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Cylindrical Triple Integrals · Tutorial 628 of 999

Radial Bounds From Coaxial Cylinders

Translate coaxial cylindrical walls into correct radial bounds and use them to set up triple integrals over solid and hollow cylindrical regions.

Intermediate 8 min read

What You'll Learn

  • Identify the radial coordinate of a cylinder centered on the z-axis.
  • Translate inside, outside, and between-cylinder conditions into radial inequalities.
  • Set up integrals over solid cylinders and hollow cylinders.
  • Represent cylindrical shells with inner and outer radial bounds.
  • Split a region into radial pieces when its bounds change.
  • Check radial limits against the original cylindrical walls.

Why Cylindrical Walls Give Radial Bounds

A cylindrical wall centered on the \(z\)-axis consists of points whose distance from that axis is constant. In cylindrical coordinates, that distance is \(r\). Consequently, the wall of radius \(R\) is described by \(r=R\), and its inside is described by \(0\le r\le R\).

This makes radial bounds especially direct when a solid is enclosed by coaxial cylinders: the cylinders have the same axis, so their radii can be compared using intervals on the \(r\)-axis. The inner and outer walls determine the radial limits. They do not, by themselves, specify how high the solid extends or what angles it occupies; those limits must come from the other boundaries of the region.

Definition: A cylinder of radius \(R\) coaxial with the \(z\)-axis is the surface \(x^2+y^2=R^2\), or equivalently \(r=R\), where \(R>0\). Its interior consists of points with \(0\le r<R\). A cylindrical shell between radii \(a\) and \(b\), where \(0<a<b\), consists of points with \(a\le r\le b\).

We use the cylindrical volume element \(r\,dz\,dr\,d\theta\). Thus, for a region whose limits are \(a\le r\le b\), \(0\le\theta\le2\pi\), and \(h_1(r,\theta)\le z\le h_2(r,\theta)\), an integral of a function \(g\) written in cylindrical coordinates is set up as

$$ \iiint_E g\,dV = \int_0^{2\pi}\int_a^b\int_{h_1(r,\theta)}^{h_2(r,\theta)} g(r,\theta,z)\,r\,dz\,dr\,d\theta. $$

The factor \(r\) belongs to the volume element; it is not an additional radial bound. The radial limits describe which distances from the axis occur in the solid. For a complete cylinder or shell, the full angular range is \(0\le\theta\le2\pi\). A restricted angular range describes only part of the cylinder, so it must be justified by the region’s other boundaries.

Deriving the Radial Inequalities

The translation from cylindrical walls to radial bounds follows from the definition \(r=\sqrt{x^2+y^2}\), with \(r\ge0\). Squaring or taking square roots of nonnegative quantities preserves the relevant order. This gives the standard conversions below.

1
Inside a cylinder.
If \(x^2+y^2\le R^2\), then \(r^2\le R^2\). Since \(r\ge0\) and \(R>0\), this is equivalent to \(0\le r\le R\).
2
Outside a cylinder.
If \(x^2+y^2\ge a^2\), where \(a>0\), then \(r^2\ge a^2\), which is equivalent to \(r\ge a\).
3
Between coaxial cylinders.
If \(a^2\le x^2+y^2\le b^2\), where \(0<a<b\), then \(a\le r\le b\). The smaller radius is the lower radial limit; the larger radius is the upper limit.

These equivalences justify the radial bounds rather than merely suggesting them. For example, when \(a\le r\le b\), squaring gives \(a^2\le r^2\le b^2\). Since \(r^2=x^2+y^2\), every point in the interval lies between the two cylinder walls. Conversely, any point between those walls satisfies the squared inequality and therefore has a radius in that same interval.

Theorem: If a region is inside the coaxial cylinder \(r=b\) and outside the coaxial cylinder \(r=a\), with \(0\le a<b\), its radial interval is \(a\le r\le b\). The value \(a=0\) gives a solid cylinder; \(a>0\) gives a hollow cylinder. The other limits of integration must still describe the region’s vertical and angular extent.

Worked Example: Setting Up an Integral Over a Solid Cylinder

Worked Example: Setting Up an Integral Over a Solid Cylinder

Let \(E\) be the complete solid cylinder of radius \(3\), between the horizontal planes \(z=-1\) and \(z=4\). Set up an integral of \(g(r,\theta,z)\) over \(E\).

The lateral wall has radius \(3\), so \(r=3\) is the outer radial boundary. The solid includes the axis, so its radial interval starts at \(r=0\). It is a complete cylinder, so \(\theta\) covers one full turn, from \(0\) to \(2\pi\). The planes give \(-1\le z\le4\).

$$ 0\le r\le3,\qquad 0\le\theta\le2\pi,\qquad -1\le z\le4. $$

Using the volume element \(r\,dz\,dr\,d\theta\), the required setup is

$$ \iiint_E g\,dV = \int_0^{2\pi}\int_0^3\int_{-1}^4 g(r,\theta,z)\,r\,dz\,dr\,d\theta. $$

Check: At the axis, \(r=0\), and at the lateral wall, \(r=3\), so the radial interval reaches exactly from the axis to the specified cylinder. The vertical endpoints are the two given planes, and the angular endpoints cover a full revolution. No radial interval outside the wall or below the axis has been included.

