-world word problem	density $\to$ mass or centre of mass	Ex 8
TWIST Exam-style twist	wrong-looking coordinate, or reversing the order to make it solvable	Ex 9

Every cell must be covered before you leave this page. Let's begin.

Ex 1 — Cartesian, straight walls (cell C1)

Worked example A slab with a slanted lid

Find the volume of the solid $E$ bounded by $0 \leq x \leq 2$ , $0 \leq y \leq 3$ , and $0 \leq z \leq x + 1$ .

Forecast: Guess before reading on — is the answer bigger or smaller than the flat box $2 \times 3 \times 1 = 6$ ? (The lid $z = x + 1$ rises from $1$ to $3$ , so the average height should be $2$ . Jot your guess.)

Step 1 — inner $z$ . Freeze $x, y$ ; shoot an arrow up. It enters at $z = 0$ and exits at the lid $z = x + 1$ . $\int_{0}^{x + 1} d z = x + 1$ Why this step? The innermost variable is the one whose top/bottom depend on the others; here the lid depends on $x$ , so $z$ goes inside.

Step 2 — middle $y$ . The shadow on the $x y$ -plane is the rectangle $0 \leq y \leq 3$ ; $y$ has constant limits. $\int_{0}^{3} (x + 1) d y = 3 (x + 1)$ Why this step? $x + 1$ carries no $y$ , so integrating over $y$ just multiplies by the width $3$ .

Step 3 — outer $x$ . Constant range $0 \to 2$ . $\int_{0}^{2} 3 (x + 1) d x = 3 [\frac{x ^{2}}{2} + x]_{0}^{2} = 3 (2 + 2) = 12$ Why this step? The outer variable is the last one standing, so it must run over the full $x$ -extent of the solid, here the constant interval $0 \leq x \leq 2$ . (In this box the extent happens to be constant; for a slanted region the outer range can be wider or narrower, but it is always the total spread of the outer variable across $E$ , with $y, z$ already integrated away.)

Verify: Average lid height over $x \in [0, 2]$ is $\frac{1}{2} \int_{0}^{2} (x + 1) d x = \frac{1}{2} \cdot 4 = 2$ (we divide by the interval length $2$ ). Base area $= 2 \times 3 = 6$ . Volume $= 6 \times 2 = 12$ . ✓ Matches — and it beat the flat box of $6$ , as forecast.

Ex 2 — Cartesian, order matters (cell C2)

Worked example Same solid, harder order

Recompute the tetrahedron $x, y, z \geq 0, x + y + z \leq 1$ but integrate in the order $d x d z d y$ . (The parent did $d z d y d x$ and got $\frac{1}{6}$ .)

Forecast: The answer must still be $\frac{1}{6}$ — the solid didn't change. The point is to see the limits re-derived cleanly.

Step 1 — inner $x$ . Freeze $y, z$ ; the constraint $x + y + z \leq 1$ gives $x$ from $0$ to $1 - y - z$ . $\int_{0}^{1 - y - z} d x = 1 - y - z$ Why this step? Whichever variable we put innermost, its exit face is the plane $x + y + z = 1$ solved for that variable.

Step 2 — middle $z$ . With $x$ gone, the shadow in the $y z$ -plane is $y, z \geq 0, y + z \leq 1$ . Freeze $y$ : $z$ runs $0 \to 1 - y$ . $\int_{0}^{1 - y} (1 - y - z) d z = [(1 - y) z - \frac{z ^{2}}{2}]_{0}^{1 - y} = \frac{( 1 - y ) ^{2}}{2}$ Why this step? Set $x = 0$ in the constraint to get the shadow's boundary $y + z = 1$ .

Step 3 — outer $y$ . $y : 0 \to 1$ . $\int_{0}^{1} \frac{( 1 - y ) ^{2}}{2} d y = \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6}$ Why this step? The outer variable runs over the whole $y$ -spread of the solid, which for this tetrahedron is the constant interval $0 \leq y \leq 1$ ; by now $x, z$ are already integrated away so only a number remains.

Verify: $\frac{1}{6}$ — identical to the parent's $d z d y d x$ answer. ✓ Lesson: order of integration never changes the value for a fixed solid; it only changes how ugly the limits look.

