Triple integrals in Cartesian, cylindrical, spherical coordinates: Visual walkthrough

We assume you have met the idea of a triple integral (adding up value × little volume) and that a warped little box needs a stretch factor — the Jacobian. Everything else we grow from scratch.

Step 1 — Name every coordinate by pointing at it

WHAT. Before any algebra, we must agree on what the three numbers $(ρ, ϕ, θ)$ mean as physical distances and turns. A point in space is located by:

$ρ$ (Greek "rho") $=$ how far the point is from the origin (a length). $ρ \geq 0$ .
$ϕ$ (Greek "phi") $=$ the polar angle, measured down from the $+ z$ axis (the "north pole" direction). $0 \leq ϕ \leq π$ .
$θ$ (Greek "theta") $=$ the azimuth, how far you have spun around the $z$ -axis, measured in the flat $x y$ -plane. $0 \leq θ < 2 π$ .

WHY these three and not $x, y, z$ . For anything round-about-a-centre (a ball, a shell), the boundary is simply " $ρ =$ constant". One clean equation replaces the nested square roots $z = \pm a^{2} - x^{2} - y^{2}$ that Cartesian would force on us.

PICTURE. Follow the arrows: the long chalk-blue arrow is $ρ$ , the pink arc curving down from the top axis is $ϕ$ , the yellow arc lying flat is $θ$ .

Figure — Triple integrals in Cartesian, cylindrical, spherical coordinates

Step 2 — Convert the angles into $x, y, z$

WHAT. We write the ordinary Cartesian coordinates of that same point using $ρ, ϕ, θ$ . The trick is to drop a perpendicular from the point to the $z$ -axis, splitting the job into two right triangles.

WHY. We can only use the Jacobian machine (the honest way to find $d V$ ) once we know $x, y, z$ as functions of $ρ, ϕ, θ$ . And right triangles are the only tool that turns an angle into a ratio of sides — that is exactly what $sin$ and $cos$ do: on a right triangle, $sin ϕ = \frac{opposite}{hypotenuse}$ and $cos ϕ = \frac{adjacent}{hypotenuse}$ .

PICTURE. In the vertical triangle the hypotenuse is $ρ$ , the angle at the top is $ϕ$ . The side along the axis (adjacent to $ϕ$ ) is the height $z$ ; the side across (opposite $ϕ$ ) is the shadow-length $r$ in the flat plane.

From that vertical right triangle: $z = hyp ρ adjacent/hyp cos ϕ, r = hyp ρ opposite/hyp sin ϕ .$

Here $r = ρ sin ϕ$ is the radius of the flat shadow circle. Now spin by $θ$ in that flat plane (a second right triangle, this time in the $x y$ -plane with hypotenuse $r$ ): $x = r cos θ = ρ sin ϕ cos θ, y = r sin θ = ρ sin ϕ sin θ .$

Step 3 — Nudge each coordinate a hair and watch what moves

WHAT. Chop the solid into tiny cells by taking small steps $d ρ$ , $d ϕ$ , $d θ$ in each coordinate separately. We ask: how far does the point physically move for each nudge? Distance, not angle, is what fills up volume.

WHY a step, and why measure distance. Volume is (length)×(length)×(length). An angle is not a length, so we cannot multiply three angles and get a volume. Each angular nudge must first be turned into an arc length — how far the point actually slides — before we can multiply. This conversion is the entire origin of the $ρ^{2} sin ϕ$ factor.

PICTURE. Three coloured arrows leave one corner point: blue for the radial nudge, pink for the $ϕ$ -nudge, yellow for the $θ$ -nudge. Notice they point in three mutually perpendicular directions.

The three edge lengths, one per nudge:

Nudge	Direction of motion	Arc length	Why
$d ρ$	straight out along $ρ$	$d ρ$	a change in distance is a distance
$d ϕ$	along a meridian (pole-to-pole circle)	$ρ d ϕ$	arc = radius × angle, and that circle has radius $ρ$
$d θ$	along a latitude circle	$ρ sin ϕ d θ$	that circle has radius $r = ρ sin ϕ$ , not $ρ$

Step 4 — Why the latitude circle is smaller: the $sin ϕ$ factor

WHAT. We zoom into the single most-forgotten fact: the $θ$ -nudge travels a shorter arc than you'd guess, because the circle you spin on is not the big circle of radius $ρ$ — it is the small latitude circle of radius $ρ sin ϕ$ .

WHY it matters. If you naively used $ρ d θ$ (as if the latitude circle were as big as the sphere), you would over-count volume near the poles. Reality: near a pole the latitude circle shrinks to a dot, so spinning $θ$ barely moves you.

PICTURE. Slice the sphere horizontally at height $z = ρ cos ϕ$ . The horizontal cut is a circle. Its radius is the flat shadow $r = ρ sin ϕ$ from Step 2 — small near the poles, largest at the equator.

$latitude radius = r = ρ 0 at poles 1 at equator sin ϕ ⟹ θ -arc = ρ sin ϕ d θ .$

Step 5 — Multiply the three edges into a volume

WHAT. The little cell is (almost exactly) a rectangular box because — from Step 3 — the three nudges point in perpendicular directions. Volume of a box = product of its three edge lengths.

WHY perpendicular lets us just multiply. For a box with square corners, volume is length × width × height with no cross-terms. (The honest Jacobian in Step 6 confirms these three directions really are orthogonal, so no correction is needed.)

PICTURE. The three edges $d ρ$ , $ρ d ϕ$ , $ρ sin ϕ d θ$ meeting at right angles, boxing off one tiny curved brick.

