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Choosing an Integration Order for Cylindrical Bounds

You will learn to read cylindrical bounds from the projected region and choose an integration order that keeps those bounds as simple as possible.

Intermediate 10 min read

What You'll Learn

  • Describe a cylindrical solid using its projection onto the r-theta plane and its vertical bounds.
  • Translate the same projected region into radial slices or angular slices.
  • Identify when one order gives a single interval but the other requires a split.
  • Derive bounds for a triangular projection in both cylindrical orders.
  • Check bounds by testing boundary points and comparing the resulting integrals.

Why the Order Matters

A cylindrical solid has a vertical extent above each point in its projection onto the \(r\)-\(\theta\) plane. To set up a triple integral with \(z\) integrated first, we need to describe that projected region in one of two ways: by fixing \(\theta\) and finding the allowed \(r\)-values, or by fixing \(r\) and finding the allowed angles.

These descriptions are two ways to slice the same region. One may give a single interval of bounds, while the other may require several pieces. Choosing the more convenient description can make an integral substantially easier to set up and interpret.

Key idea: For the orders \(dz\,dr\,d\theta\) and \(dz\,d\theta\,dr\), first find the projection onto the \(r\)-\(\theta\) plane. Then describe that projection using radial slices or angular slices, respectively.

Two Ways to Describe the Projection

Let \(R\) be the projection of a solid onto the \(r\)-\(\theta\) plane. Suppose its vertical bounds are \(g(r,\theta)\le z\le h(r,\theta)\). The cylindrical volume element is \(r\,dz\,dr\,d\theta\), so an integral of a function \(f\) can be written in either order below, provided the bounds describe the same region and the integrand is integrable.

$$ \int_{\alpha}^{\beta} \int_{a(\theta)}^{b(\theta)} \int_{g(r,\theta)}^{h(r,\theta)} f(r,\theta,z)\,r\,dz\,dr\,d\theta $$
$$ \int_{c}^{d} \int_{p(r)}^{q(r)} \int_{g(r,\theta)}^{h(r,\theta)} f(r,\theta,z)\,r\,dz\,d\theta\,dr $$

In the first expression, \(\theta\) ranges over an outer interval, and each fixed angle gives a radial interval \(a(\theta)\le r\le b(\theta)\). In the second, \(r\) ranges over an outer interval, and each fixed radius gives an angular interval \(p(r)\le\theta\le q(r)\). If a fixed radius intersects the projection in separate angular pieces, the second description must be split into separate integrals.

Order reminder: In \(dz\,dr\,d\theta\), \(z\) is integrated first, then \(r\), then \(\theta\). In \(dz\,d\theta\,dr\), \(z\) is still integrated first, but \(\theta\) is integrated before \(r\). Changing the order does not change the solid or its vertical bounds.

Why Rewriting the Bounds Works

The radial description groups points of \(R\) by their angle: at each fixed \(\theta\), it lists the radii that belong to \(R\). The angular description groups those same points by their radius: at each fixed \(r\), it lists the angles that belong to \(R\). These are alternate descriptions of the same projected set, not different regions.

For a continuous function on a bounded solid, the order of integration may be changed when the bounds are rewritten to describe the same region. The reason is that integration sums the function over that region; grouping the points by angle and radius in a different order does not change the points being summed. In cylindrical coordinates, the factor \(r\) must remain in the integrand in either order. It accounts for the cylindrical volume element and is not changed by the choice of slice.

Practical test: To check a proposed set of bounds, fix an allowed outer variable and ask whether the inner intervals include exactly the points of the projection on that slice. If a slice has two separated pieces, one interval cannot describe it without also including points outside the region.

Worked Example: An Annular Sector

Worked Example: An Annular Sector

Consider the solid above the annular sector \(1\le r\le 3\), \(0\le\theta\le\pi/3\), and below the plane \(z=2\), with bottom \(z=0\). Set up and evaluate its volume in both orders.