Ex 3 — Cylindrical, axis symmetry (cell CYL)

Figure — Triple integrals in Cartesian, cylindrical, spherical coordinates

Worked example Volume under a paraboloid, above the plane

Find the volume of the region $E$ inside the paraboloid $z = 4 - x^{2} - y^{2}$ and above $z = 0$ .

Forecast: The paraboloid caps a bowl of radius $2$ (set $z = 0$ : $x^{2} + y^{2} = 4$ ) and height $4$ . Guess: is it bigger or smaller than the cylinder $π \cdot 2^{2} \cdot 4 = 16 π$ that contains it? (A dome sits below its bounding cylinder.)

In the figure: the cyan curve is the paraboloid $z = 4 - r^{2}$ seen edge-on; the white line at the bottom is the floor $z = 0$ ; the amber vertical arrow at $r = 1$ shows the inner- $z$ integration path — it starts on the floor and stops on the cyan lid, exactly what "shoot an arrow up" means.

Step 1 — switch to cylindrical. Put $x = r cos θ, y = r sin θ$ . Then $x^{2} + y^{2} = r^{2}$ , so the lid is $z = 4 - r^{2}$ and $d V = r d r d θ d z$ . We will integrate in the order $d z d r d θ$ : innermost $z$ (hold $r, θ$ fixed), then $r$ (hold $θ$ fixed), then outer $θ$ . Why the factor $r$ ? A polar "pixel" is a little wedge of width $d r$ and arc length $r d θ$ , so its base area is $r d r d θ$ (not $d r d θ$ ); times the height $d z$ gives $d V = r d r d θ d z$ . Bigger $r$ ⇒ wider wedge ⇒ more volume. That $r$ is the cylindrical Jacobian. Note $d V$ groups the differentials; the order we actually integrate in is our free choice, stated above. Why cylindrical at all? The region is round about the $z$ -axis; every horizontal slice is a disk. Cylindrical turns the ugly $x^{2} + y^{2}$ into a single $r^{2}$ .

Step 2 — inner $z$ . Hold $r, θ$ fixed and follow the amber arrow: it enters at $z = 0$ , exits at $z = 4 - r^{2}$ . $\int_{0}^{4 - r^{2}} r d z = r (4 - r^{2})$ Why this step? With $r, θ$ frozen, the height at that spot is exactly $4 - r^{2}$ . Keep the $r$ from $d V$ along for the ride.

Step 3 — middle $r$ . Now hold $θ$ fixed; the shadow disk has radius $2$ , so $r : 0 \to 2$ . $\int_{0}^{2} r (4 - r^{2}) d r = \int_{0}^{2} (4 r - r^{3}) d r = [2 r^{2} - \frac{r ^{4}}{4}]_{0}^{2} = 8 - 4 = 4$ Why this step? $r$ ranges over the projected shadow, which is the disk of radius $2$ .

Step 4 — outer $θ$ . Full turn: $θ : 0 \to 2 π$ . $\int_{0}^{2 π} 4 d θ = 8 π$ Why this step? Nothing left depends on $θ$ , and the region wraps all the way around the axis, so $θ$ sweeps a full $2 π$ .

Verify: $8 π \approx 25.1$ . It is exactly half the bounding cylinder $16 π$ — a paraboloid dome fills half its box, a known fact. ✓ Smaller than the cylinder, as forecast.

Ex 4 — Spherical, full round (cell SPH)

Worked example Mass with radial density

A ball of radius $a$ has density $δ (x, y, z) = x^{2} + y^{2} + z^{2}$ (denser near the surface). Find its total mass. (We write the density as $δ$ , not $ρ$ , to keep it clearly separate from the spherical radial coordinate $ρ$ .)

Forecast: Density grows outward, so mass should be more than "uniform density $\times$ volume" if we used the average density. Just guess the form: it will look like (constant) $\times a^{5}$ . Why $a^{5}$ ? Density has units of length $^{2}$ , volume length $^{3}$ — product length $^{5}$ . Note that guess.

In the figure: the cyan wireframe is the sphere $ρ = a$ ; the amber radial spoke is one value of $ρ$ pointing out to the surface; $ϕ$ is measured down from the $+ z$ axis and $θ$ swings around it — the three coordinates that build every point of the ball.

