Triple Integrals in Spherical Coordinates

Triple integrals in spherical coordinates describe each point by its distance ρ from the origin and two angles, θ and φ. You write x = ρ sin φ cos θ, y = ρ sin φ sin θ and z = ρ cos φ, and replace dV with ρ² sin φ dρ dφ dθ. Balls, spherical shells and cones then get constant limits.

Before this lesson: Triple Integrals and Triple Integrals in Cylindrical Coordinates

What spherical coordinates are

A point P has three spherical coordinates. ρ (rho) is the distance from the origin to P. θ is the same angle as in cylindrical coordinates, measured on the floor from the positive x-axis. φ (phi) is the angle from the positive z-axis down to the segment OP.

The right triangle in the picture gives the formulas. The height of P is z = ρ cos φ. The shadow has length r = ρ sin φ, and polar coordinates on the floor turn that into x and y. We always have ρ ≥ 0 and 0 ≤ φ ≤ π: φ = 0 points straight up, φ = π/2 is the floor, and φ = π points straight down.

Simple shapes have simple equations. The sphere x² + y² + z² = 9 is ρ = 3. The cone z = √(x² + y²) is φ = π/4. The upper half of space is 0 ≤ φ ≤ π/2.

Triple integral in spherical coordinates formula

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

\iiint_E f(x, y, z)\,dV = \int_{\alpha}^{\beta} \int_{c}^{d} \int_{a}^{b} f(\rho\sin\varphi\cos\theta, \rho\sin\varphi\sin\theta, \rho\cos\varphi)\,\rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta

A small spherical box has sides dρ, ρ dφ and ρ sin φ dθ. Their product gives the volume element dV = ρ² sin φ dρ dφ dθ.

1. Describe the solid with limits on ρ, φ and θ. Spheres give ρ limits, cones give φ limits and half-planes give θ limits.
2. Rewrite the integrand: x² + y² + z² = ρ² and z = ρ cos φ.
3. Multiply by ρ² sin φ.
4. Integrate in ρ, then φ, then θ. With constant limits, you can split the integral into a product of three single integrals.

Triple integrals in spherical coordinates examples

Example 1: Volume of a ball

Use spherical coordinates to find the volume of a ball of radius 3.

0 \le \rho \le 3, \quad 0 \le \varphi \le \pi, \quad 0 \le \theta \le 2\pi

V = \int_0^{2\pi} \int_0^{\pi} \int_0^3 \rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta

\int_0^3 \rho^2\sin\varphi\,d\rho = \sin\varphi\Big[\dfrac{\rho^3}{3}\Big]_0^3

= 9\sin\varphi

\int_0^{\pi} 9\sin\varphi\,d\varphi = 9\Big[-\cos\varphi\Big]_0^{\pi}

= 9(1 + 1) = 18

V = \int_0^{2\pi} 18\,d\theta = 36\pi

Answer: 36π ≈ 113.10. This matches the formula (4/3)π(3)³ = 36π.

Example 2: Integrating the squared distance

Evaluate \iiint_B (x^2 + y^2 + z^2)\,dV, where B is the ball of radius 2 centered at the origin.

(x^2 + y^2 + z^2)\,dV = \rho^2 \cdot \rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta

= \rho^4\sin\varphi\,d\rho\,d\varphi\,d\theta

\int_0^2 \rho^4\,d\rho = \dfrac{32}{5}

\int_0^{\pi} \sin\varphi\,d\varphi = 2

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

\dfrac{32}{5} \cdot 2 \cdot 2\pi = \dfrac{128\pi}{5}

Answer: 128π/5 ≈ 80.42.

Example 3: An ice cream cone

Find the volume of the solid inside the sphere x² + y² + z² = 4 and above the cone z = √(x² + y²).

\rho = 2

\rho\cos\varphi = \rho\sin\varphi

\tan\varphi = 1 \Rightarrow \varphi = \dfrac{\pi}{4}

V = \int_0^{2\pi} \int_0^{\pi/4} \int_0^2 \rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta

\int_0^2 \rho^2\sin\varphi\,d\rho = \dfrac{8}{3}\sin\varphi

\int_0^{\pi/4} \dfrac{8}{3}\sin\varphi\,d\varphi = \dfrac{8}{3}\Big[-\cos\varphi\Big]_0^{\pi/4}

= \dfrac{8}{3}\Big(1 - \dfrac{\sqrt{2}}{2}\Big)

V = 2\pi \cdot \dfrac{8}{3}\Big(1 - \dfrac{\sqrt{2}}{2}\Big)

= \dfrac{16\pi}{3} - \dfrac{8\sqrt{2}\,\pi}{3}

\approx 4.907

Answer: (16π − 8√2 π)/3 ≈ 4.91.

