Triple Integrals in Cylindrical and Spherical Coordinates

In this lesson you’ll learn how to swap out Cartesian coordinates for cylindrical or spherical coordinates when setting up triple integrals. The payoff is huge: integrals that would be nightmarish in $x, y, z$ become clean and manageable when you pick the coordinate system that matches the shape of your region.

In the last lesson, every triple integral used Cartesian coordinates with $dx \, dy \, dz$. That works fine for boxes and regions with flat sides. But try integrating over a sphere or a cylinder in Cartesian and you’ll quickly run into ugly square roots and complicated limits.

The fix is the same idea from Lesson 10 (polar coordinates for double integrals), just extended to 3D. If the region has circular symmetry, use coordinates that speak that language.

Cylindrical coordinates are polar coordinates in the $xy$-plane plus a regular $z$-axis. Every point in 3D is described by three numbers:

• r is the distance from the z-axis (same as polar r)
• theta is the angle in the $xy$-plane measured from the positive x-axis
• z is the same old height

The conversion formulas are

$$x = r\cos\theta, \quad y = r\sin\theta, \quad z = z$$

The volume element picks up an extra factor of $r$ (just like polar area had $r \, dr \, d\theta$):

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

That $r$ out front is the Jacobian of the coordinate transformation. It accounts for the fact that a small change in $\theta$ sweeps out more area when $r$ is large than when $r$ is small.

Use cylindrical coordinates when the region has rotational symmetry around the $z$-axis: cylinders, cones, paraboloids, anything where $x^2 + y^2$ shows up naturally.

Spherical coordinates describe a point by its distance from the origin and two angles:

• $\rho$ (rho) is the distance from the origin to the point
• $\varphi$ (phi) is the angle down from the positive z-axis (0 at the north pole, $\pi$ at the south pole)
• theta is the same azimuthal angle as in cylindrical (angle in the $xy$-plane)

The symbol $\rho$ is the Greek letter rho. The symbol $\varphi$ is the Greek letter phi. Think of $\rho$ as “how far from the center” and $\varphi$ as “how far down from the top.”

The conversion formulas are

$$x = \rho\sin\varphi\cos\theta, \quad y = \rho\sin\varphi\sin\theta, \quad z = \rho\cos\varphi$$

The volume element in spherical coordinates is

$$dV = \rho^2 \sin\varphi \, d\rho \, d\varphi \, d\theta$$

That $\rho^2 \sin\varphi$ factor is the Jacobian. It accounts for the fact that volume elements get bigger as you move farther from the origin (the $\rho^2$ part) and are largest at the equator where sin phi = 1 (the sin phi part).

Use spherical coordinates when the region involves $x^2 + y^2 + z^2$ or is bounded by spheres, cones, or other surfaces centered at the origin.

• Cartesian ($dx \, dy \, dz$): boxes, rectangular regions, flat boundaries
• Cylindrical ($r \, dr \, d\theta \, dz$): cylinders, cones, regions symmetric about the $z$-axis
• Spherical ($\rho^2 \sin\varphi \, d\rho \, d\varphi \, d\theta$): spheres, balls, regions defined by distance from the origin

Example 1: Volume of a cylinder using cylindrical coordinates

Find the volume of the solid cylinder where $x^2 + y^2 \leq 4$ and $0 \leq z \leq 5$

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This is a cylinder of radius 2 and height 5 centered on the z-axis. In cylindrical coordinates, the boundary $x^2 + y^2 = 4$ becomes simply r = 2. The limits are:

• r from 0 to 2
• theta from 0 to $2\pi$ (full revolution)
• z from 0 to 5

$$V = \int_0^{2\pi} \int_0^2 \int_0^5 r \, dz \, dr \, d\theta$$

Innermost integral, over z, treating r and theta as constants

$$\int_0^5 r \, dz = 5r$$

Middle integral, over r

$$\int_0^2 5r \, dr = 5 \cdot \frac{r^2}{2}\bigg|_0^2 = 5 \cdot 2 = 10$$

Outermost integral, over theta

$$\int_0^{2\pi} 10 \, d\theta = 20\pi$$

The volume is $20\pi \approx 62.83$ cubic units. Sanity check: the formula for a cylinder gives $\pi r^2 h = \pi(4)(5) = 20\pi,$ and it matches.

