Multivariable Calculus · Undergraduate

Triple Integrals in Cylindrical and Spherical Coordinates

Quick answer

Cylindrical coordinates (r, θ, z) are polar coordinates in the xy-plane with the height z added, and they suit solids with an axis of symmetry: dV = r dz dr dθ. Spherical coordinates (ρ, φ, θ) measure the distance from the origin and two angles, and they suit balls and cones: dV = ρ² sin φ dρ dφ dθ. In each system, the right solids become boxes with constant limits, and the volume element records how the small pieces grow with distance.

What you'll learn

  • Convert points between rectangular, cylindrical and spherical coordinates
  • Choose the coordinate system that fits a solid
  • Integrate with dV = r dz dr dθ and dV = ρ² sin φ dρ dφ dθ
  • Compute volumes, masses and centroids of symmetric solids

Two new coordinate systems

Polar coordinates turned disks into rectangles. Space has two versions of the idea.

Cylindrical coordinates (r,θ,z)(r, \theta, z) keep the height zz and use polar coordinates for the point’s shadow on the xyxy-plane:

x=rcos⁡θ,y=rsin⁡θ,z=z,r2=x2+y2x = r\cos\theta, \quad y = r\sin\theta, \quad z = z, \quad r^2 = x^2 + y^2

The surface r=ar = a is a cylinder around the zz-axis, which gives the system its name.

Spherical coordinates (ρ,φ,θ)(\rho, \varphi, \theta) use the distance ρ\rho from the origin, the angle φ\varphi measured down from the positive zz-axis, with 0≤φ≤π0 \le \varphi \le \pi, and the same θ\theta as before:

x=ρsin⁡φcos⁡θ,y=ρsin⁡φsin⁡θ,z=ρcos⁡φ,ρ2=x2+y2+z2x = \rho\sin\varphi\cos\theta, \quad y = \rho\sin\varphi\sin\theta, \quad z = \rho\cos\varphi, \quad \rho^2 = x^2 + y^2 + z^2

The surface ρ=a\rho = a is a sphere, and φ=α\varphi = \alpha is a cone with its vertex at the origin. The quantity ρsin⁡φ\rho\sin\varphi is the distance from the zz-axis, the cylindrical rr.

Why the volume elements look the way they do

In cylindrical coordinates, a small piece is a polar rectangle of area r Δr Δθr\,\Delta r\,\Delta\theta lifted through a height Δz\Delta z. Its volume is r Δz Δr Δθr\,\Delta z\,\Delta r\,\Delta\theta, so

dV=r dz dr dθdV = r\,dz\,dr\,d\theta

In spherical coordinates, a small piece is bounded by two spheres, two cones and two half-planes. Its edges are close to perpendicular. Along the radius, the edge has length Δρ\Delta\rho. Changing φ\varphi moves the point along a circle of radius ρ\rho, so that edge has length ρ Δφ\rho\,\Delta\varphi. Changing θ\theta moves it around a circle of radius ρsin⁡φ\rho\sin\varphi, the distance to the zz-axis, so that edge has length ρsin⁡φ Δθ\rho\sin\varphi\,\Delta\theta. Multiplying the three edges:

dV=ρ2sin⁡φ dρ dφ dθdV = \rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta

Each factor is the length of one edge, so the volume element grows like ρ2\rho^2 far from the origin and shrinks near the zz-axis, where sin⁡φ\sin\varphi is small.

Worked examples

The bowl z = x² + y² capped by the plane z = 4 The paraboloid z = x² + y², opening upward from the origin, and the horizontal plane z = 4 that caps it. The paraboloid meets the plane in the circle of radius 2 at height 4. The solid inside the bowl and below the plane has volume 8π. 1 1 1 2 2 3 2 4 x y z z = x² + y²
The bowl z = x² + y² capped by the plane z = 4

Common mistakes

Practice problems

  1. Use cylindrical coordinates to find the volume of a cylinder of radius 22 and height 33.

    Answer

    12π12\pi

    Full solution

    ∫02π∫02∫03r dz dr dθ=2π⋅2⋅3=12π\int_0^{2\pi} \int_0^2 \int_0^3 r\,dz\,dr\,d\theta = 2\pi \cdot 2 \cdot 3 = 12\pi, matching πr2h\pi r^2 h.

  2. Evaluate ∭E(x2+y2) dV\iiint_E (x^2 + y^2)\,dV over the cylinder x2+y2≤1x^2 + y^2 \le 1, 0≤z≤20 \le z \le 2.

    Answer

    π\pi

    Full solution

    The integrand is r2r^2, so the integral is ∫02π∫01∫02r3 dz dr dθ=2π⋅14⋅2=π\int_0^{2\pi} \int_0^1 \int_0^2 r^3\,dz\,dr\,d\theta = 2\pi \cdot \tfrac{1}{4} \cdot 2 = \pi.

  3. Write the point (0,2,0)(0, 2, 0) in spherical coordinates.

    Answer

    (2,π2,π2)\left(2, \tfrac{\pi}{2}, \tfrac{\pi}{2}\right)

    Full solution

    ρ=2\rho = 2. The point lies in the xyxy-plane, a right angle from the zz-axis, so φ=π2\varphi = \tfrac{\pi}{2}. Its shadow lies on the positive yy-axis, so θ=π2\theta = \tfrac{\pi}{2}.