Example 4: A function over a hemisphere

Evaluate \iiint_H z\,dV, where H is the upper half of the ball of radius 1.

0 \le \rho \le 1, \quad 0 \le \varphi \le \dfrac{\pi}{2}, \quad 0 \le \theta \le 2\pi

z\,dV = \rho\cos\varphi \cdot \rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta

\int_0^1 \rho^3\,d\rho = \dfrac{1}{4}

\int_0^{\pi/2} \sin\varphi\cos\varphi\,d\varphi = \Big[\dfrac{\sin^2\varphi}{2}\Big]_0^{\pi/2}

= \dfrac{1}{2}

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

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

Answer: π/4 ≈ 0.785. Dividing by the volume 2π/3 of the hemisphere gives its balance height, 3/8, as in center of mass problems.

Example 5: Mass of a display ball

A museum display ball has radius 10 cm. It is densest at the center: its density is δ = 9 − 0.5ρ grams per cubic centimeter, where ρ is the distance from the center. Find its mass.

m = \int_0^{2\pi} \int_0^{\pi} \int_0^{10} (9 - 0.5\rho)\rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta

(9 - 0.5\rho)\rho^2 = 9\rho^2 - 0.5\rho^3

\int_0^{10} (9\rho^2 - 0.5\rho^3)\,d\rho = \Big[3\rho^3 - \dfrac{\rho^4}{8}\Big]_0^{10}

= 3000 - 1250

= 1750

1750 \int_0^{\pi} \sin\varphi\,d\varphi = 3500

m = 2\pi \cdot 3500

= 7000\pi \approx 21{,}991

Answer: about 21,991 g, or about 22.0 kg. As a check, the density runs from 9 at the center down to 4 at the surface, and the average is 7000π ÷ (4000π/3) = 5.25 g/cm³. That is closer to 4 than to 9, because most of a ball’s volume lies near its surface.

Common mistakes

Related methods

Converting points

From spherical to rectangular, use the three formulas above. From rectangular to spherical, use ρ = √(x² + y² + z²), cos φ = z/ρ, and the same θ as in polar coordinates. For example, (ρ, θ, φ) = (2, 0, π/2) is the point (2, 0, 0) on the x-axis.

Spherical or cylindrical?

Choose spherical coordinates when the solid is part of a ball or a cone with its tip at the origin, or when the integrand contains x² + y² + z². Choose cylindrical coordinates when the solid is round about the z-axis but has flat tops or paraboloids.

Other changes of variables

Cylindrical and spherical coordinates are two cases of a general change of variables. The extra factors r and ρ² sin φ are Jacobians, the topic of the next lesson.

Practice problems

1. Evaluate \int_0^{2\pi} \int_0^{\pi} \int_0^1 \rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta. What does it measure?
2. Evaluate \iiint_B \sqrt{x^2 + y^2 + z^2}\,dV, where B is the ball of radius 2 centered at the origin.
3. Find the volume of the upper half of the ball of radius 3.
4. Find the volume of the spherical shell between the spheres ρ = 1 and ρ = 2.
5. Evaluate \iiint_E z\,dV, where E lies inside the sphere ρ = 2 and above the cone φ = π/3.
6. Evaluate \iiint_B e^{(x^2 + y^2 + z^2)^{3/2}}\,dV, where B is the ball of radius 1.
7. Find the volume of the part of the ball of radius 4 that lies in the first octant.
8. Convert the spherical point (ρ, θ, φ) = (4, π/2, π/6) to rectangular coordinates.
9. Convert the point (x, y, z) = (1, 1, √2) to spherical coordinates.
10. A dome is a concrete half-shell. Its inner radius is 9 m and its outer radius is 10 m. Find the volume of concrete.

