Chapter 12 · Triple integrals: Cartesian, cylindrical, spherical

The hook

One ball, three coordinate systems

A CT scanner does not photograph your insides — it reconstructs them. It fires X-rays through your body from every angle, measures how much each ray is absorbed, and from those shadows rebuilds the density at every point. The total mass it recovers is one number: $\displaystyle\iiint_T \delta\,dV$ , the density $\delta$ summed over the solid region $T$ . That triple integral is the machine this chapter builds — and the first lesson is that the coordinate system you choose can turn a page of algebra into three lines.

Here is the cleanest possible demonstration. The volume of a ball of radius $a$ is $\tfrac{4}{3}\pi a^3$ — the formula everyone memorises and almost no one can derive. Watch it fall out of spherical coordinates, then watch Cartesian drown in square roots.

Same solid. Left: spherical coordinates slice it into onion shells of constant

\rho

— three clean integrals. Right: Cartesian coordinates chop it into slabs whose every limit is a nested square root

\pm\sqrt{a^2-x^2-y^2}

— a page of work for the same number.

In spherical coordinates the ball is just $0 \le \rho \le a$ , $0 \le \phi \le \pi$ , $0 \le \theta \le 2\pi$ , the volume element is $dV = \rho^2\sin\phi\,d\rho\,d\phi\,d\theta$ , and the whole computation is:

\int_0^{2\pi}\!\!\int_0^{\pi}\!\!\int_0^{a}\rho^2\sin\phi\,d\rho\,d\phi\,d\theta = 2\pi\cdot 2\cdot \frac{a^3}{3} = \frac{4}{3}\pi a^3.

In Cartesian coordinates the very same volume is

\int_{-a}^{a}\!\int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\! \int_{-\sqrt{a^2-x^2-y^2}}^{\sqrt{a^2-x^2-y^2}} 1\,dz\,dy\,dx,

which is correct, doable, and miserable — three trigonometric substitutions deep. The mathematics is identical; only the coordinates differ. Choosing the coordinate system is the skill this chapter trains.

Why this matters

Single integrals measure along a line, double integrals over a flat region. The triple integral is how you total anything spread through a three-dimensional body: its volume, its mass, its centre of mass, the heat stored in it, the dose a tumour absorbs. Every one of those is $\iiint_T (\text{something})\,dV$ . The hard part is almost never the calculus — it is setting up the limits, and picking the coordinate system in which the limits are simple. Get that habit and the integrals do themselves.

The method that always works

The 3-D ritual, and Cartesian triple integrals

Chapter 10 gave you a ritual for double integrals: sketch → describe the region → read off the limits → integrate. Triple integrals extend it by exactly one move. Here is the full ritual, and you never skip it:

The 3-D ritual

Sketch the solid $T$ (roughly — you are not drawing for an art class).
Shadow. Project $T$ straight down onto a coordinate plane (usually $xy$ ). That silhouette is a 2-D region — and it supplies the outer double integral's limits.
Slice. Over each point of the shadow, run a vertical line up through the solid. It enters through a floor surface and leaves through a roof surface — those two surfaces are the inner $z$ -limits.
Limits, then integrate from the inside out: $z$ first (floor to roof), then the shadow's double integral.

One honest sentence before you panic about "seeing" a solid in your head: nobody pictures a 3-D region whole. Every competent person projects it to a flat shadow and slices it — that is what the ritual is. "I can't visualise in 3-D" names a thing literally no one does; it is not a deficit, it is the universal condition, and shadow-and-slice is the workaround everyone uses.

The Cartesian triple integral, by shadow and slice

\iiint_T f\,dV = \iint_{\text{shadow}}\!\left[\int_{\text{floor}(x,y)}^{\text{roof}(x,y)} f(x,y,z)\,dz\right]dA.

Worked example · WE-1 · pure setup, the tetrahedron

Find the volume of the tetrahedron $T$ : $x, y, z \ge 0$ and $x + y + z \le 1$ .

Shadow. Set $z = 0$ : the solid's footprint on the $xy$ -plane is the triangle $0 \le x \le 1$ , $0 \le y \le 1 - x$ . Slice. Above a point $(x, y)$ of that triangle, $z$ runs from the floor $z = 0$ up to the roof $z = 1 - x - y$ (the slanted plane). So

V = \int_0^1\!\!\int_0^{1-x}\!\!\int_0^{1-x-y} 1\,dz\,dy\,dx = \int_0^1\!\!\int_0^{1-x}(1-x-y)\,dy\,dx = \int_0^1 \frac{(1-x)^2}{2}\,dx = \frac{1}{6}.