In the projection, each angle from \(0\) to \(\pi/3\) has the same radial interval, from \(1\) to \(3\). Thus the order \(dz\,dr\,d\theta\) gives

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

Integrating first with respect to \(z\) gives \(\int_0^2 r\,dz=2r\). Therefore

$$ V= \int_0^{\pi/3}\int_1^3 2r\,dr\,d\theta = \int_0^{\pi/3}\left[r^2\right]_1^3d\theta = \int_0^{\pi/3}8\,d\theta = \frac{8\pi}{3}. $$

For \(dz\,d\theta\,dr\), each radius from \(1\) to \(3\) has the same angular interval, from \(0\) to \(\pi/3\). Hence

$$ V= \int_1^3\int_0^{\pi/3}\int_0^2 r\,dz\,d\theta\,dr = \int_1^3\int_0^{\pi/3}2r\,d\theta\,dr = \frac{\pi}{3}\left[r^2\right]_1^3 = \frac{8\pi}{3}. $$

Both orders give the same volume. Here neither order needs to be split because every radial slice and every angular slice is a single interval.

Worked Example: A Triangular Projection

Worked Example: A Triangular Projection

Let the projection be the triangle \(x\ge0\), \(y\ge0\), \(x+y\le2\), and let the solid have \(0\le z\le2\). Find the cylindrical bounds and volume in both orders. In cylindrical coordinates, \(x=r\cos\theta\) and \(y=r\sin\theta\), so the projected boundary is \(r(\cos\theta+\sin\theta)=2\).

For a fixed angle in the first quadrant, the radius runs from \(0\) to the boundary. This gives one radial interval for each angle:

$$ 0\le\theta\le\frac{\pi}{2}, \qquad 0\le r\le\frac{2}{\cos\theta+\sin\theta}. $$

Since the height is \(2\), integrating \(r\) over \(z\) gives \(2r\). The volume in the order \(dz\,dr\,d\theta\) is

$$ V= \int_0^{\pi/2}\int_0^{2/(\cos\theta+\sin\theta)}2r\,dr\,d\theta = \int_0^{\pi/2}\frac{4}{(\cos\theta+\sin\theta)^2}\,d\theta. $$

To evaluate the remaining integral, set \(u=\tan\theta\), so \(d\theta=du/(1+u^2)\), \(\cos\theta+\sin\theta=(1+u)/\sqrt{1+u^2}\), and the limits become \(0\) and \(+\infty\). Thus

$$ V= \int_0^\infty \frac{4}{(1+u)^2}\,du = \left[-\frac{4}{1+u}\right]_0^\infty = 0-(-4) = 4. $$

Now fix \(r\) and find the allowed angles. Because \(\cos\theta+\sin\theta\le\sqrt{2}\) throughout the first quadrant, every angle is allowed when \(0\le r\le\sqrt{2}\). For \(\sqrt{2}<r\le2\), the condition is \(\cos\theta+\sin\theta\le2/r\). The allowed angles lie in two separate pieces, one near each axis.

Let \(\theta_0(r)\) be the endpoint of the piece near \(\theta=0\). Solving \(\cos\theta_0+\sin\theta_0=2/r\), using \(\cos\theta+\sin\theta=\sqrt{2}\sin(\theta+\pi/4)\), gives

$$ \theta_0(r)=\arcsin\left(\frac{\sqrt{2}}{r}\right)-\frac{\pi}{4}. $$

Thus, for \(\sqrt{2}<r<2\), the angular intervals are \(0\le\theta\le\theta_0(r)\) and \(\pi/2-\theta_0(r)\le\theta\le\pi/2\). The second order must be split to keep only those two pieces:

$$ V= \int_0^{\sqrt{2}}\int_0^{\pi/2}\int_0^2 r\,dz\,d\theta\,dr + \int_{\sqrt{2}}^2 \left( \int_0^{\theta_0(r)}\int_0^2 r\,dz\,d\theta + \int_{\pi/2-\theta_0(r)}^{\pi/2}\int_0^2 r\,dz\,d\theta \right)dr. $$

The first integral evaluates to \(\int_0^{\sqrt{2}}\pi r\,dr=\pi\). The two angular pieces in the second integral have total width \(2\theta_0(r)\), so that part is \(\int_{\sqrt{2}}^2 4r\theta_0(r)\,dr\). To evaluate it, set \(a=\sqrt{2}\). Integration by parts gives

$$ \int r\arcsin\left(\frac{a}{r}\right)dr = \frac{r^2}{2}\arcsin\left(\frac{a}{r}\right) + \frac{a}{2}\sqrt{r^2-a^2}. $$

At \(r=2\), the right-hand side is \(\pi/2+1\); at \(r=\sqrt{2}\), it is \(\pi/2\). Therefore the contribution from the arcsine term is \(4((\pi/2+1)-\pi/2)=4\). The subtracted \(\pi/4\) term contributes \(\pi\int_{\sqrt{2}}^2 r\,dr=\pi\). The second part is \(4-\pi\), and the total is \(\pi+(4-\pi)=4\), matching the first order.

