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Cylindrical Triple Integrals · Tutorial 633 of 999

Volumes of Circular and Annular Cylinders

Learn to represent cylindrical solids with radial, angular, and height bounds and verify their volumes using familiar geometric formulas.

Intermediate 9 min read

What You'll Learn

  • Set up a triple integral for the volume of a full circular cylinder.
  • Use angular bounds to represent a cylindrical sector.
  • Use inner and outer radii to describe an annular cylinder.
  • Integrate the cylindrical volume element with the correct radial factor.
  • Check triple-integral results against geometric volume formulas.

Why Cylindrical Solids Are Natural for Triple Integrals

A circular cylinder is described by a fixed radius around an axis and a range of heights. In cylindrical coordinates, those features become direct bounds: the radius varies between its inner and outer limits, the angle sweeps around the axis, and the height runs between two horizontal planes.

This makes cylindrical triple integrals a natural way to calculate the volume of full cylinders, sectors of cylinders, and hollow cylinders. The key is to use the cylindrical volume element \(r\,dz\,dr\,d\theta\), rather than treating a small piece of the solid as though it had rectangular volume.

Key idea: For a solid described by cylindrical bounds, its volume is the triple integral of \(1\) over the solid. In cylindrical coordinates, the volume element is \(r\,dz\,dr\,d\theta\).

Bounds and Volume Formulas

Let a cylindrical solid have constant lower and upper heights \(z_0\) and \(z_1\), with \(z_1>z_0\). Its height is \(h=z_1-z_0\). Suppose its radial bounds are \(a\le r\le R\), where \(0\le a<R\), and its angular bounds are \(\alpha\le\theta\le\beta\), where \(0<\beta-\alpha\le 2\pi\). The angular width is \(\Delta\theta=\beta-\alpha\).

The volume is found by integrating \(1\) across these bounds. The factor \(r\) in the integrand comes from the cylindrical volume element: a thin piece at radius \(r\) has greater circumferential width than an otherwise identical piece near the axis.

$$ V = \int_{\alpha}^{\beta}\int_a^R\int_{z_0}^{z_1} r\,dz\,dr\,d\theta. $$

Evaluating the height integral gives the factor \(h\). The radial integral gives half the difference of the squared radii, and the angular integral gives the angular width. Thus the general formula is:

Theorem: A cylindrical solid with constant height \(h\), radial bounds \(a\le r\le R\), and angular width \(\Delta\theta\) has volume $$ V=\frac{\Delta\theta}{2}(R^2-a^2)h. $$ This assumes the bounds describe the solid once, with no overlapping angular sweep.

Several familiar cases follow from this formula. A full cylinder has \(a=0\) and \(\Delta\theta=2\pi\). A sector-shaped cylinder has \(a=0\) and an angular width smaller than \(2\pi\). A full annular cylinder has \(a>0\) and \(\Delta\theta=2\pi\).

Deriving the Formula from the Triple Integral

The formula follows by integrating the volume element across the bounds. Since the bounds are constant in each coordinate, the integral can be evaluated one variable at a time. The following calculation also shows exactly where the height, angular width, and squared radii enter.

1
Integrate through the height.
Because the integrand \(r\) is constant with respect to \(z\), integration from \(z_0\) to \(z_1\) multiplies \(r\) by \(z_1-z_0=h\).
2
Integrate across the radial interval.
The antiderivative of \(r\) is \(r^2/2\), so the radial contribution is \(h(R^2-a^2)/2\).
3
Integrate across the angle.
The remaining expression does not depend on \(\theta\), so integrating from \(\alpha\) to \(\beta\) multiplies it by \(\beta-\alpha=\Delta\theta\).
$$ \begin{aligned} V &=\int_{\alpha}^{\beta}\int_a^R\int_{z_0}^{z_1} r\,dz\,dr\,d\theta\\ &=\int_{\alpha}^{\beta}\int_a^R r(z_1-z_0)\,dr\,d\theta\\ &=h\int_{\alpha}^{\beta}\left[\frac{r^2}{2}\right]_a^R\,d\theta\\ &=\int_{\alpha}^{\beta}\frac{h}{2}(R^2-a^2)\,d\theta\\ &=\frac{\Delta\theta}{2}(R^2-a^2)h. \end{aligned} $$

This derivation is also a check on the role of the Jacobian factor \(r\). Without it, the radial integration would not produce the difference of squared radii required for circular cross-sections.