The whole skill was the setup; the integration was three easy steps. Notice the structure of the limits — outer constant, each inner limit depending only on the variables further out. That nesting is the single most-tested correctness check in the chapter.

Worked example · WE-2 · the integrand rides along

Over the same tetrahedron, evaluate $\displaystyle\iiint_T x\,dV$ (think of $x$ as a density that grows away from the $yz$ -plane).

\iiint_T x\,dV = \int_0^1\!\!\int_0^{1-x}\!\!\int_0^{1-x-y} x\,dz\,dy\,dx = \int_0^1 x\,\frac{(1-x)^2}{2}\,dx = \frac{1}{24}.

The point worth carrying: the integrand changed nothing about the limits. Volume, mass, a moment — same region, same shadow, same slice; only the function inside changes. Set the region up once and every quantity over it is within reach.

When there is an axis of symmetry

Cylindrical coordinates: polar, with a z-axis

Cylindrical coordinates are nothing new — they are Chapter 10's polar coordinates with the $z$ -axis carried along unchanged:

x = r\cos\theta, \qquad y = r\sin\theta, \qquad z = z, \qquad \boxed{\,dV = r\,dr\,d\theta\,dz\,}.

That $r$ in the volume element is exactly the polar $r$ you were told in Chapter 10 never to drop — and Chapter 11 named it for what it is: the Jacobian of the polar map. Same factor, same reason, now with a height. (We do not re-derive it here; that was Chapter 11's whole job.)

When to reach for cylindrical

Three tells, any one of which is enough: an axis of rotational symmetry; a circular (or annular) shadow in the $xy$ -plane; or an integrand built from $x^2 + y^2$ (which becomes a clean $r^2$ ). When you see a cylinder, a cone, a paraboloid, or "rotated about the $z$ -axis," reach for $r\,dr\,d\theta\,dz$ .

Worked example · WE-3 · the load-bearing r

Find the volume of the solid above the paraboloid $z = x^2 + y^2$ and below the plane $z = 4$ .

Shadow. The two surfaces meet where $x^2 + y^2 = 4$ , so the footprint is the disk $r \le 2$ . Slice. Above a point at radius $r$ , the solid runs from the floor $z = r^2$ (the paraboloid, since $x^2+y^2 = r^2$ ) up to the roof $z = 4$ . With $dV = r\,dr\,d\theta\,dz$ :

V = \int_0^{2\pi}\!\!\int_0^{2}\!\!\int_{r^2}^{4} r\,dz\,dr\,d\theta = \int_0^{2\pi}\!\!\int_0^{2} r(4 - r^2)\,dr\,d\theta = 2\pi\Big[2r^2 - \tfrac{r^4}{4}\Big]_0^2 = 2\pi\cdot 4 = 8\pi.

Watch the $r$ . Drop it from the element — integrate $(4 - r^2)$ alone — and you get $\tfrac{32\pi}{3} \approx 33.5$ instead of $8\pi \approx 25.1$ : a wrong answer that looks perfectly reasonable. There is no sign it is wrong except that you skipped the Jacobian. This exact slip returns, dressed up, in the gauntlet.

End of sitting 1 — you can now run the shadow-and-slice ritual in Cartesian and in cylindrical coordinates, and you know the cylindrical $r$ is the polar Jacobian, never optional. After the break: spherical coordinates and their famous $\sin\phi$ ; then the Coordinate Lab to drill the one decision that matters — which system — and the exam-shaped practice set.

When everything radiates from the origin

Spherical coordinates, and the $\sin\phi$ everyone drops

Spherical coordinates pin a point by its distance from the origin and two angles:

Spherical coordinates (φ measured from the +z-axis)

x = \rho\sin\phi\cos\theta, \quad y = \rho\sin\phi\sin\theta, \quad z = \rho\cos\phi, \qquad \boxed{\,dV = \rho^2\sin\phi\,d\rho\,d\phi\,d\theta\,}.