Worked Example: Vertical Bounds Can Depend on Radius

Worked Example: Vertical Bounds Can Depend on Radius

Suppose \(1\le r\le2\), \(0\le\theta\le\pi/4\), and the solid lies between \(z=0\) and \(z=r\). The projection is a rectangular region in the \(r\)-\(\theta\) plane, so both orders use single intervals. The upper vertical bound depends on \(r\), but it does not change when the order of \(r\) and \(\theta\) is exchanged.

For \(dz\,dr\,d\theta\), integrate \(r\) over \(z\), including the cylindrical factor:

$$ V= \int_0^{\pi/4}\int_1^2\int_0^r r\,dz\,dr\,d\theta = \int_0^{\pi/4}\int_1^2 r^2\,dr\,d\theta = \frac{\pi}{4}\left[\frac{r^3}{3}\right]_1^2 = \frac{7\pi}{12}. $$

For \(dz\,d\theta\,dr\), the same projection is described by \(1\le r\le2\) and \(0\le\theta\le\pi/4\):

$$ V= \int_1^2\int_0^{\pi/4}\int_0^r r\,dz\,d\theta\,dr = \int_1^2\frac{\pi}{4}r^2\,dr = \frac{\pi}{4}\left[\frac{r^3}{3}\right]_1^2 = \frac{7\pi}{12}. $$

The matching results confirm that the vertical bound remains \(z=r\) in both setups and that the projection is unchanged.

A Common Bounds Mistake

For the triangular projection in the second example, a tempting but incorrect setup is to use \(0\le\theta\le\pi/2\) and \(0\le r\le2\). At \(\theta=\pi/4\) and \(r=2\), however, \(x+y=2\sqrt{2}>2\), so that point is outside the triangle. These bounds include points not in the projected region.

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

This incorrect setup gives \(2\pi\), not the correct volume \(4\). The correction is to use the actual boundary \(r=2/(\cos\theta+\sin\theta)\) for radial slices, or to split the angular slices when \(r>\sqrt{2}\). A reliable check is to substitute a proposed boundary point into the original constraint before accepting the bounds.

Choosing Between the Orders

When the projection is irregular, trace a few slices in both directions before writing the integral. In particular, check where slice endpoints change, where a slice becomes empty, and whether a fixed-radius slice has disconnected pieces. A split is required whenever one interval would include points outside the projection or leave points inside it unaccounted for.

1
Project the solid.
Write the constraints that describe its footprint in the \(r\)-\(\theta\) plane.
2
Test radial slices.
Fix \(\theta\) and determine the allowed radii. Note any changes in the radial endpoints as \(\theta\) varies.
3
Test angular slices.
Fix \(r\) and determine the allowed angles. Look for radii at which the angular interval changes or separates into pieces.
4
Choose the clearer description.
Prefer the order with fewer pieces and simpler endpoints, then verify that the bounds describe precisely the original projection.

The best order is not determined by a universal preference for radial or angular integration. It depends on the geometry of the projected region. An order with a single integral is often simpler, but a split may be unavoidable if slices are disconnected or their endpoints change.

Check Your Understanding

For each question, focus on how the projected region determines the bounds.

  1. In \(dz\,dr\,d\theta\), which variable describes the outermost slices of the projection?
  2. What geometric feature of a fixed-radius slice tells you that the \(dz\,d\theta\,dr\) setup needs to be split?
  3. For the projection \(x\ge0\), \(y\ge0\), \(x+y\le2\), why does a radial slice have upper bound \(2/(\cos\theta+\sin\theta)\)?
  4. For that same triangle, why are all angles allowed when \(0\le r\le\sqrt{2}\), but not for every larger radius?
  5. When switching between the two orders, what happens to the vertical bounds and the cylindrical factor \(r\)?

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