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Reducing Rotationally Symmetric Integrals to One Dimension

Use rotational symmetry to eliminate the angular variable and simplify cylindrical volume, mass, and moment integrals.

Intermediate 9 min read

What You'll Learn

  • Identify when a solid and integrand are independent of the angular coordinate
  • Derive the factor of two pi from integration over a full revolution
  • Reduce cylindrical triple integrals to radial and vertical integrations
  • Apply the reduction to volume, variable-density mass, and axial moment
  • Recognize when angular reduction is invalid

Why Rotational Symmetry Simplifies an Integral

A cylindrical triple integral often includes three integrations: one in the radial coordinate, one in the angular coordinate, and one in the vertical coordinate. If both the solid and the quantity being integrated are unchanged by rotation about the \(z\)-axis, every angular position contributes the same amount. The angular integration can then be performed immediately, leaving an integral over the radial and vertical coordinates.

This reduction is useful for volume, mass, and moments of inertia. Its central factor is \(2\pi\), the total angle in one complete revolution. The factor applies only when the region covers the full rotation and the integrand has no angular dependence.

Rotational Symmetry in Cylindrical Coordinates

In cylindrical coordinates, a point is described by \(r\), \(\theta\), and \(z\), where \(r\) is distance from the \(z\)-axis and \(\theta\) is the angle around that axis. The volume element is \(dV=r\,dr\,d\theta\,dz\). The factor \(r\) is the Jacobian factor in the cylindrical volume element.

Definition: A solid is rotationally symmetric about the \(z\)-axis if rotating it by any angle about that axis leaves the solid unchanged. A scalar integrand is rotationally symmetric about the \(z\)-axis if its cylindrical-coordinate expression is independent of \(\theta\).

For an integral to reduce by angular symmetry, the solid must include the full circle of angles at each radial and vertical location in its projection. Its bounds may vary with \(r\) and \(z\), but they must not select only a wedge or sector. The integrand must also have no \(\theta\)-dependence. These conditions allow us to describe the solid by a region \(D\) in the \(rz\)-plane and use the full angular interval \(0\leq\theta\leq2\pi\).

Theorem: Suppose \(E\) is a rotationally symmetric solid described in cylindrical coordinates by \((r,z)\in D\) and \(0\leq\theta\leq2\pi\). If \(g(r,z)\) is integrable on \(E\) and is independent of \(\theta\), then $$ \iiint_E g(r,z)\,dV = 2\pi\iint_D g(r,z)\,r\,dr\,dz. $$ The region \(D\) can be integrated in either order when the corresponding bounds describe it correctly.

The reduction removes the angular integration; it does not automatically remove both of the other integrations. In general, the result is a double integral in \(r\) and \(z\). If one of those integrations can be evaluated directly from the bounds, the remaining calculation may be a single integral.

Why the Angular Factor Is \(2\pi\)

Start with the cylindrical-coordinate expression for the integral. Since \(g\) is independent of \(\theta\), and the angular bounds cover a complete revolution, the integrand and the other bounds do not change as \(\theta\) varies. The iterated integral therefore separates into an angular factor and an integral over \(D\).

$$ \begin{aligned} \iiint_E g(r,z)\,dV &= \iint_D\int_0^{2\pi}g(r,z)\,r\,d\theta\,dr\,dz\\ &= \iint_D g(r,z)\,r \left(\int_0^{2\pi}d\theta\right)\,dr\,dz\\ &= 2\pi\iint_D g(r,z)\,r\,dr\,dz. \end{aligned} $$

This argument uses the cylindrical volume element and the stated symmetry conditions. The \(2\pi\) comes from the angular interval, while the \(r\) remains because it is part of the volume element. These are distinct factors with distinct sources.

