Triple Integrals in Cylindrical Coordinates

Convert a triple integral over a solid cylinder to cylindrical coordinates and evaluate it.

The idea. Cylindrical coordinates are just polar coordinates with a z-axis bolted on. You already know two-thirds of this from polar integrals — convert x, y to r, θ exactly as in polar, leave z alone:

x = r\cos\theta, \qquad y = r\sin\theta, \qquad z = z, \qquad x^2 + y^2 = r^2

Cylindrical substitutions

$r, \theta$ polar coordinates in the xy-plane (distance from the z-axis, angle).
$z$ height, unchanged from Cartesian.
$x^2+y^2=r^2$ collapses circular integrands.

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The volume element:

dV = r\, dz\, dr\, d\theta

Volume element in cylindrical coordinates

$dV = r\,dz\,dr\,d\theta$ volume element; the r is the stretch factor, not optional.

The r is back, and it's the same trap as polar. The volume element carries an r — it's r dz dr dθ, never dz dr dθ. Same geometric reason: a cylindrical wedge is wider the farther out you go. If you internalized the forgotten-r from polar integrals, you already have this; it just rides along in 3D.

When to reach for cylindrical — the tell: the solid has circular symmetry around the z-axis. Cylinders, cones, paraboloids, anything round in the xy-plane that extends up and down. Or the integrand has x² + y² in it, which collapses to r². Same signal as polar, plus a vertical dimension.

The general form:

\iiint_E f\, dV = \int_\alpha^\beta \int_{h_1(\theta)}^{h_2(\theta)} \int_{z_1(r,\theta)}^{z_2(r,\theta)} f(r\cos\theta, r\sin\theta, z)\; r\, dz\, dr\, d\theta

Triple integral in cylindrical coordinates

$z_1(r,\theta), z_2(r,\theta)$ innermost bounds, may depend on r and θ.

The bounds nest inside-out just like rectangular triples: the z-bounds (innermost) can depend on r and θ; the r-bounds can depend on θ; the θ-bounds are constants.

Worked example

\iiint_E z\, dV, \qquad E: x^2+y^2 \le 1,\; 0 \le z \le 2

$x^2+y^2 \le 1,\; 0 \le z \le 2 \;\Longrightarrow\; 0 \le \theta \le 2\pi,\; 0 \le r \le 1,\; 0 \le z \le 2$
Convert the region. Circular in the xy-plane, radius 1: θ from 0 to 2π, r from 0 to 1. The z bounds are given: 0 to 2. Convert the integrand: f = z has no x or y, so it stays z.
$\int_0^{2\pi} \int_0^1 \int_0^2 z \cdot r\, dz\, dr\, d\theta$
Set up — attach the r.
$\int_0^2 z\, dz = \frac{z^2}{2}\bigg|_0^2 = 2 \quad\Longrightarrow\quad \int_0^{2\pi}\int_0^1 2r\, dr\, d\theta$
Innermost, over z.
$\int_0^1 2r\, dr = r^2\big|_0^1 = 1 \quad\Longrightarrow\quad \int_0^{2\pi} 1\, d\theta$
Middle, over r.
$\int_0^{2\pi} 1\, d\theta = 2\pi$
Outermost, over θ. Notice how the r from dV made the middle integral 2r, not 2 — miss it and you'd get a wrong answer that still looks plausible.

Answer:

2\pi

How to solve one of these

1.
Spot the circular symmetry around the $z$-axis, or $x^2+y^2$ in the integrand. That's the signal to use cylindrical.
2.
Convert the $xy$-part to polar. Find the $\theta$ sweep and the $r$ bounds exactly as in ordinary polar coordinates. Leave $z$ as $z$ .
3.
Find the $z$-bounds. Often given directly, or the surfaces the solid sits between — they may depend on $r$ .
4.
Convert the integrand. Every $x \to r\cos\theta$ , $y \to r\sin\theta$ , $x^2+y^2 \to r^2$ ; $z$ stays.
5.
Attach the $r$. Write $r\,dz\,dr\,d\theta$ . Multiply it in before integrating.
6.
Integrate inside-out. $z$ , then $r$ , then $\theta$ . Substitute bounds at each step.

\iiint_E f\, dV = \int_\alpha^\beta \int_{h_1(\theta)}^{h_2(\theta)} \int_{z_1(r,\theta)}^{z_2(r,\theta)} f(r\cos\theta, r\sin\theta, z)\; r\, dz\, dr\, d\theta

Convert x and y to r and θ exactly as in polar, leave z alone, integrate innermost to outermost — z, then r, then θ — and multiply by r dz dr dθ.

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