Triple Integrals in Cylindrical Coordinates

Triple integrals in cylindrical coordinates use r, θ and z in place of x, y and z. You write x = r cos θ and y = r sin θ, keep z, and replace dV with r dz dr dθ. This makes integrals over cylinders, cones and paraboloids much easier, because their limits become constants or simple functions of r.

Before this lesson: Triple Integrals, Double Integrals in Polar Coordinates and Polar Coordinates

What cylindrical coordinates are

Cylindrical coordinates are polar coordinates on the floor plus a height. A point P has a distance r from the z-axis, an angle θ measured from the positive x-axis, and the same height z as before.

Shapes that are round about the z-axis have simple equations in these coordinates. The cylinder x² + y² = 4 is just r = 2. The paraboloid z = x² + y² is z = r². The cone z = √(x² + y²) is z = r.

Triple integral in cylindrical coordinates formula

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

\iiint_E f(x, y, z)\,dV = \int_{\alpha}^{\beta} \int_{h_1(\theta)}^{h_2(\theta)} \int_{u_1(r, \theta)}^{u_2(r, \theta)} f(r\cos\theta, r\sin\theta, z)\,r\,dz\,dr\,d\theta

The extra factor r is the same one you met in double integrals in polar coordinates. A small cylindrical box has sides dr, r dθ and dz, so its volume is dV = r dz dr dθ.

1. Find the bottom surface z = u₁ and the top surface z = u₂, written in r and θ.
2. Find the shadow D of the solid on the xy-plane, and describe it in polar form: θ from α to β, r from h₁(θ) to h₂(θ).
3. Replace x² + y² with r², x with r cos θ and y with r sin θ.
4. Multiply by r, then integrate in z, then r, then θ.

Triple integrals in cylindrical coordinates examples

Example 1: A function over a cylinder

Evaluate \iiint_E z\,dV, where E is the solid cylinder x² + y² ≤ 4, 0 ≤ z ≤ 3.

0 \le \theta \le 2\pi, \quad 0 \le r \le 2, \quad 0 \le z \le 3

\iiint_E z\,dV = \int_0^{2\pi} \int_0^2 \int_0^3 z\,r\,dz\,dr\,d\theta

\int_0^3 z\,r\,dz = r\Big[\dfrac{z^2}{2}\Big]_0^3

= \dfrac{9r}{2}

\int_0^2 \dfrac{9r}{2}\,dr = \Big[\dfrac{9r^2}{4}\Big]_0^2

= 9

\int_0^{2\pi} 9\,d\theta = 18\pi

Answer: 18π ≈ 56.55.

Example 2: Volume under a paraboloid

Find the volume of the solid under the paraboloid z = 9 − x² − y² and above the xy-plane.

z = 9 - r^2

9 - r^2 = 0 \Rightarrow r = 3

V = \int_0^{2\pi} \int_0^3 \int_0^{9 - r^2} r\,dz\,dr\,d\theta

\int_0^{9 - r^2} r\,dz = (9 - r^2)r

= 9r - r^3

\int_0^3 (9r - r^3)\,dr = \Big[\dfrac{9r^2}{2} - \dfrac{r^4}{4}\Big]_0^3

= \dfrac{81}{2} - \dfrac{81}{4}

= \dfrac{81}{4}

V = 2\pi \cdot \dfrac{81}{4}

= \dfrac{81\pi}{2}

Answer: the volume is 81π/2 ≈ 127.23.

Example 3: Converting a rectangular integral

Evaluate \int_{-1}^{1} \int_{-\sqrt{1 - x^2}}^{\sqrt{1 - x^2}} \int_0^2 \sqrt{x^2 + y^2}\,dz\,dy\,dx.

-\sqrt{1 - x^2} \le y \le \sqrt{1 - x^2}, \quad -1 \le x \le 1

0 \le \theta \le 2\pi, \quad 0 \le r \le 1

\sqrt{x^2 + y^2} = r

\int_0^{2\pi} \int_0^1 \int_0^2 r \cdot r\,dz\,dr\,d\theta

\int_0^2 r^2\,dz = 2r^2

\int_0^1 2r^2\,dr = \Big[\dfrac{2r^3}{3}\Big]_0^1

= \dfrac{2}{3}

\int_0^{2\pi} \dfrac{2}{3}\,d\theta = \dfrac{4\pi}{3}

Answer: 4π/3 ≈ 4.19.