$\rho \ge 0$ is the straight-line distance from the origin ( $0 \to \infty$ ).
$\phi$ is the angle down from the $+z$ -axis (the colatitude), $0 \to \pi$ : $\phi = 0$ points straight up, $\phi = \pi$ straight down, $\phi = \tfrac{\pi}{2}$ is the equator.
$\theta$ is the same azimuth as before, going around, $0 \to 2\pi$ . Note $r = \rho\sin\phi$ ties spherical back to cylindrical.

Notation convention — check the angle names

In this chapter, $\theta$ is the azimuth in the $xy$ -plane with $0\le\theta\le2\pi$ , while $\phi$ is measured down from the positive $z$ -axis with $0\le\phi\le\pi$ . Some books interchange the names $\theta$ and $\phi$ . That is only a relabelling: keep the stated ranges and attach $\sin(\text{polar angle})$ to the angle measured from the $z$ -axis.

The headline mark-loser: the dropped

\sin\phi

Why is the element $\rho^2\sin\phi$ and not just $\rho^2$ ? Picture the little box cut out by nudging each coordinate: its sides are $d\rho$ (outward), $\rho\,d\phi$ (north–south), and $\rho\sin\phi\,d\theta$ (east–west). That last side is short near the poles and longest at the equator, because the circles of constant latitude shrink to a point at the poles — and $\sin\phi$ is exactly that shrink factor. Multiply the three sides: $d\rho \cdot \rho\,d\phi \cdot \rho\sin\phi\,d\theta = \rho^2\sin\phi\,d\rho\,d\phi\,d\theta$ . Forget the $\sin\phi$ and you are pretending the Earth's longitude lines stay equally far apart all the way to the pole. This is the three-dimensional cousin of Chapter 11's dropped $r$ , and the single most common lost mark in the chapter.

When to reach for spherical

Reach for it when the geometry radiates from the origin: spheres, balls, cones with their apex at the origin, regions between two concentric spheres, or an integrand built from $x^2 + y^2 + z^2$ (which becomes a clean $\rho^2$ ). If the boundary is " $\rho =$ constant," spherical makes the outer limit a single number.

Worked example · WE-4 · the sin φ does visible work

Evaluate $\displaystyle\iiint_B (x^2 + y^2 + z^2)\,dV$ over the unit ball $B$ : $\rho \le 1$ .

The integrand is $x^2 + y^2 + z^2 = \rho^2$ , and $dV = \rho^2\sin\phi\,d\rho\,d\phi\, d\theta$ , so the whole thing is $\rho^2 \cdot \rho^2\sin\phi = \rho^4\sin\phi$ :

\int_0^{2\pi}\!\!\int_0^{\pi}\!\!\int_0^{1}\rho^4\sin\phi\,d\rho\,d\phi\,d\theta = \underbrace{\int_0^{2\pi}\!d\theta}_{2\pi}\cdot \underbrace{\int_0^{\pi}\!\sin\phi\,d\phi}_{2}\cdot \underbrace{\int_0^{1}\!\rho^4\,d\rho}_{1/5} = 2\pi\cdot 2\cdot\tfrac15 = \frac{4\pi}{5}.

The $\int_0^\pi \sin\phi\,d\phi = 2$ is unmissable here. Drop the $\sin\phi$ and you would compute $2\pi\cdot\pi\cdot\tfrac15 = \tfrac{2\pi^2}{5} \approx 3.95$ instead of $\tfrac{4\pi}{5}\approx 2.51$ — wrong, and again with nothing to flag it but the missing factor.

Worked example · WE-5 · application: the centroid of a hemisphere

Where is the centre of mass of a uniform solid hemisphere of radius $a$ ( $z \ge 0$ , $\rho \le a$ )? By symmetry $\bar x = \bar y = 0$ ; only $\bar z$ is in question. We need the volume and the $z$ -moment.

Volume (a hemisphere, $\phi$ only $0 \to \tfrac{\pi}{2}$ ):

V = \int_0^{2\pi}\!\!\int_0^{\pi/2}\!\!\int_0^{a}\rho^2\sin\phi\,d\rho\,d\phi\,d\theta = 2\pi\cdot 1\cdot\frac{a^3}{3} = \frac{2}{3}\pi a^3.