Step 1 — spherical everything. Distance from the origin is the radial coordinate $ρ$ , so $x^{2} + y^{2} + z^{2} = ρ^{2}$ and hence the density is $δ = ρ^{2}$ . The boundary is the single equation $ρ = a$ , and $d V = ρ^{2} sin ϕ d ρ d ϕ d θ$ . Why the factor $ρ^{2} sin ϕ$ ? The little spherical box has three perpendicular edges: the radial edge $d ρ$ , the meridian arc $ρ d ϕ$ , and the latitude-circle arc of radius $ρ sin ϕ$ giving length $ρ sin ϕ d θ$ . Multiply the three lengths: $d ρ \cdot (ρ d ϕ) \cdot (ρ sin ϕ d θ) = ρ^{2} sin ϕ d ρ d ϕ d θ$ . The $sin ϕ$ shrinks to $0$ at the poles because latitude circles collapse to points there. That product is the spherical Jacobian. $Mass = \int_{0}^{2 π} \int_{0}^{π} \int_{0}^{a} density δ ρ^{2} \cdot d V ρ^{2} sin ϕ d ρ d ϕ d θ$ Why spherical at all? Both the integrand ( $x^{2} + y^{2} + z^{2}$ ) and the region (a ball) are round in all directions — the textbook flag for spherical.

Step 2 — factor the integral. Limits are all constants and the integrand splits into a $ρ$ -part and a $ϕ$ -part: $= 2 π \int_{0}^{2 π} d θ \cdot 2 \int_{0}^{π} sin ϕ d ϕ \cdot a^{5} /5 \int_{0}^{a} ρ^{4} d ρ$ Why this step? Fubini lets us multiply three 1D integrals when everything separates — the fastest route.

Step 3 — multiply. $Mass = 2 π \cdot 2 \cdot \frac{a ^{5}}{5} = \frac{4 π a ^{5}}{5}$

Verify: Units check — density (length $^{2}$ ) $\times$ volume (length $^{3}$ ) = length $^{5}$ : matches $a^{5}$ . ✓ Compare to a uniform ball with the boundary density $a^{2}$ : that would give $a^{2} \cdot \frac{4}{3} π a^{3} = \frac{4}{3} π a^{5} = \frac{20}{15} π a^{5}$ , and our answer $\frac{4}{5} π a^{5} = \frac{12}{15} π a^{5}$ is smaller — correct, since density is smaller than $a^{2}$ everywhere inside. ✓

Related to Center of mass and moments of inertia and Spherical coordinates geometry.

Ex 5 — Degenerate / zero region (cell DEG)

Worked example The trap where the answer is exactly zero

"Find the volume of the solid $E$ given by $0 \leq z \leq 0$ , $0 \leq x \leq 1$ , $0 \leq y \leq 1$ ." What is it?

Forecast: The $z$ -limits are both $0$ . Guess: is this a real solid, or a flat square with no thickness?

Step 1 — inner $z$ . Arrow up from $z = 0$ to $z = 0$ — it never leaves the floor. $\int_{0}^{0} d z = 0$ Why this step? When the lower and upper limits of any integral coincide, that integral is $0$ regardless of everything outside it.

Step 2 — the rest doesn't matter. Once one factor is $0$ , the whole product is $0$ : $\int_{0}^{1} \int_{0}^{1} 0 d y d x = 0$ Why this step? Multiplying $0$ by the finite base area $1 \times 1$ still gives $0$ .

Verify: A flat square has no thickness, so it encloses zero volume. ✓ This is the sanity check you run whenever your limits collapse: if $z_{1} (x, y) = z_{2} (x, y)$ on the whole region, stop — the answer is $0$ , and any nonzero result means an algebra slip.

Ex 6 — Limiting behaviour (cell LIM)

Worked example A cone whose tip angle shrinks

Find the volume of the ice-cream cone $0 \leq ρ \leq a$ , $0 \leq ϕ \leq α$ (all $θ$ ), where $α$ is the half-angle from the north pole. Then take the limits $α \to 0$ and $α \to π$ .