Example 4: The solid between a cone and a paraboloid

Find the volume of the solid that lies above the cone z = √(x² + y²) and below the paraboloid z = 2 − x² − y².

z = r \quad \text{and} \quad z = 2 - r^2

r = 2 - r^2

r^2 + r - 2 = 0

(r + 2)(r - 1) = 0

r = 1

V = \int_0^{2\pi} \int_0^1 \int_r^{2 - r^2} r\,dz\,dr\,d\theta

\int_r^{2 - r^2} r\,dz = r(2 - r^2 - r)

= 2r - r^3 - r^2

\int_0^1 (2r - r^3 - r^2)\,dr = \Big[r^2 - \dfrac{r^4}{4} - \dfrac{r^3}{3}\Big]_0^1

= 1 - \dfrac{1}{4} - \dfrac{1}{3}

= \dfrac{5}{12}

V = 2\pi \cdot \dfrac{5}{12}

= \dfrac{5\pi}{6}

Answer: the volume is 5π/6 ≈ 2.62. This solid looks like an ice cream cone with a rounded top.

Example 5: Mass of a log

A log stands upright. It is a solid cylinder of radius 0.5 m and height 2 m. The wood is denser near the bark: its density is ρ = 600 + 400r kilograms per cubic meter, where r is the distance from the center line. Find the mass of the log.

m = \int_0^{2\pi} \int_0^{0.5} \int_0^2 (600 + 400r)\,r\,dz\,dr\,d\theta

(600 + 400r)r = 600r + 400r^2

\int_0^2 (600r + 400r^2)\,dz = 1200r + 800r^2

\int_0^{0.5} (1200r + 800r^2)\,dr = \Big[600r^2 + \dfrac{800r^3}{3}\Big]_0^{0.5}

= 600(0.25) + \dfrac{800(0.125)}{3}

= 150 + \dfrac{100}{3}

= \dfrac{550}{3}

m = 2\pi \cdot \dfrac{550}{3}

= \dfrac{1100\pi}{3} \approx 1151.9

Answer: the log has a mass of about 1152 kg. As a check, the volume is π(0.5)²(2) ≈ 1.571 m³, so the average density is about 733 kg/m³, between 600 and 800 as expected.

Common mistakes

Related methods

Converting points

To go from cylindrical to rectangular, use x = r cos θ and y = r sin θ. To go the other way, use r = √(x² + y²) and tan θ = y/x, choosing θ in the right quadrant. For example, (r, θ, z) = (2, π/3, 5) is the point (1, √3, 5).

When to use cylindrical coordinates

Use them when the solid is round about the z-axis or the integrand contains x² + y². Use rectangular coordinates for boxes and slanted planes, as in triple integrals. If the solid is round about a different axis, such as the y-axis, let r and θ describe the xz-plane instead.

Spheres

Cylindrical coordinates work for spheres, but the z limits become ±√(a² − r²). Spherical coordinates, the next lesson, often make sphere problems simpler.

Practice problems

1. Evaluate \int_0^{2\pi} \int_0^1 \int_0^4 r\,dz\,dr\,d\theta. What solid does it measure?
2. Evaluate \iiint_E (x^2 + y^2)\,dV, where E is the cylinder x² + y² ≤ 4, 0 ≤ z ≤ 1.
3. Find the volume of the solid under z = 1 − x² − y² and above the xy-plane.
4. Evaluate \iiint_E z\,dV, where E lies above the xy-plane, below the cone z = √(x² + y²) and inside the cylinder x² + y² = 4.
5. Convert to cylindrical coordinates and evaluate: \int_{-2}^{2} \int_{-\sqrt{4 - x^2}}^{\sqrt{4 - x^2}} \int_0^3 1\,dz\,dy\,dx.
6. Find the volume of the solid inside both the sphere x² + y² + z² = 4 and the cylinder x² + y² = 1.
7. Evaluate \iiint_E x\,dV, where E is the part of the cylinder x² + y² ≤ 4, 0 ≤ z ≤ 3, in the first octant.
8. Find the volume of the solid between the paraboloids z = x² + y² and z = 8 − x² − y².
9. Evaluate \iiint_E \sqrt{x^2 + y^2}\,dV, where E lies above the cone z = √(x² + y²) and below the plane z = 2.
10. A concrete pipe stands on end. Its inner radius is 1 m, its outer radius is 1.5 m and its height is 2 m. Concrete has a density of 2400 kg/m³. Find the volume of concrete and the mass of the pipe.