In Cartesian, this same integral would require limits like $x$ from $-2$ to $2$, $y$ from $-\sqrt{4 - x^2}$ to $\sqrt{4 - x^2}$, and you’d be wrestling with square roots the whole time. Cylindrical coordinates made it trivial.

Example 2: Volume of the unit ball using spherical coordinates

Find the volume of the unit ball where $x^2 + y^2 + z^2 \leq 1$

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The unit ball is the set of all points within distance 1 of the origin. In spherical coordinates, the boundary $x^2 + y^2 + z^2 = 1$ becomes simply rho = 1. Recall that rho is the distance from the origin, phi is the angle down from the north pole, and theta is the angle around the z-axis. The limits are:

• rho from 0 to 1 (distance from origin to the surface)
• phi from 0 to $\pi$ (top of the sphere to the bottom)
• theta from 0 to $2\pi$ (full revolution around the z-axis)

$$V = \int_0^{2\pi} \int_0^{\pi} \int_0^1 \rho^2 \sin\varphi \, d\rho \, d\varphi \, d\theta$$

The integrand factors nicely because the limits are all constants. We can split this into three separate integrals:

$$V = \left(\int_0^{2\pi} d\theta\right) \left(\int_0^{\pi} \sin\varphi \, d\varphi\right) \left(\int_0^1 \rho^2 \, d\rho\right)$$

Each piece:

$$\int_0^{2\pi} d\theta = 2\pi$$ $$\int_0^{\pi} \sin\varphi \, d\varphi = [-\cos\varphi]_0^{\pi} = -\cos\pi + \cos 0 = 1 + 1 = 2$$ $$\int_0^1 \rho^2 \, d\rho = \frac{\rho^3}{3}\bigg|_0^1 = \frac{1}{3}$$

Multiply them together:

$$V = 2\pi \cdot 2 \cdot \frac{1}{3} = \frac{4\pi}{3}$$

The volume is $\frac{4\pi}{3} \approx 4.19$ cubic units. That’s the classic sphere volume formula $\frac{4}{3}\pi r^3$ with r = 1. Spherical coordinates turned this into three easy one-variable integrals.

Example 3: Mass of a ball with varying density

Find the total mass of a ball of radius 2 whose density at any point equals the distance from the center. So the density function in spherical coordinates is simply f = rho (since rho already measures distance from the origin).

The limits cover the full ball of radius 2. Again, rho is distance from the origin, phi is the angle down from the north pole, and theta sweeps around the z-axis:

• rho from 0 to 2
• phi from 0 to $\pi$
• theta from 0 to $2\pi$

$$M = \int_0^{2\pi} \int_0^{\pi} \int_0^2 \rho \cdot \rho^2 \sin\varphi \, d\rho \, d\varphi \, d\theta$$

The density rho times the volume element $\rho^2 \sin\varphi$ gives $\rho^3 \sin\varphi$ as the integrand. Again it separates:

$$M = \left(\int_0^{2\pi} d\theta\right) \left(\int_0^{\pi} \sin\varphi \, d\varphi\right) \left(\int_0^2 \rho^3 \, d\rho\right)$$ $$\int_0^{2\pi} d\theta = 2\pi$$ $$\int_0^{\pi} \sin\varphi \, d\varphi = 2$$ $$\int_0^2 \rho^3 \, d\rho = \frac{\rho^4}{4}\bigg|_0^2 = \frac{16}{4} = 4$$ $$M = 2\pi \cdot 2 \cdot 4 = 16\pi$$

Total mass is 16 pi. Notice how the density being a function of rho alone made this clean. If the density depended on x and y separately, Cartesian coordinates would have been a mess, but spherical coordinates kept everything factored.

Cylindrical and spherical coordinates show up everywhere in science and engineering:

• Physics uses spherical coordinates for gravitational and electric fields, which radiate outward from a point. Computing the total gravitational potential energy of a star requires integrating density over a sphere.
• Engineering uses cylindrical coordinates for heat flow through pipes, pressure distribution in cylindrical tanks, and stress analysis in shafts and axles.
• Game development uses spherical coordinates for explosion damage falloff (damage decreases with distance from the blast center), 3D sound attenuation, and rendering spherical particle effects.
• Medical imaging uses these coordinates when analyzing CT scans of roughly cylindrical body parts (limbs) or spherical structures (the skull, eyeballs).