Worked Example: A Full Circular Cylinder

Worked Example: A Full Circular Cylinder

Find the volume of the solid with radius \(3\), extending from \(z=1\) to \(z=6\), and verify the result with the geometric cylinder formula.

The height is \(h=6-1=5\). For a full cylinder, the radial bounds are \(0\le r\le3\), and the angle makes one complete turn, \(0\le\theta\le2\pi\). Set up the volume integral:

$$ V=\int_0^{2\pi}\int_0^3\int_1^6 r\,dz\,dr\,d\theta. $$

First integrate with respect to \(z\):

$$ \int_1^6 r\,dz=r(6-1)=5r. $$

Then integrate with respect to \(r\):

$$ \int_0^3 5r\,dr =5\left[\frac{r^2}{2}\right]_0^3 =5\left(\frac{9}{2}-0\right) =\frac{45}{2}. $$

Finally, integrate with respect to \(\theta\):

$$ V=\int_0^{2\pi}\frac{45}{2}\,d\theta =\frac{45}{2}(2\pi) =45\pi. $$

Geometrically, the base area is \(\pi(3)^2=9\pi\), and the height is \(5\). Therefore \(V=9\pi\cdot5=45\pi\), matching the triple integral.

Worked Example: A Sector-Shaped Cylinder

Worked Example: A Sector-Shaped Cylinder

Find the volume of a cylinder sector of radius \(4\), height \(3\), and angular bounds \(\pi/6\le\theta\le\pi/2\). Verify the answer using the area of a circular sector.

The radial bounds are \(0\le r\le4\), and the height bounds can be taken as \(0\le z\le3\). The angular width is:

$$ \Delta\theta=\frac{\pi}{2}-\frac{\pi}{6} =\frac{3\pi}{6}-\frac{\pi}{6} =\frac{\pi}{3}. $$

Set up and evaluate the triple integral:

$$ \begin{aligned} V &=\int_{\pi/6}^{\pi/2}\int_0^4\int_0^3 r\,dz\,dr\,d\theta\\ &=\int_{\pi/6}^{\pi/2}\int_0^4 3r\,dr\,d\theta\\ &=\int_{\pi/6}^{\pi/2}3\left[\frac{r^2}{2}\right]_0^4\,d\theta\\ &=\int_{\pi/6}^{\pi/2}3\left(\frac{16}{2}\right)\,d\theta\\ &=\int_{\pi/6}^{\pi/2}24\,d\theta\\ &=24\left(\frac{\pi}{2}-\frac{\pi}{6}\right) =24\left(\frac{\pi}{3}\right) =8\pi. \end{aligned} $$

As a geometric check, the base sector has area \(\frac12(4)^2(\pi/3)=8\pi/3\). Multiplying by height \(3\) gives \((8\pi/3)(3)=8\pi\), the same volume.

Worked Example: A Full Annular Cylinder

Worked Example: A Full Annular Cylinder

A hollow cylindrical solid has inner radius \(1\), outer radius \(3\), and height \(2\). Find its volume using a triple integral and verify it by subtracting the volume of the hollow center.

