Triple Integrals in Cylindrical Coordinates

Main idea

Cylindrical coordinates are polar coordinates on the floor with an ordinary height $z$ on top, made for solids that are round about an axis. The volume element is $dV = r\,dz\,dr\,d\theta$ : the polar $r\,dr\,d\theta$ times a height $dz$ .

\iiint_E f\,dV = \int_\alpha^\beta\!\int_{r_1(\theta)}^{r_2(\theta)}\!\int_{h_1(r,\theta)}^{h_2(r,\theta)} f\,{\htmlData{sym=0}{r\,dz\,dr\,d\theta}}

What the symbols mean

$r\,dz\,dr\,d\theta$
Cylindrical volume element
The polar area element r dr dθ times the height dz.

Explanation

Cylindrical coordinates simply describe a point by three things: how far it is from the $z$ -axis ( $r$ ), which direction it's in around that axis ( $\theta$ ) and how high up it is ( $z$ ). They get their name because $r = 2$ is a cylinder of radius 2 around the $z$ -axis, the same way $x = 2$ is a flat wall in Cartesian coordinates. Nothing here is new: it's the polar coordinates from the last few lessons with a $z$ added. The volume element $r\,dz\,dr\,d\theta$ is a small polar patch ( $r\,dr\,d\theta$ ) stacked $dz$ high.

When to use it: cylinders, cones, paraboloids, anything round about the $z$ -axis, or integrands with $x^2 + y^2$ . In these coordinates the cone $z = \sqrt{x^2 + y^2}$ is just $z = r$ , and the paraboloid $z = 4 - x^2 - y^2$ is $z = 4 - r^2$ .

Setting up. It's the same plan as any triple integral. For each point on the floor, $z$ runs from the bottom surface to the top surface, both written in $r$ and $\theta$ . Then describe the shadow on the $xy$ -plane in polar coordinates. Notice that for a solid that's round about the axis, the $\theta$ -integral usually just contributes a factor of $2\pi$ .

A common mistake. It's tempting to drop the $r$ because there's already a $z$ doing the third-dimension work. The $r$ isn't about height; it's the same stretch factor as in polar double integrals, and leaving it out gives the wrong volume for even a plain cylinder.

If you've done volumes by shells, this will look familiar: $2\pi r\,h(r)\,dr$ is exactly what's left after the $z$ - and $\theta$ -integrals in $\int_0^{2\pi}\!\int\!\int_0^{h(r)} r\,dz\,dr\,d\theta$ .

See it

The dome z = 4 − r², with a cylinder of radius R cutting out its middle. The volume inside the cylinder is π(4R² − R⁴/2). Notice that at R = 1 it's already 3.5π, almost half of the dome's total 8π, even though R = 1 is only half the way out: the dome is tallest near the axis. At R = 2, where the dome meets the floor, you have all 8π.

Worked examples

Example 1

Find the volume under the paraboloid $z = 4 - r^2$ and above the plane $z = 0$ .

1
For each point on the floor, $z$ runs from $0$ to $4 - r^2$ . The dome meets the floor where $4 - r^2 = 0$ , at $r = 2$ , so the shadow is the disk $0 \le r \le 2$ .

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

Use cylindrical coordinates to find the volume of a cylinder of radius 2 and height 5.

Problem 2+5 XP

Find the volume under $z = 4 - r^2$ above $z = 0$ .

Problem 3+5 XP

Evaluate $\iiint_E z\,dV$ over the cylinder $r \le 1$ , $0 \le z \le 2$ .

Problem 4+5 XP

Find the volume between the cone $z = r$ and the plane $z = 1$ .

Problem 5+5 XP

Find the mass of the cylinder $r \le 1$ , $0 \le z \le 1$ with density equal to the distance from the axis.

Problem 6+5 XP

A cylindrical grain silo has radius $3$ m and height $10$ m. The grain packs more densely toward the bottom: $\rho = 800 - 20z$ kg/m³ at height $z$ . Find the total mass of grain when the silo is full.

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

Find the volume of the region inside the sphere $x^2 + y^2 + z^2 = 4$ and above the cone $z = \sqrt{x^2 + y^2}$ : an ice cream cone.

Where it's used

Pressure vessels, drive shafts, and the moment of inertia of flywheels and wheels are computed in cylindrical coordinates.

Triple Integrals Triple Integrals in Spherical Coordinates