$z$ -moment (with $z = \rho\cos\phi$ , the integrand is $\rho\cos\phi\cdot\rho^2\sin\phi = \rho^3\cos\phi\sin\phi$ ):

\iiint z\,dV = \int_0^{2\pi}\!\!\int_0^{\pi/2}\!\!\int_0^{a}\rho^3\cos\phi\sin\phi\, d\rho\,d\phi\,d\theta = 2\pi\cdot\underbrace{\tfrac12}_{\int_0^{\pi/2}\cos\phi\sin\phi\,d\phi}\cdot \frac{a^4}{4} = \frac{\pi a^4}{4}.

\bar z = \frac{\iiint z\,dV}{V} = \frac{\pi a^4/4}{\tfrac23\pi a^3} = \frac{3a}{8}.

The centre of mass sits at $\tfrac{3a}{8}$ up the axis — not halfway, because there is more material near the flat base than near the dome. $\tfrac{3a}{8}$ is a classic, memorable result, and a favourite exam number.

Interactive playground

The Coordinate Lab

The skill this chapter trains is a decision: which coordinate system, and therefore which volume element. So commit to both before the Lab tells you anything. Pick a solid, choose a coordinate system, choose its volume element $dV$ — and only then does the Lab reveal whether your system was the clean one and whether your element was right. A wrong element that names the exact factor you dropped is the whole point; recognising the right setup after being shown it is not the same skill as choosing it.

The Coordinate Lab

Solid:

the solid

Step 1 · commit a coordinate system

Step 2 · commit the volume element dV

Two of the five element chips are booby-traps — the dropped $r$ and the dropped $\sin\phi$ . They are never the right element, in any system. The set-up integral and its value appear only when both your commitments are correct.

What to notice

The solid's symmetry chooses the system. Origin-centred and round → spherical; an axis with circular cross-sections → cylindrical; flat faces at right angles → Cartesian.
A correct element in the wrong system still "works" — and is a nightmare. You can integrate the ball in Cartesian with $dx\,dy\,dz$ ; the Lab grants it is correct, then points at the three clean lines spherical would have given.
The traps are the marks you lose. Choosing cylindrical and " $dr\, d\theta\,dz$ " drops the $r$ ; choosing spherical and " $\rho^2\,d\rho\,d\phi\,d\theta$ " drops the $\sin\phi$ . Neither is ever the volume element.

Retrieval gauntlet

Commit to an answer before reading the feedback. These interleave Cartesian, cylindrical and spherical on purpose — knowing which system a solid wants is half the skill.

Quiz 1 · Cartesian setup — which integral is the tetrahedron?

Which iterated integral correctly gives the volume of the tetrahedron $x, y, z \ge 0$ , $x + y + z \le 1$ ?

Quiz 2 · cylindrical, and the r

The solid above the paraboloid $z = x^2 + y^2$ and below $z = 4$ has volume:

Quiz 3 · spherical, and the sin φ

The "ice-cream" solid inside the sphere $\rho = 1$ and inside the cone $\phi \le \tfrac{\pi}{3}$ has volume:

Predict the next step. You have the ice-cream wedge sketched and you know the element is

\rho^2\sin\phi

. Before revealing: what are the three limits, and which integral produces the

\tfrac12

Limits. Inside the sphere $\rho = 1$ : $0 \le \rho \le 1$ . Inside the cone measured from the axis: $0 \le \phi \le \tfrac{\pi}{3}$ . All the way around: $0 \le \theta \le 2\pi$ . So

V = \int_0^{2\pi}\!\!\int_0^{\pi/3}\!\!\int_0^{1}\rho^2\sin\phi\,d\rho\,d\phi\,d\theta = 2\pi\cdot\Big[-\cos\phi\Big]_0^{\pi/3}\cdot\frac13 = 2\pi\cdot\big(1 - \tfrac12\big)\cdot\frac13 = \frac{\pi}{3}.

The $\tfrac12$ is the $\phi$ -integral: $\int_0^{\pi/3}\sin\phi\,d\phi = 1 - \cos\tfrac{\pi}{3} = 1 - \tfrac12 = \tfrac12$ . Use $\sin$ of the angle instead of this, and you land on the $\tfrac{\sqrt3\,\pi}{3}$ distractor.

Quiz 4 · choose the system — do not integrate

Match each solid to the coordinate system with the cleanest limits: (i) the ball $x^2+y^2+z^2 \le 9$ ; (ii) the cylinder $x^2+y^2 \le 1,\ 0 \le z \le 5$ ; (iii) the box $0 \le x \le 2,\ 0 \le y \le 3,\ 0 \le z \le 1$ .