The radial bounds are \(1\le r\le3\), the angular bounds are \(0\le\theta\le2\pi\), and the height bounds are \(0\le z\le2\). Integrate:

$$ \begin{aligned} V &=\int_0^{2\pi}\int_1^3\int_0^2 r\,dz\,dr\,d\theta\\ &=\int_0^{2\pi}\int_1^3 2r\,dr\,d\theta\\ &=\int_0^{2\pi}2\left[\frac{r^2}{2}\right]_1^3\,d\theta\\ &=\int_0^{2\pi}(9-1)\,d\theta\\ &=\int_0^{2\pi}8\,d\theta\\ &=8(2\pi)=16\pi. \end{aligned} $$

For the geometric check, the cross-sectional area is the outer disk area minus the inner disk area: \(9\pi-1\pi=8\pi\). Multiplying by height \(2\) gives \(8\pi\cdot2=16\pi\), in agreement with the integral.

One Formula, Three Geometric Cases

The general volume formula makes it easy to distinguish the common cases. The bounds—not a new integration method—encode whether the solid is full, sector-shaped, or hollow.

Solid Radial bounds Angular bounds Volume
Full cylinder \(0\le r\le R\) Width \(2\pi\) \(\pi R^2h\)
Cylinder sector \(0\le r\le R\) Width \(\Delta\theta\) \(\frac12\Delta\theta R^2h\)
Full annular cylinder \(a\le r\le R\) Width \(2\pi\) \(\pi(R^2-a^2)h\)
Annular sector \(a\le r\le R\) Width \(\Delta\theta\) \(\frac12\Delta\theta(R^2-a^2)h\)

For a full cylinder, setting \(a=0\) and \(\Delta\theta=2\pi\) gives \(\frac{2\pi}{2}R^2h=\pi R^2h\). For a full annular cylinder, keeping \(a>0\) gives \(\pi(R^2-a^2)h\), which is the difference between the outer and inner cylinder volumes.

A Common Mistake: Omitting the Radial Factor

A frequent error is to integrate \(1\) over \(dz\,dr\,d\theta\) instead of using the cylindrical volume element \(r\,dz\,dr\,d\theta\). For the full cylinder of radius \(3\) and height \(5\), that incorrect calculation would give:

$$ \int_0^{2\pi}\int_0^3\int_1^6 1\,dz\,dr\,d\theta =(5)(3)(2\pi)=30\pi. $$

This is wrong: it treats the radial coordinate as if equal radial increments represented equal-sized pieces all around the circle. But a thin ring at radius \(r\) has circumference proportional to \(r\), so its volume contribution must include the factor \(r\). The corrected calculation is:

$$ \int_0^{2\pi}\int_0^3\int_1^6 r\,dz\,dr\,d\theta =45\pi, $$

which agrees with the geometric formula \(\pi(3)^2(5)=45\pi\). A useful check is therefore to compare the integral result with the expected area of the circular or annular cross-section multiplied by the height.

Checking Units and Bounds

When the radii and height are measured in units of length, \(R^2-a^2\) has units of length squared. The angular width is dimensionless, so the formula \(\frac12\Delta\theta(R^2-a^2)h\) has units of length cubed, as a volume should.

Before evaluating, check that the radial bounds have the inner radius first and outer radius second, the height bounds run from the lower plane to the upper plane, and the angular interval covers exactly the intended sector. For a complete revolution its width is \(2\pi\); a smaller interval describes only part of the circular sweep.

Takeaway: Write the bounds from the geometry, include the factor \(r\), and integrate \(1\) over the solid. Then check the result against the appropriate geometric cross-sectional area times height.

Check Your Understanding

Use the cylindrical volume element and, where possible, check each result with a geometric volume formula.

  1. Set up a triple integral for a full cylinder of radius \(2\) and height \(7\). What volume does it give?
  2. A cylindrical sector has radius \(5\), height \(2\), and angular width \(\pi/4\). Find its volume.
  3. Write the radial, angular, and height bounds for a full annular cylinder with inner radius \(2\), outer radius \(4\), and height \(3\).
  4. Explain why integrating \(1\) over \(dr\,d\theta\,dz\) does not give the volume in cylindrical coordinates.
  5. For an annular sector, identify which part of the formula accounts for the hollow center and which part accounts for the angular restriction.

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