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Projecting Solids Into the r-theta Plane

Turn a solid’s height constraints into a precise \(r\)-\(\theta\) projection, then use that projected region to set the two outer bounds of a cylindrical integral.

Intermediate 9 min read

What You'll Learn

  • Define the projection of a solid onto the \(r\)-\(\theta\) parameter plane.
  • Find where lower and upper height constraints allow at least one \(z\)-value.
  • Determine disk, annular, and angle-dependent projections from solid constraints.
  • Use the projection to identify the outer bounds in an iterated cylindrical integral.
  • Preserve whether projection boundaries are included or excluded.

Why Project a Solid?

A three-dimensional solid can be described by listing its possible horizontal locations and, above each location, the heights that belong to the solid. In cylindrical coordinates, a horizontal location is recorded by \(r\) and \(\theta\), while \(z\) records height.

When an integral is written in the order \(dz\,dr\,d\theta\), the \(z\)-values are considered first. The remaining two integrations range over all horizontal locations for which the solid contains at least one point. Finding those locations is the task of projecting the solid into the \(r\)-\(\theta\) plane.

This projection is a parameter region: its points are pairs \((r,\theta)\), not points in the original \(xy\)-plane. The cylindrical-coordinate relationships \(x=r\cos\theta\) and \(y=r\sin\theta\) connect the two descriptions. Since \(r\) is a distance from the axis, \(r\ge 0\).

Definition: The projection of a solid \(E\) into the \(r\)-\(\theta\) plane is the set of pairs \((r,\theta)\) for which there exists at least one height \(z\) such that the corresponding point is in \(E\). In symbols, the projected region is \(P=\{(r,\theta):\text{ there exists }z\text{ with }(r,\theta,z)\in E\}\).

Projection as an Existence Test

Suppose a solid’s height is constrained by \(L(r,\theta)\le z\le U(r,\theta)\). At a fixed pair \((r,\theta)\), the solid contains a point above that location exactly when at least one \(z\) satisfies both inequalities. This is possible precisely when the proposed lower height does not exceed the proposed upper height.

$$ L(r,\theta)\le U(r,\theta). $$

Thus the projection is found by combining this overlap condition with all other restrictions on \(r\) and \(\theta\), such as cylinder constraints or angular limits. The projection does not ask how much height is available; it asks only whether any height is available.

Theorem: If, at each fixed \((r,\theta)\), the height constraints are \(L_i(r,\theta)\le z\) for every lower constraint and \(z\le U_j(r,\theta)\) for every upper constraint, then a permissible \(z\) exists exactly when
$$ \max_i L_i(r,\theta)\le \min_j U_j(r,\theta). $$

To justify this criterion, first suppose a permissible \(z\) exists. Every lower bound is at most \(z\), so their maximum is at most \(z\); every upper bound is at least \(z\), so \(z\) is at most their minimum. Therefore the displayed inequality holds. Conversely, suppose the inequality holds. Choose \(z=\max_i L_i(r,\theta)\). This value satisfies every lower constraint by the definition of a maximum, and it satisfies every upper constraint because it is no greater than their minimum. Hence a permissible height exists. This argument assumes non-strict inequalities; with strict inequalities, endpoint inclusion must be checked separately.

Worked Example: A Disk Projection from Two Height Surfaces

Worked Example: A Disk Projection from Two Height Surfaces

Project the solid described by \(r^2\le z\le 9-r^2\), with \(r\ge 0\) and no additional angular restriction.

At a fixed \((r,\theta)\), the lower height is \(r^2\) and the upper height is \(9-r^2\). A height exists exactly when the lower bound is no greater than the upper bound:

$$ r^2\le 9-r^2. $$

Add \(r^2\) to both sides, then divide by \(2\):

$$ 2r^2\le 9, \qquad r^2\le \frac{9}{2}. $$

Since \(r\ge 0\), taking square roots gives \(0\le r\le 3/\sqrt{2}\). There is no angular dependence, so \(\theta\) can range over any chosen full interval of length \(2\pi\), for example \(0\le\theta\le 2\pi\). The projection is therefore a disk in the \(r\)-\(\theta\) description, not an annulus: it includes the axis \(r=0\).