Forecast: As $α \to 0$ the cone becomes a needle — volume should $\to 0$ . As $α \to π$ it swallows the whole ball — volume should $\to \frac{4}{3} π a^{3}$ . Predict the formula that does both.

In the figure: the cyan arc is the spherical cap $ρ = a$ ; the two white lines from the origin are the cone's slanted sides; the amber vertical line is the $+ z$ axis, and the half-angle $α$ is the opening measured from that axis down to the cone wall.

Step 1 — set up in spherical. The wedge in the figure has $ϕ$ swept only from $0$ to $α$ . Recall from Ex 4 that $d V = ρ^{2} sin ϕ d ρ d ϕ d θ$ (radial edge $d ρ$ , meridian arc $ρ d ϕ$ , latitude arc $ρ sin ϕ d θ$ ): $V (α) = \int_{0}^{2 π} \int_{0}^{α} \int_{0}^{a} ρ^{2} sin ϕ d ρ d ϕ d θ$ Why this step? A cone about the $z$ -axis is bounded by a fixed polar angle $ϕ = α$ — spherical describes it with one constant, nothing else can.

Step 2 — factor and integrate. $= 2 π \int_{0}^{2 π} d θ \cdot 1 - c o s α \int_{0}^{α} sin ϕ d ϕ \cdot a^{3} /3 \int_{0}^{a} ρ^{2} d ρ$ Why this step? $\int_{0}^{α} sin ϕ d ϕ = [- cos ϕ]_{0}^{α} = 1 - cos α$ — the only piece that depends on $α$ .

Step 3 — assemble. $V (α) = \frac{2 π a ^{3}}{3} (1 - cos α)$ Why this step? Multiply the three factors from Step 2 in any order (they're just numbers now): $2 π \cdot (1 - cos α) \cdot \frac{a ^{3}}{3}$ . Collecting them gives one tidy formula in the single parameter $α$ , which is exactly what we need to take limits in.

Verify (the whole point):

$α \to 0$ : $cos α \to 1$ , so $V \to 0$ . Needle has no volume. ✓
$α = \frac{π}{2}$ (hemisphere): $cos \frac{π}{2} = 0$ , $V = \frac{2 π a ^{3}}{3} = \frac{1}{2} \cdot \frac{4}{3} π a^{3}$ — exactly half a ball. ✓
$α \to π$ : $cos π = - 1$ , $V \to \frac{2 π a ^{3}}{3} \cdot 2 = \frac{4}{3} π a^{3}$ — the full ball. ✓

One clean formula passes all three limits — that is how you know it's right.

Ex 7 — Negative coordinates & a sign-changing integrand (cell SIGN)

Worked example When the integrand goes positive then negative

Evaluate $I = ∭_{E} x d V$ over the box $E : - 1 \leq x \leq 2, 0 \leq y \leq 1, 0 \leq z \leq 1$ . The region crosses $x = 0$ , and the integrand $x$ is negative for $x < 0$ , positive for $x > 0$ .

Forecast: For $x < 0$ the term $x d V$ is negative, for $x > 0$ it is positive. The box reaches farther into positive $x$ (out to $2$ ) than negative $x$ (only to $- 1$ ), so guess: net result positive.

Step 1 — separate the integral. Limits are all constants and $f = x$ carries no $y, z$ , so it factors: $I = (\int_{- 1}^{2} x d x) (\int_{0}^{1} d y) (\int_{0}^{1} d z)$ Why this step? When the box has constant limits and $f$ depends on only one variable, Fubini turns the triple integral into a product of 1D integrals.

Step 2 — the $x$ -integral, watching the sign. Split at $x = 0$ so you see the cancellation: $\int_{- 1}^{2} x d x = - 1/2 (negative part) \int_{- 1}^{0} x d x + + 2 (positive part) \int_{0}^{2} x d x = - \frac{1}{2} + 2 = \frac{3}{2}$ Why this step? Splitting at the sign change makes the physics explicit: the small negative slab ( $- \frac{1}{2}$ ) is outweighed by the larger positive slab ( $+ 2$ ). (Directly, $[\frac{x ^{2}}{2}]_{- 1}^{2} = \frac{4}{2} - \frac{1}{2} = \frac{3}{2}$ .)