Quiz 5 · spherical, integrand and element both convert

Converting $\displaystyle\iiint_B \sqrt{x^2+y^2+z^2}\,dV$ over the unit ball $B$ to spherical coordinates and evaluating gives:

Predict the next step. The trap in Q5 is the power of

\rho

. Before revealing: what does

\sqrt{x^2+y^2+z^2}

become, and what power of

\rho

does the integrand carry once you include the element?

Integrand. $\sqrt{x^2+y^2+z^2} = \sqrt{\rho^2} = \rho$ — the first power. Multiply by the element $\rho^2\sin\phi$ : the integrand is $\rho\cdot\rho^2\sin\phi = \rho^3\sin\phi$ . Then

\int_0^{2\pi}\!\!\int_0^{\pi}\!\!\int_0^{1}\rho^3\sin\phi\,d\rho\,d\phi\,d\theta = 2\pi\cdot 2\cdot\frac14 = \pi.

Two separate $\rho$ 's are in play: one from the integrand ( $\rho^1$ ) and two from the element ( $\rho^2$ ). Lose track of either — call the root $\rho^2$ , or drop the element's $\rho^2$ — and you land on a wrong-but-plausible $\tfrac{4\pi}{5}$ or $2\pi$ .

Exam translator

What the exam will actually ask

Exam shapes

In standard Indian engineering-math exams, and likely in MCC201A unless your class notes differ, triple integrals wear four costumes — each one you have now seen worked:

"Evaluate $\iiint_T f\,dV$ over ⟨solid⟩" → run the ritual: sketch, shadow, slice (WE-1, WE-2).
"Find the volume / mass / centroid" → volume $\iiint 1\,dV$ ; mass $\iiint \delta\,dV$ ; centroid $\bar z = \dfrac{\iiint z\,\delta\,dV}{\text{mass}}$ (WE-5).
"Evaluate using cylindrical / spherical coordinates" → match the element. The dropped $r$ and dropped $\sin\phi$ are the classic lost marks (WE-3, WE-4).
"Change to a convenient coordinate system" → the question is the decision: the system whose symmetry matches the solid (Beat 1 thesis, Q4).

Worked solutions are complete; nothing points off this page. Do them as listed — interleaved on purpose — not blocked by method. Two of them, PR5 and PR7, open with a faded bridge: part of the work is set out, the key step left for you to predict.

PR1 · Cartesian · a tetrahedron

Shape · coordinate-plane tetrahedron ⇒ Cartesian shadow + slice

PR1. Find the volume of the tetrahedron bounded by the coordinate planes and the plane $x + 2y + z = 4$ .

Shadow. Set $z = 0$ : the footprint is $x + 2y \le 4$ , $x, y \ge 0$ , so $0 \le x \le 4$ , $0 \le y \le \tfrac{4-x}{2}$ . Slice. $z$ from $0$ to the roof $z = 4 - x - 2y$ :

V = \int_0^4\!\!\int_0^{(4-x)/2}\!\!\int_0^{4-x-2y} dz\,dy\,dx = \int_0^4 \frac{(4-x)^2}{4}\,dx = \frac{16}{3}.

Quick check: the plane has intercepts $4, 2, 4$ , and the coordinate-plane tetrahedron of intercepts $a, b, c$ has volume $\tfrac{abc}{6} = \tfrac{4\cdot 2\cdot 4}{6} = \tfrac{16}{3}$ . ✓

PR2 · cylindrical · under a paraboloid

Shape · circular shadow + axis symmetry ⇒ cylindrical; dV = r dr dθ dz

PR2. Find the volume of the solid under $z = 9 - x^2 - y^2$ and above $z = 0$ .

Shadow. The paraboloid meets $z = 0$ where $x^2 + y^2 = 9$ , the disk $r \le 3$ . Slice. $z$ from $0$ to $9 - r^2$ ; keep the $r$ :

V = \int_0^{2\pi}\!\!\int_0^{3}\!\!\int_0^{9-r^2} r\,dz\,dr\,d\theta = 2\pi\int_0^3 r(9 - r^2)\,dr = 2\pi\Big[\tfrac{9r^2}{2} - \tfrac{r^4}{4}\Big]_0^3 = 2\pi\cdot\frac{81}{4} = \frac{81\pi}{2}.