  4. Find the volume of the upper half of the unit ball, z≥0z \ge 0.

    Answer

    2π3\tfrac{2\pi}{3}

    Full solution

    The half ball is 0≤φ≤π20 \le \varphi \le \tfrac{\pi}{2}. ∫02π∫0π/2∫01ρ2sin⁡φ dρ dφ dθ=2π⋅1⋅13=2π3\int_0^{2\pi} \int_0^{\pi/2} \int_0^1 \rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta = 2\pi \cdot 1 \cdot \tfrac{1}{3} = \tfrac{2\pi}{3}.

  5. Find the centroid of the upper half of the unit ball.

    Hint

    By symmetry, xˉ=yˉ=0\bar{x} = \bar{y} = 0. For zˉ\bar{z}, use z=ρcos⁡φz = \rho\cos\varphi.

    Answer

    (0,0,38)\left(0, 0, \tfrac{3}{8}\right)

    Full solution

    ∭z dV=∫02π∫0π/2∫01ρ3cos⁡φsin⁡φ dρ dφ dθ=2π⋅12⋅14=π4\iiint z\,dV = \int_0^{2\pi} \int_0^{\pi/2} \int_0^1 \rho^3\cos\varphi\sin\varphi\,d\rho\,d\varphi\,d\theta = 2\pi \cdot \tfrac{1}{2} \cdot \tfrac{1}{4} = \tfrac{\pi}{4}. Dividing by the volume 2π3\tfrac{2\pi}{3} gives zˉ=38\bar{z} = \tfrac{3}{8}.

  6. Evaluate ∭Be(x2+y2+z2)3/2 dV\iiint_B e^{(x^2 + y^2 + z^2)^{3/2}}\,dV over the unit ball.

    Answer

    4π(e−1)3≈7.198\tfrac{4\pi(e - 1)}{3} \approx 7.198

    Full solution

    The integrand is eρ3e^{\rho^3}. The angles give 2π⋅2=4π2\pi \cdot 2 = 4\pi, and with u=ρ3u = \rho^3, ∫01ρ2eρ3 dρ=13(e−1)\int_0^1 \rho^2 e^{\rho^3}\,d\rho = \tfrac{1}{3}(e - 1).

  7. Find the volume of the solid inside the sphere x2+y2+z2=4x^2 + y^2 + z^2 = 4 and outside the cylinder x2+y2=1x^2 + y^2 = 1.

    Hint

    Use cylindrical coordinates. At radius rr, a column runs from z=−4−r2z = -\sqrt{4 - r^2} to z=4−r2z = \sqrt{4 - r^2}.

    Answer

    43 π≈21.774\sqrt{3}\,\pi \approx 21.77

    Full solution

    V=∫02π∫1224−r2 r dr dθV = \int_0^{2\pi} \int_1^2 2\sqrt{4 - r^2}\,r\,dr\,d\theta. With u=4−r2u = 4 - r^2, ∫122r4−r2 dr=∫03u du=23⋅33/2=23\int_1^2 2r\sqrt{4 - r^2}\,dr = \int_0^3 \sqrt{u}\,du = \tfrac{2}{3} \cdot 3^{3/2} = 2\sqrt{3}. So V=2π⋅23=43 πV = 2\pi \cdot 2\sqrt{3} = 4\sqrt{3}\,\pi.

  8. A student finds the volume of the unit ball as ∫02π∫0π∫01dρ dφ dθ=2π2\int_0^{2\pi} \int_0^{\pi} \int_0^1 d\rho\,d\varphi\,d\theta = 2\pi^2. What went wrong?

    Answer

    The volume element ρ2sin⁡φ\rho^2\sin\varphi is missing; the answer is 4π3\tfrac{4\pi}{3}.

    Full solution

    Spherical boxes are not all the same size: they grow like ρ2\rho^2 and shrink near the zz-axis. With dV=ρ2sin⁡φ dρ dφ dθdV = \rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta, the integral is 2π⋅2⋅13=4π32\pi \cdot 2 \cdot \tfrac{1}{3} = \tfrac{4\pi}{3}, as in Example 4.

Frequently asked questions

What are cylindrical coordinates?

Polar coordinates r and θ for the point's shadow on the xy-plane, together with its height z. So x = r cos θ, y = r sin θ and z = z.

What are spherical coordinates?

ρ is the distance from the origin, φ is the angle down from the positive z-axis, between 0 and π, and θ is the same angle as in cylindrical coordinates. Then x = ρ sin φ cos θ, y = ρ sin φ sin θ and z = ρ cos φ.

What is dV in cylindrical coordinates?

dV = r dz dr dθ: the polar area element r dr dθ times the height dz.

What is dV in spherical coordinates?

dV = ρ² sin φ dρ dφ dθ. A small spherical box has sides dρ, ρ dφ and ρ sin φ dθ.

When should you use each system?

Use cylindrical coordinates for solids around the z-axis bounded by cylinders, paraboloids or planes. Use spherical coordinates for balls, spherical shells and cones with vertex at the origin, or when the integrand contains x² + y² + z².

What to learn next

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