\int_0^1 \rho^2\,d\rho = \dfrac{1}{3}

\int_0^{\pi} \sin\varphi\,d\varphi = 2

\dfrac{1}{3} \cdot 2 \cdot 2\pi = \dfrac{4\pi}{3}

\sqrt{x^2 + y^2 + z^2}\,dV = \rho \cdot \rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta

\int_0^2 \rho^3\,d\rho = 4

4 \cdot 2 \cdot 2\pi = 16\pi

\int_0^3 \rho^2\,d\rho = 9

\int_0^{\pi/2} \sin\varphi\,d\varphi = 1

9 \cdot 1 \cdot 2\pi = 18\pi

\int_1^2 \rho^2\,d\rho = \dfrac{8 - 1}{3} = \dfrac{7}{3}

\dfrac{7}{3} \cdot 2 \cdot 2\pi = \dfrac{28\pi}{3}

z\,dV = \rho^3\cos\varphi\sin\varphi\,d\rho\,d\varphi\,d\theta

\int_0^2 \rho^3\,d\rho = 4

\int_0^{\pi/3} \sin\varphi\cos\varphi\,d\varphi = \dfrac{\sin^2(\pi/3)}{2} = \dfrac{3}{8}

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

e^{(x^2 + y^2 + z^2)^{3/2}} = e^{\rho^3}

\int_0^1 e^{\rho^3}\rho^2\,d\rho = \Big[\dfrac{e^{\rho^3}}{3}\Big]_0^1 = \dfrac{e - 1}{3}

\dfrac{e - 1}{3} \cdot 2 \cdot 2\pi = \dfrac{4\pi(e - 1)}{3}

0 \le \varphi \le \dfrac{\pi}{2}, \quad 0 \le \theta \le \dfrac{\pi}{2}

\int_0^4 \rho^2\,d\rho = \dfrac{64}{3}

\dfrac{64}{3} \cdot 1 \cdot \dfrac{\pi}{2} = \dfrac{32\pi}{3}

x = 4\sin\dfrac{\pi}{6}\cos\dfrac{\pi}{2} = 4 \cdot \dfrac{1}{2} \cdot 0 = 0

y = 4\sin\dfrac{\pi}{6}\sin\dfrac{\pi}{2} = 4 \cdot \dfrac{1}{2} \cdot 1 = 2

z = 4\cos\dfrac{\pi}{6} = 4 \cdot \dfrac{\sqrt{3}}{2} = 2\sqrt{3}

\rho = \sqrt{1 + 1 + 2} = 2

\cos\varphi = \dfrac{\sqrt{2}}{2} \Rightarrow \varphi = \dfrac{\pi}{4}

\tan\theta = \dfrac{1}{1} \Rightarrow \theta = \dfrac{\pi}{4}

V = \int_0^{2\pi} \int_0^{\pi/2} \int_9^{10} \rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta

\int_9^{10} \rho^2\,d\rho = \dfrac{1000 - 729}{3} = \dfrac{271}{3}

\dfrac{271}{3} \cdot 1 \cdot 2\pi = \dfrac{542\pi}{3} \approx 567.6

Triple integrals in spherical coordinates FAQ

Why is dV equal to ρ² sin φ dρ dφ dθ?

A small spherical box has edges dρ, ρ dφ and ρ sin φ dθ. The last edge is an arc of a circle of radius ρ sin φ, the distance to the z-axis. Multiplying the three edges gives ρ² sin φ dρ dφ dθ.

What are the limits for a sphere in spherical coordinates?

For a ball of radius a centered at the origin, use 0 ≤ ρ ≤ a, 0 ≤ φ ≤ π and 0 ≤ θ ≤ 2π. For the upper half, stop φ at π/2.

Why does φ only go from 0 to π?

φ = 0 points straight up and φ = π points straight down. Together with a full turn in θ, that already reaches every direction.

When should I use spherical coordinates?

Use them for balls, spherical shells and cones with their tip at the origin, or when the integrand contains x² + y² + z².

Summary

• Spherical coordinates: x = ρ sin φ cos θ, y = ρ sin φ sin θ, z = ρ cos φ.
• ρ is the distance from the origin, φ runs from 0 to π down from the z-axis, and θ runs around the z-axis.
• The volume element is dV = ρ² sin φ dρ dφ dθ.
• Spheres are ρ = a, and cones are φ = constant.
• With constant limits, split the integral into a product of three single integrals.

Previous lesson: Triple Integrals in Cylindrical Coordinates

Next lesson: Jacobians and Change of Variables

Back to the Calculus 3 Index