PR3 · spherical · a spherical cap inside a cone

Shape · sphere + cone from the origin ⇒ spherical; dV = ρ² sin φ dρ dφ dθ

PR3. Find the volume of the solid inside the sphere $\rho = 2$ and above the cone $\phi \le \tfrac{\pi}{4}$ .

Limits. $0 \le \rho \le 2$ , $0 \le \phi \le \tfrac{\pi}{4}$ , $0 \le \theta \le 2\pi$ :

V = \int_0^{2\pi}\!\!\int_0^{\pi/4}\!\!\int_0^{2}\rho^2\sin\phi\,d\rho\,d\phi\,d\theta = 2\pi\cdot\Big(1 - \cos\tfrac{\pi}{4}\Big)\cdot\frac{8}{3} = \frac{8\pi\,(2 - \sqrt2)}{3} \approx 4.907.

The $\phi$ -integral $\int_0^{\pi/4}\sin\phi\,d\phi = 1 - \tfrac{\sqrt2}{2}$ is where the $\sqrt2$ enters — and where the $\sin\phi$ earns its keep.

PR4 · decision-only · name the system

Shape · no integration — choose the system and say why in one line

PR4. For each, name the coordinate system and a one-line reason — do not integrate. (a) $\iiint (x^2+y^2)\,dV$ over a finite cylinder; (b) $\iiint dV$ over the region between two concentric spheres; (c) $\iiint f\,dV$ over a rectangular brick.

(a) Cylindrical. The integrand $x^2 + y^2 = r^2$ and the circular cross-section both point at $r\,dr\,d\theta\,dz$ .

(b) Spherical. Both boundaries are " $\rho =$ constant," so the $\rho$ -limits are just the inner and outer radii — spherical turns the shell into a rectangle.

(c) Cartesian. Six flat faces at right angles ⇒ six constant limits. Anything fancier only drags in roots.

PR5 · faded bridge · cylindrical mass

Shape · cylinder + density ⇒ cylindrical; mass = ∭ δ dV

PR5. Find the mass of the cylinder $r \le 1$ , $0 \le z \le 2$ with density $\delta = z$ (denser toward the top).

Bridge: mass is $\iiint_T \delta\,dV$ with $\delta = z$ and $dV = r\,dr\,d\theta\,dz$ , so the integrand is $z\,r$ . Before revealing, predict the next step: what are the three limits, and which factor is the Jacobian you must keep?

Set up. $0 \le r \le 1$ , $0 \le \theta \le 2\pi$ , $0 \le z \le 2$ ; integrand $z$ , element $r\,dr\,d\theta\,dz$ :

m = \int_0^{2\pi}\!\!\int_0^{1}\!\!\int_0^{2} z\,r\,dz\,dr\,d\theta = 2\pi\cdot\underbrace{\tfrac12}_{\int_0^1 r\,dr}\cdot \underbrace{2}_{\int_0^2 z\,dz} = 2\pi.

The $r$ here is the Jacobian, not part of the density — keep both straight: density is $z$ , the $r$ comes from $dV$ .

PR6 · spherical mass

Shape · half-ball + radial density ⇒ spherical; mass = ∭ δ dV

PR6. Find the mass of the upper half-ball $z \ge 0$ , $\rho \le 1$ with density $\delta = \rho$ (proportional to distance from the centre).

Set up. Half-ball ⇒ $\phi$ from $0$ to $\tfrac{\pi}{2}$ . Integrand $\delta = \rho$ times element $\rho^2\sin\phi$ gives $\rho^3\sin\phi$ :

m = \int_0^{2\pi}\!\!\int_0^{\pi/2}\!\!\int_0^{1}\rho^3\sin\phi\,d\rho\,d\phi\,d\theta = 2\pi\cdot\underbrace{1}_{\int_0^{\pi/2}\sin\phi\,d\phi}\cdot \underbrace{\tfrac14}_{\int_0^1\rho^3\,d\rho} = \frac{\pi}{2}.

PR7 · faded bridge · a second moment

Shape · z² over the ball ⇒ spherical; z = ρ cos φ

PR7. Evaluate $\displaystyle\iiint_B z^2\,dV$ over the unit ball $\rho \le 1$ .