Volume of a Rotationally Symmetric Solid

Volume is the triple integral of the constant function \(1\). Thus, for a solid satisfying the theorem’s conditions, set \(g(r,z)=1\). The reduced formula is

$$ \operatorname{Vol}(E)=2\pi\iint_D r\,dr\,dz. $$

For instance, if the solid is described by \(0\leq r\leq R\) and \(a\leq z\leq b\), then its radial and vertical bounds form a rectangle in the \(rz\)-plane. The integral can be evaluated in either order.

Worked Example: Volume of a Cylinder

Find the volume of the cylinder \(0\leq r\leq2\), \(0\leq z\leq3\), with the full angular range.

The solid is rotationally symmetric, and the volume integrand \(1\) is independent of \(\theta\). Apply the reduced volume formula:

$$ \begin{aligned} \operatorname{Vol}(E) &=2\pi\int_0^3\int_0^2 r\,dr\,dz\\ &=2\pi\int_0^3\left[\frac{r^2}{2}\right]_0^2dz\\ &=2\pi\int_0^3\left(\frac{4}{2}-0\right)dz\\ &=2\pi\int_0^3 2\,dz\\ &=4\pi[z]_0^3\\ &=4\pi(3-0)\\ &=12\pi. \end{aligned} $$

As a check, the base disk has area \(\pi(2)^2=4\pi\), and the height is \(3\), so the geometric volume is \(4\pi\cdot3=12\pi\), matching the integral.

Mass and Moment Integrals

If a solid has density \(\rho(r,z)\) independent of \(\theta\), its mass is the integral of density over the solid. Under the same symmetry conditions, this becomes

$$ M=\iiint_E\rho(r,z)\,dV =2\pi\iint_D\rho(r,z)\,r\,dr\,dz. $$

For a moment of inertia about the \(z\)-axis, each mass element is weighted by the square of its distance from that axis. In cylindrical coordinates, that distance is \(r\), so the weighting factor is \(r^2\). If the density and solid are rotationally symmetric, the resulting formula is

$$ I_z=\iiint_E r^2\rho(r,z)\,dV =2\pi\iint_D r^2\rho(r,z)\,r\,dr\,dz =2\pi\iint_D r^3\rho(r,z)\,dr\,dz. $$

The \(r^3\) in the last expression combines two factors: \(r^2\) from the squared distance in the definition of \(I_z\), and \(r\) from the cylindrical volume element. Keeping those sources separate helps prevent errors.

Worked Example: Mass With Radially Varying Density

A solid occupies \(1\leq r\leq2\), \(0\leq z\leq3\), and \(0\leq\theta\leq2\pi\). Its density is \(\rho(r,z)=2+r^2\). Find its mass.

The region covers a full revolution, and the density is independent of \(\theta\), so use the reduced mass formula:

$$ \begin{aligned} M &=2\pi\int_0^3\int_1^2(2+r^2)r\,dr\,dz\\ &=2\pi\int_0^3\int_1^2(2r+r^3)\,dr\,dz\\ &=2\pi\int_0^3\left[r^2+\frac{r^4}{4}\right]_1^2dz\\ &=2\pi\int_0^3\left(\left(4+\frac{16}{4}\right) -\left(1+\frac{1}{4}\right)\right)dz\\ &=2\pi\int_0^3\left(8-\frac{5}{4}\right)dz\\ &=2\pi\int_0^3\frac{27}{4}\,dz\\ &=2\pi\cdot\frac{27}{4}(3-0)\\ &=\frac{81\pi}{2}. \end{aligned} $$

To check the integration, differentiating \(r^2+\frac{r^4}{4}\) gives \(2r+r^3=(2+r^2)r\), the radial mass integrand. The value is positive, as expected for a positive density and a solid of positive volume.

Worked Example: Moment of Inertia About the Axis

A solid cylinder has radius \(1\), height \(2\), and constant density \(\rho=3\). Find its moment of inertia about the \(z\)-axis.