Step 3 — multiply by the trivial $y, z$ factors. $I = \frac{3}{2} \cdot 1 \cdot 1 = \frac{3}{2}$ Why this step? Each of $\int_{0}^{1} d y$ and $\int_{0}^{1} d z$ equals $1$ , so they don't change the value — they only confirm the cross-section is a unit square.

Verify: Positive, as forecast. Sanity check via the mean: the average of $x$ over $[- 1, 2]$ is the midpoint $\frac{- 1 + 2}{2} = \frac{1}{2}$ ; times the interval length $3$ gives $\frac{1}{2} \cdot 3 = \frac{3}{2}$ . ✓ Lesson: a sign-changing integrand does not mean split into pieces is required — but splitting shows you why the answer is not zero. If the box were symmetric ( $- 1 \leq x \leq 1$ ), the two halves would cancel and $I = 0$ .

Ex 8 — Real-world word problem (cell WORD)

Worked example Where is the balance point of a cone?

A solid cone of uniform density sits with its tip at the origin, opening upward: it is the region $E$ with $0 \leq z \leq h$ , and at height $z$ its radius is $\frac{R}{h} z$ (radius grows linearly to $R$ at the top). Find the $z$ -coordinate of its centre of mass, $\overset{z}{ˉ} = \frac{∭ _{E} z d V}{∭ _{E} d V}$ .

Forecast: More material sits near the wide top than the narrow tip, so the balance point should be above the midpoint $h /2$ . Guess a fraction of $h$ .

Step 1 — describe in cylindrical. At height $z$ the cross-section is a disk of radius $\frac{R}{h} z$ , so $r$ runs $0 \to \frac{R}{h} z$ , $θ$ full turn, $z : 0 \to h$ . Recall $d V = r d r d θ d z$ — the extra $r$ is the polar-wedge area factor. We integrate in the order $d r d z d θ$ : innermost $r$ (hold $z, θ$ fixed), then $z$ , then $θ$ . Why this step? The cone is round about the $z$ -axis; slicing by height makes each slice a disk whose radius is a known function of $z$ .

Step 2 — denominator (volume). $V = \int_{0}^{2 π} \int_{0}^{h} \int_{0}^{R z / h} r d r d z d θ = 2 π \int_{0}^{h} \frac{1}{2} (\frac{R z}{h})^{2} d z = π \frac{R ^{2}}{h ^{2}} \cdot \frac{h ^{3}}{3} = \frac{π R ^{2} h}{3}$ Why this step? The famous cone volume $\frac{1}{3} π R^{2} h$ appears — a built-in check that our limits are right.

Step 3 — numerator ( $∭ z d V$ ). Insert the extra $z$ : $\int_{0}^{2 π} \int_{0}^{h} \int_{0}^{R z / h} z r d r d z d θ = 2 π \int_{0}^{h} z \cdot \frac{1}{2} (\frac{R z}{h})^{2} d z = π \frac{R ^{2}}{h ^{2}} \int_{0}^{h} z^{3} d z = π \frac{R ^{2}}{h ^{2}} \cdot \frac{h ^{4}}{4} = \frac{π R ^{2} h ^{2}}{4}$ Why this step? The moment about $z = 0$ weights each slice by its height $z$ ; the $z^{3}$ (one $z$ from the moment, $z^{2}$ from the growing radius) is what tilts the balance upward.

Step 4 — divide. $\overset{z}{ˉ} = \frac{π R ^{2} h ^{2} /4}{π R ^{2} h /3} = \frac{3 h}{4}$ Why this step? The centre of mass is the moment divided by the total mass (here uniform density cancels top and bottom), leaving a pure geometric ratio.

Verify: $\overset{z}{ˉ} = \frac{3}{4} h$ , which is above the midpoint $\frac{1}{2} h$ — exactly as forecast, because the wide end at the top holds more mass. Standard textbook result for a solid cone: centre of mass is $\frac{3}{4}$ of the way from tip to base. ✓ Units: $\overset{z}{ˉ}$ has units of length. ✓

Related to Center of mass and moments of inertia.