Bridge: in spherical, $z = \rho\cos\phi$ , so $z^2 = \rho^2\cos^2\phi$ ; times the element $\rho^2\sin\phi$ the integrand is $\rho^4\cos^2\phi\sin\phi$ . Before revealing, predict the next step: what is $\int_0^{\pi}\cos^2\phi\sin\phi\,d\phi$ ?

The $\phi$ -integral is a $u = \cos\phi$ substitution: $\int_0^{\pi}\cos^2\phi\sin\phi\,d\phi = \big[-\tfrac{\cos^3\phi}{3}\big]_0^{\pi} = \tfrac13 + \tfrac13 = \tfrac23$ . Then

\iiint_B z^2\,dV = \int_0^{2\pi}\!\!\int_0^{\pi}\!\!\int_0^{1}\rho^4\cos^2\phi\sin\phi\,d\rho\,d\phi\,d\theta = 2\pi\cdot\frac23\cdot\frac15 = \frac{4\pi}{15}.

Pre-flight checklist — run it before you integrate

\textbf{1.}\ \text{Sketch drawn? Shadow and slice identified, limits read off them?}

\textbf{2.}\ \text{Outer limits constant? Each inner limit depends only on variables further out?}

\textbf{3.}\ \text{Cylindrical: is the } r \text{ present in } dV = r\,dr\,d\theta\,dz?

\textbf{4.}\ \text{Spherical: is the } \sin\phi \text{ present in } dV = \rho^2\sin\phi\,d\rho\,d\phi\,d\theta?

\textbf{5.}\ \text{Did the coordinate system match the solid's symmetry — or did you fight it?}

Three things to carry out of this chapter

The ritual is shadow then slice. Project the solid to a flat shadow for the outer limits, slice from floor to roof for the inner limit. Nobody pictures the whole solid; everybody projects and slices.
The coordinate system is a decision, and it is the skill. Origin-radial → spherical; an axis with circular cross-sections → cylindrical; flat right-angled faces → Cartesian. Match the symmetry to the surface that becomes a constant.
Never drop the Jacobian. $dV = r\,dr\,d\theta\,dz$ (cylindrical) and $dV = \rho^2\sin\phi\,d\rho\,d\phi\,d\theta$ (spherical). The dropped $r$ and the dropped $\sin\phi$ are the marks the chapter is engineered to make you stop losing.

The anchor, one more time

Back to the CT scanner. Reconstructing a body's total mass from its density is exactly $\iiint_T \delta\,dV$ — and a real scanner does it in whichever coordinates make the patient simplest, slice by slice. You have just learned the three systems it chooses between, and the one habit (keep the Jacobian) that separates a right answer from a plausible wrong one. Re-do PR5 and PR7 from a blank page tomorrow — not re-read, re-do. Keeping the $r$ and the $\sin\phi$ is muscle memory, built with the pencil.

What's next. Chapter 13 leaves volumes behind for line integrals — adding up a force along a path — and Green's theorem, which trades a boundary walk for an area integral. The "total something over a region" idea continues; the region just becomes a curve and the area it encloses.

Same material, another voice

If a different explanation would help, these map one-to-one onto this chapter — free, from MIT OpenCourseWare:

Read: MIT 18.02 (Auroux), Lecture Notes — Unit IV, "Triple integrals in rectangular, cylindrical and spherical coordinates."
Watch: MIT 18.02 Video Lectures (Auroux) — Lecture 25 (triple integrals, rectangular & cylindrical) and Lecture 26 (spherical coordinates).

✓ Chapter complete. Your progress, and every quiz answer, is saved on this computer — revisit any time.

One ball, three coordinate systems

The 3-D ritual, and Cartesian triple integrals

Cylindrical coordinates: polar, with a z-axis

Spherical coordinates, and the sin⁡ϕ\sin\phisinϕ everyone drops

The Coordinate Lab

Retrieval gauntlet

What the exam will actually ask

PR1 · Cartesian · a tetrahedron

PR2 · cylindrical · under a paraboloid

PR3 · spherical · a spherical cap inside a cone

PR4 · decision-only · name the system

PR5 · faded bridge · cylindrical mass

PR6 · spherical mass

PR7 · faded bridge · a second moment

Three things to carry out of this chapter

Same material, another voice

Spherical coordinates, and the $\sin\phi$ everyone drops