$d V = radial edge (d ρ) \cdot meridian arc (ρ d ϕ) \cdot latitude arc (ρ sin ϕ d θ) = ρ^{2} sin ϕ d ρ d ϕ d θ .$

Count the $ρ$ 's: one from the meridian arc, one from the latitude arc → $ρ^{2}$ . The lone $sin ϕ$ comes only from the latitude arc.

Step 6 — Confirm it with the honest Jacobian

WHAT. The geometric argument assumed the three edges are perpendicular. Let the algebra prove it. The Jacobian $∣ J ∣$ is the exact stretch factor between the coordinate box $d ρ d ϕ d θ$ and the real volume — no "almost" about it.

WHY do this too. The determinant automatically handles any skew between edges. If our picture were wrong (non-perpendicular edges), the determinant would differ from $ρ^{2} sin ϕ$ . It doesn't — so the picture was right.

PICTURE. The $3 \times 3$ table of partial derivatives, each column being one edge-direction vector from Step 3.

\begin{vmatrix} \sin\phi\cos\theta & \rho\cos\phi\cos\theta & -\rho\sin\phi\sin\theta\\ \sin\phi\sin\theta & \rho\cos\phi\sin\theta & \rho\sin\phi\cos\theta\\ \cos\phi & -\rho\sin\phi & 0 \end{vmatrix}=\rho^2\sin\phi.$$ Each column is a nudge-direction: column 1 is "increase $\rho$", column 2 is "increase $\phi$", column 3 is "increase $\theta$". The determinant of edge-vectors *is* the box volume — matching Step 5 exactly. We take $|J|=\rho^2\sin\phi$ (with $0\le\phi\le\pi$, $\sin\phi\ge0$, so no absolute-value drama). --- ## Step 7 — Degenerate cases: check the formula at the edges **WHAT.** A formula is only trustworthy if it behaves sensibly at its extremes. We test the three places where something vanishes. **WHY.** These are exactly where a wrong formula would blow up or double-count. If $dV$ stays finite and shrinks where it should, we believe it everywhere in between. **PICTURE.** North pole ($\phi=0$) and south pole ($\phi=\pi$): the latitude circle is a single point, so the yellow $\theta$-arc has zero length. The origin ($\rho=0$): all edges collapse. ![[deepdives/dd-maths-4.4.20-d2-s07.png]] - **Poles, $\phi=0$ or $\phi=\pi$:** $\sin\phi=0\Rightarrow dV=0$. Correct — the latitude circle has shrunk to a dot, so a $\theta$-step sweeps *no* volume. Without $\sin\phi$ we'd fabricate volume out of nothing here. - **Origin, $\rho=0$:** $\rho^2=0\Rightarrow dV=0$. Correct — everything piles into one point; no volume. - **Equator, $\phi=\pi/2$:** $\sin\phi=1$, so $dV=\rho^2\,d\rho\,d\phi\,d\theta$ — the latitude circle is at full size $\rho$, the fattest cells live here. > [!recall]- Why doesn't $dV=0$ at the poles wreck an integral? > A single line (the polar axis) has measure zero — it contributes nothing to the total. The $\sin\phi\to0$ correctly starves those cells; it never makes a real volume vanish. ::: The poles are a set of measure zero, so shrinking their cells to nothing is exactly right — real solids fill the space *around* the axis. --- ## The one-picture summary **WHAT.** One diagram carrying the whole derivation: the tiny spherical brick with its three labelled edges, each traced back to its origin. ![[deepdives/dd-maths-4.4.20-d2-s08.png]] $$\underbrace{d\rho}_{\text{radial}}\times\underbrace{\rho\,d\phi}_{\text{meridian}}\times\underbrace{\rho\sin\phi\,d\theta}_{\text{latitude}}=\rho^2\sin\phi\,d\rho\,d\phi\,d\theta.$$ > [!mnemonic] Two rho's and a sine > "**One $\rho$ from swinging down ($\phi$), one $\rho$ from spinning round ($\theta$), and a $\sin\phi$ that kills the poles.**" > [!recall]- Feynman retelling — explain the whole walkthrough to a friend > Picture a globe. To find the volume of a tiny brick inside it, I measure its three sides. Side one: step a little further from the centre — that side is just $d\rho$, a plain distance. Side two: tilt a little from the north pole toward the equator — I'm walking along a big pole-to-pole circle of radius $\rho$, so that arc is $\rho\,d\phi$. Side three: spin a little around the axis — but here's the catch, I'm not walking on the big circle, I'm walking on the *horizontal ring* through my point, which is smaller. Its radius is $\rho\sin\phi$: fat at the equator, zero at the poles. So that arc is $\rho\sin\phi\,d\theta$. Multiply the three perpendicular sides and I get $\rho^2\sin\phi\,d\rho\,d\phi\,d\theta$. The two $\rho$'s came from the two curved sides; the $\sin\phi$ is nature reminding me that spinning near a pole barely moves you. The Jacobian determinant then double-checks the arithmetic and agrees. --- ## Connections - [[Triple integrals in Cartesian, cylindrical, spherical coordinates]] — the parent that *states* this result; here we *earn* it. - [[Jacobian and change of variables]] — Step 6 is a worked Jacobian; this is the general stretch-factor machine. - [[Spherical coordinates geometry]] — the naming and right triangles of Steps 1–2. - [[Double integrals & polar coordinates]] — cylindrical/polar's single $r$ is the flat cousin of this $\rho^2\sin\phi$. - [[Center of mass and moments of inertia]] — every such integral over a ball uses this $dV$.