Worked Example: Setting Up an Integral Over a Hollow Cylinder

Worked Example: Setting Up an Integral Over a Hollow Cylinder

Let \(E\) be the region inside the cylinder \(x^2+y^2=25\), outside the cylinder \(x^2+y^2=4\), and between \(z=0\) and \(z=3\). Set up an integral of \(g(r,\theta,z)\) over \(E\).

First translate each radial condition. Being inside the outer wall means \(x^2+y^2\le25\), so \(r^2\le25\), and hence \(r\le5\). Being outside the inner wall means \(x^2+y^2\ge4\), so \(r^2\ge4\), and hence \(r\ge2\). Combining these gives \(2\le r\le5\). The region is complete around the axis, and its vertical interval is \(0\le z\le3\).

$$ 2\le r\le5,\qquad 0\le\theta\le2\pi,\qquad 0\le z\le3. $$

Therefore, the integral is

$$ \iiint_E g\,dV = \int_0^{2\pi}\int_2^5\int_0^3 g(r,\theta,z)\,r\,dz\,dr\,d\theta. $$

Check: Squaring the proposed radial limits gives \(4\le r^2\le25\). Since \(r^2=x^2+y^2\), this is exactly the condition of being outside the inner cylinder and inside the outer cylinder. In particular, points with \(r<2\) are excluded, as the hollow center requires.

Shells and Radial Intervals That Change

A cylindrical shell is not necessarily thin. The term describes a region with an inner cylindrical wall and an outer cylindrical wall; the radial thickness is \(b-a\). A thin shell is the special case in which this difference is small. In either case, the radial limits are the inner radius followed by the outer radius.

Sometimes different parts of a solid have different radial limits or different vertical extent. Then a single set of bounds may include points that are not in the solid, or miss points that are. One remedy is to divide the region into pieces and set up one integral for each piece. The pieces should cover the region without overlapping except possibly along boundaries, which do not affect the triple integral.

Worked Example: Splitting a Region Into Radial Pieces

Suppose \(E\) is the union of two complete cylindrical regions. The first has \(0\le r\le2\) and \(0\le z\le1\). The second has \(2\le r\le4\) and \(0\le z\le3\). Set up an integral of \(g(r,\theta,z)\) over \(E\).

The first piece reaches from the axis to radius \(2\), and its height runs from \(0\) to \(1\). The second starts at radius \(2\), reaches radius \(4\), and extends from \(z=0\) to \(z=3\). Both pieces go through a full turn. Their common boundary at \(r=2\) has no volume, so adding the two integrals counts the region without double-counting any volume.

$$ \iiint_E g\,dV = \int_0^{2\pi}\int_0^2\int_0^1 g(r,\theta,z)\,r\,dz\,dr\,d\theta + \int_0^{2\pi}\int_2^4\int_0^3 g(r,\theta,z)\,r\,dz\,dr\,d\theta. $$

Check: For \(0\le r<2\), only the first piece is included, with \(0\le z\le1\). For \(2<r\le4\), only the second is included, with \(0\le z\le3\). For \(r>4\), neither piece contributes. Thus the two radial intervals and their corresponding vertical limits describe exactly the stated union.

A Common Mistake: Filling the Hollow Center

When a region lies between two coaxial walls, using \(0\) as the lower radial limit incorrectly fills the hole. For the hollow cylinder from the previous example, the following setup is wrong:

$$ \int_0^{2\pi}\int_0^5\int_0^3 g(r,\theta,z)\,r\,dz\,dr\,d\theta. $$

This expression includes every radius from \(0\) to \(5\), including \(0\le r<2\). Those points lie inside the inner wall, not outside it. The corrected radial interval is \(2\le r\le5\), giving

$$ \int_0^{2\pi}\int_2^5\int_0^3 g(r,\theta,z)\,r\,dz\,dr\,d\theta. $$

A reliable check is to test the radial limits against the walls: the lower endpoint must match the inner radius when a central hole is present, and the upper endpoint must match the outer radius. For a solid cylinder that includes its axis, the lower radial endpoint is instead \(0\).

Why the Bounds Matter

Correct radial limits encode which distances from the axis belong to the region. If the lower bound is too small, the integral includes an unintended central portion. If the upper bound is too large, it extends beyond the outer wall. If the limits are reversed, they do not describe the intended radial interval. The walls are geometric constraints, so the bounds should be read from the geometry before writing the integral.

When identifying those limits, keep the roles of the boundaries distinct:

  • A coaxial cylindrical wall gives a constant radial value.
  • A horizontal plane gives a vertical value for \(z\).
  • A restriction to only part of a full cylinder requires angular limits that match that restriction.

For the complete cylinders and shells here, the angular interval is \(0\le\theta\le2\pi\). The next decision is not implied by the cylindrical walls: the vertical boundaries and any angular restrictions must be read from the rest of the solid.

Check Your Understanding

For each question, identify the radial interval before considering any other bounds.

  1. A solid cylinder is inside the wall \(x^2+y^2=36\) and includes its axis. What are its radial limits?
  2. A region lies outside \(x^2+y^2=9\) and inside \(x^2+y^2=49\). What are its radial limits?
  3. Why does the radial interval \(0\le r\le5\) fail to describe the region between radii \(2\) and \(5\)?
  4. A complete shell has radii \(a\) and \(b\), with \(0<a<b\), and lies between two horizontal planes. Which boundaries determine its radial limits, and which determine its vertical limits?
  5. A solid has one vertical interval for \(0\le r\le2\) and another for \(2\le r\le4\). What is one way to set up its triple integral without forcing one vertical interval to apply everywhere?

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