Ex 9 — Exam-style twist (cell TWIST)

Worked example The integral that only works after you reverse it

Evaluate $I = \int_{0}^{1} \int_{x}^{1} \int_{0}^{1 - y} 1 d z d y d x .$ As written it's fine, but the twist is: what solid is this, and can we confirm the answer geometrically by reversing the order?

Forecast: This is a pure volume ( $f = 1$ ). The $y$ -limits $x \leq y \leq 1$ and $z$ -limit $z \leq 1 - y$ describe some curved-based prism. Guess whether the answer is a "nice" fraction.

Step 1 — inner $z$ . Hold $x, y$ fixed. $\int_{0}^{1 - y} d z = 1 - y$ Why this step? Freeze $x, y$ ; the roof is $z = 1 - y$ , floor $z = 0$ .

Step 2 — middle $y$ , from $x$ to $1$ . $\int_{x}^{1} (1 - y) d y = [y - \frac{y ^{2}}{2}]_{x}^{1} = (1 - \frac{1}{2}) - (x - \frac{x}{2}) = \frac{1}{2} - x + \frac{x}{2}$ Why this step? The lower curve $y = x$ is what makes the base region curved — it's the parabola $x = y^{2}$ seen sideways.

Step 3 — outer $x : 0 \to 1$ . Now integrate term by term, using $\int_{0}^{1} x d x = \frac{2}{3}$ and $\int_{0}^{1} x d x = \frac{1}{2}$ : $I = \int_{0}^{1} (\frac{1}{2} - x + \frac{x}{2}) d x = \frac{1}{2} - \frac{2}{3} + \frac{1}{4} = \frac{6 - 8 + 3}{12} = \frac{1}{12}$ Why this step? The outer variable runs over the whole $x$ -spread of the base region, here $0 \leq x \leq 1$ ; with $y, z$ already gone, only a number remains, so we simply integrate the three terms.

Verify — reverse the order. The base region in the $x y$ -plane is ${0 \leq x \leq y^{2}, 0 \leq y \leq 1}$ (from $x \leq y^{2}$ , i.e. $y \geq x$ ). Redo with $x$ inside: $I = \int_{0}^{1} \int_{0}^{y^{2}} \int_{0}^{1 - y} d z d x d y = \int_{0}^{1} y^{2} (1 - y) d y = \frac{1}{3} - \frac{1}{4} = \frac{1}{12} .$ Both orders give $\frac{1}{12}$ . ✓ Lesson: when limits look awkward, redrawing the base region and swapping the order is a full, independent check — and here it also reveals the twist: the solid has a parabolic base $x = y^{2}$ , invisible until you flip the order.

Coverage check

Recall Did we hit every cell of the matrix?

C1 → Ex 1 · C2 → Ex 2 · CYL → Ex 3 · SPH → Ex 4 · DEG → Ex 5 · LIM → Ex 6 · SIGN → Ex 7 · WORD → Ex 8 · TWIST → Ex 9. Every scenario class is covered.

Recall Quick self-test

Volume under $z = 4 - r^{2}$ over its shadow disk ::: $8 π$ Mass of ball radius $a$ with density $x^{2} + y^{2} + z^{2}$ ::: $\frac{4 π a ^{5}}{5}$ Cone wedge volume $V (α)$ for half-angle $α$ ::: $\frac{2 π a ^{3}}{3} (1 - cos α)$ $∭ x d V$ over the box $- 1 \leq x \leq 2, 0 \leq y, z \leq 1$ ::: $\frac{3}{2}$ Centre of mass height of a solid cone ::: $\frac{3}{4} h$ above the tip Value of the reversed-order integral $I$ ::: $\frac{1}{12}$

Connections

Triple integrals in Cartesian, cylindrical, spherical coordinates — the parent; this page is its full worked-example companion.
Double integrals & polar coordinates — Ex 3 and Ex 8 reuse $r d r d θ$ from the 2D case.
Jacobian and change of variables — every $d V$ factor ( $r$ , $ρ^{2} sin ϕ$ ) is a Jacobian in disguise.
Spherical coordinates geometry — Ex 4 and Ex 6 lean on the $ρ, ϕ, θ$ picture.
Center of mass and moments of inertia — Ex 8 is a direct centre-of-mass computation.
Divergence theorem — later, these volume integrals become the "inside" of flux calculations.