\int_0^4 r\,dz = 4r

\int_0^1 4r\,dr = 2

\int_0^{2\pi} 2\,d\theta = 4\pi

(x^2 + y^2)\,dV = r^2 \cdot r\,dz\,dr\,d\theta

\int_0^1 r^3\,dz = r^3

\int_0^2 r^3\,dr = \Big[\dfrac{r^4}{4}\Big]_0^2 = 4

\int_0^{2\pi} 4\,d\theta = 8\pi

V = \int_0^{2\pi} \int_0^1 \int_0^{1 - r^2} r\,dz\,dr\,d\theta

\int_0^{1 - r^2} r\,dz = r - r^3

\int_0^1 (r - r^3)\,dr = \dfrac{1}{2} - \dfrac{1}{4}

= \dfrac{1}{4}

V = 2\pi \cdot \dfrac{1}{4} = \dfrac{\pi}{2}

\int_0^{2\pi} \int_0^2 \int_0^{r} z\,r\,dz\,dr\,d\theta

\int_0^{r} z\,r\,dz = r \cdot \dfrac{r^2}{2} = \dfrac{r^3}{2}

\int_0^2 \dfrac{r^3}{2}\,dr = \Big[\dfrac{r^4}{8}\Big]_0^2 = 2

\int_0^{2\pi} 2\,d\theta = 4\pi

0 \le \theta \le 2\pi, \quad 0 \le r \le 2

\int_0^{2\pi} \int_0^2 \int_0^3 r\,dz\,dr\,d\theta

\int_0^3 r\,dz = 3r

\int_0^2 3r\,dr = 6

\int_0^{2\pi} 6\,d\theta = 12\pi

-\sqrt{4 - r^2} \le z \le \sqrt{4 - r^2}, \quad 0 \le r \le 1

\int_{-\sqrt{4 - r^2}}^{\sqrt{4 - r^2}} r\,dz = 2r\sqrt{4 - r^2}

\int_0^1 2r\sqrt{4 - r^2}\,dr = \Big[-\dfrac{2}{3}(4 - r^2)^{3/2}\Big]_0^1

= -\dfrac{2}{3}(3)^{3/2} + \dfrac{2}{3}(4)^{3/2}

= \dfrac{16}{3} - 2\sqrt{3}

V = 2\pi\Big(\dfrac{16}{3} - 2\sqrt{3}\Big) = \dfrac{32\pi}{3} - 4\sqrt{3}\,\pi

0 \le \theta \le \dfrac{\pi}{2}

x\,dV = r\cos\theta \cdot r\,dz\,dr\,d\theta

\int_0^3 r^2\cos\theta\,dz = 3r^2\cos\theta

\int_0^2 3r^2\cos\theta\,dr = 8\cos\theta

\int_0^{\pi/2} 8\cos\theta\,d\theta = 8

r^2 = 8 - r^2 \Rightarrow r = 2

\int_{r^2}^{8 - r^2} r\,dz = 8r - 2r^3

\int_0^2 (8r - 2r^3)\,dr = 16 - 8 = 8

V = 2\pi \cdot 8 = 16\pi

r \le z \le 2, \quad 0 \le r \le 2

\int_r^2 r \cdot r\,dz = r^2(2 - r) = 2r^2 - r^3

\int_0^2 (2r^2 - r^3)\,dr = \dfrac{16}{3} - 4 = \dfrac{4}{3}

2\pi \cdot \dfrac{4}{3} = \dfrac{8\pi}{3}

V = \int_0^{2\pi} \int_1^{1.5} \int_0^2 r\,dz\,dr\,d\theta

\int_0^2 r\,dz = 2r

\int_1^{1.5} 2r\,dr = 1.5^2 - 1^2 = 1.25

V = 2\pi(1.25) = 2.5\pi \approx 7.854

m = 2400 \cdot 2.5\pi = 6000\pi \approx 18850

Triple integrals in cylindrical coordinates FAQ

Why is there an extra r in cylindrical coordinates?

A small piece of the solid has sides dr, r dθ and dz. The arc side is r dθ because arcs farther from the z-axis are longer, so dV = r dz dr dθ.

When should I use cylindrical coordinates instead of rectangular?

Use them when the solid is a cylinder, cone or paraboloid around the z-axis, or when the integrand contains x² + y². Then the limits are much simpler.

What is the difference between cylindrical and spherical coordinates?

Cylindrical coordinates use the distance r from the z-axis, the angle θ and the height z. Spherical coordinates use the distance ρ from the origin and two angles, which suits spheres and cones.

How do you convert a triple integral to cylindrical coordinates?

Describe the shadow D in polar form, write the bottom and top surfaces in r and θ, replace x and y with r cos θ and r sin θ, and multiply by r.

Summary

• Cylindrical coordinates: x = r cos θ, y = r sin θ, z = z, and x² + y² = r².
• The volume element is dV = r dz dr dθ.
• Describe the shadow D in polar form, then integrate z from the bottom surface to the top surface.
• Cylinders become r = a, cones become z = kr, and paraboloids become z = a ± r².
• Use them for solids that are round about the z-axis.

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Next lesson: Triple Integrals in Spherical Coordinates

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