For this moment, the integrand is \(r^2\rho=3r^2\). Including the cylindrical volume factor \(r\), the reduced integral is

$$ \begin{aligned} I_z &=2\pi\int_0^2\int_0^1 3r^2\,r\,dr\,dz\\ &=6\pi\int_0^2\int_0^1 r^3\,dr\,dz\\ &=6\pi\int_0^2\left[\frac{r^4}{4}\right]_0^1dz\\ &=6\pi\int_0^2\frac{1}{4}\,dz\\ &=\frac{3\pi}{2}[z]_0^2\\ &=\frac{3\pi}{2}(2-0)\\ &=3\pi. \end{aligned} $$

As a check, the mass is \(3\cdot\pi(1)^2\cdot2=6\pi\). The mean value of \(r^2\) over a uniform disk of radius \(1\) is \(1/2\), so the corresponding axial moment is \(6\pi\cdot\frac12=3\pi\), agreeing with the integral.

When the Reduction Does Not Apply

The factor \(2\pi\) is not a general shortcut for every cylindrical integral. It requires both a full angular range and an integrand independent of \(\theta\). If either condition fails, the angular dependence must be handled rather than replaced automatically by \(2\pi\).

  • A sector or wedge: If the angular interval is \(\alpha\leq\theta\leq\beta\) and the integrand is independent of \(\theta\), the angular factor is \(\beta-\alpha\), not \(2\pi\).
  • An angularly varying integrand: If the integrand contains \(\theta\), \(\cos\theta\), or another angular factor, it must be integrated over the actual angular interval.
  • A non-rotationally symmetric solid: If the allowed angles change with \(r\) or \(z\), the angular bounds cannot simply be treated as a full revolution.

Common Mistake: Using \(2\pi\) for a Wedge

Consider a wedge with \(0\leq r\leq1\), \(0\leq z\leq1\), and \(0\leq\theta\leq\frac{\pi}{2}\). A mistaken calculation treats it as a full rotation:

$$ \text{Incorrect: }\quad 2\pi\int_0^1\int_0^1 r\,dr\,dz =2\pi\cdot\frac12\cdot1 =\pi. $$

But the solid covers only a quarter-turn. The correct angular integral is \(\int_0^{\pi/2}d\theta=\frac{\pi}{2}\), so its volume is

$$ \begin{aligned} \operatorname{Vol}(E) &=\int_0^1\int_0^1\int_0^{\pi/2}r\,d\theta\,dr\,dz\\ &=\frac{\pi}{2}\int_0^1\int_0^1r\,dr\,dz\\ &=\frac{\pi}{2}\cdot\frac12\cdot1\\ &=\frac{\pi}{4}. \end{aligned} $$

The correction is to use the actual angular width. The incorrect answer is four times the correct one because it counted four quarter-turns instead of one.

A Reliable Reduction Checklist

Before removing the angular integration, verify the geometry and the integrand separately. This short check makes the method safe to use even when the radial and vertical bounds are complicated.

1
Check the angular range.
Confirm that the solid includes \(0\leq\theta\leq2\pi\) at each location represented in its \(rz\)-region.
2
Check the integrand.
Rewrite it in cylindrical coordinates and confirm that it contains no \(\theta\)-dependence.
3
Keep the volume factor.
After integrating over angle, retain the \(r\) from \(dV=r\,dr\,d\theta\,dz\).
4
Integrate over the \(rz\)-region.
Use radial and vertical bounds that describe the solid, then evaluate the resulting double integral.

Check Your Understanding

Use the symmetry conditions, not just the appearance of cylindrical coordinates, to decide how each integral should be handled.

  1. What two conditions permit an angular integral to be replaced by a factor of \(2\pi\)?
  2. Why does the factor \(r\) remain after the angular integration?
  3. A rotationally symmetric solid has density \(\rho(r,z)=1+r\). Write its reduced mass integral over an \(rz\)-region \(D\).
  4. For an axial moment of inertia, explain why the reduced integrand includes \(r^3\rho(r,z)\).
  5. What angular factor should replace \(2\pi\) for an integrand independent of \(\theta\) over \(\alpha\leq\theta\leq\beta\)?

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