Check the limiting radii in the original height constraints. At \(r=0\), they become \(0\le z\le 9\), which permits heights. At \(r=3/\sqrt{2}\), both bounds equal \(9/2\), so \(z=9/2\) is permitted. For \(r>3/\sqrt{2}\), the lower bound exceeds the upper bound, so no height is possible. The projection is \(0\le r\le 3/\sqrt{2}\), with the stated angular range.

Using the Projection for the Outer Bounds

For an integral in the order \(dz\,dr\,d\theta\), the inner integration follows the solid vertically, while the two outer integrations cover its projection. Once \(P\) has been found, its description tells you the \(\theta\)-limits and the \(r\)-limits, which may depend on \(\theta\). In cylindrical coordinates the volume element is \(r\,dz\,dr\,d\theta\), so the factor \(r\) is included in the integrand. The projection determines the outer region; it does not by itself determine the full limits for \(z\).

$$ \iiint_E f(r,\theta,z)\,r\,dz\,dr\,d\theta = \iint_P \left(\int_{\text{permitted heights at }(r,\theta)} f(r,\theta,z)\,r\,dz\right)dr\,d\theta. $$

The notation on the right emphasizes the roles of the two regions: the inner integral uses the heights permitted above a fixed \((r,\theta)\), and the outer integrals range over \(P\). To write the two outer integrals as nested one-variable integrals, express \(P\) using an angular interval and radial limits for each angle.

Worked Example: An Annular Projection from Cylinders and Height Surfaces

Worked Example: An Annular Projection from Cylinders and Height Surfaces

Find the projection of the solid constrained by \(1\le r\le 3\), \(r^2\le z\le 10-r^2\), and \(0\le\theta\le 2\pi\).

The cylinder constraints already require \(1\le r\le 3\). But a point in the solid must also satisfy both height inequalities. Their overlap condition is

$$ r^2\le 10-r^2. $$

Rearranging gives \(2r^2\le 10\), so \(r^2\le 5\). Because \(r\ge 0\), this is \(r\le\sqrt{5}\). Intersecting that condition with the cylinder limits gives

$$ 1\le r\le 3 \quad\text{and}\quad r\le\sqrt{5} \quad\Longrightarrow\quad 1\le r\le\sqrt{5}. $$

The projection is the annular region \(1\le r\le\sqrt{5}\), \(0\le\theta\le2\pi\). It is an annulus because the inner cylinder excludes all radii below \(1\), including the axis.

Check the limiting radii. At \(r=1\), the height interval is \(1\le z\le9\), so it is nonempty. At \(r=\sqrt{5}\), both height bounds equal \(5\), so exactly one height is allowed. At \(r=3\), which the cylinder alone would permit, the bounds would require \(9\le z\le1\), which is impossible. Thus the outer radius is cut down by the height-overlap condition, and the projection includes both radial boundaries.

Angle-Dependent Height Constraints

Sometimes the height overlap depends on direction as well as distance from the axis. Then the projection’s radial extent may vary with \(\theta\), and some angles may not occur at all. A useful way to proceed is to keep \(r\ge0\), derive the overlap condition, and then identify which angles make that condition possible.

Worked Example: A Projection with Angle-Dependent Radius

Project the solid defined by \(0\le z\le 4\cos\theta-r^2\), with \(r\ge0\). Use the representative angular interval \(-\pi\le\theta\le\pi\).

For a permissible height, the upper bound must be at least the lower bound \(0\):

$$ 0\le 4\cos\theta-r^2, \qquad r^2\le4\cos\theta. $$

Because \(r^2\ge0\), this inequality requires \(\cos\theta\ge0\). Within \([-\pi,\pi]\), that holds for \(-\pi/2\le\theta\le\pi/2\). For each such angle, taking nonnegative square roots gives

$$ 0\le r\le2\sqrt{\cos\theta}. $$

Therefore the projection is described by \(-\pi/2\le\theta\le\pi/2\) and \(0\le r\le2\sqrt{\cos\theta}\). At \(\theta=\pm\pi/2\), the radial upper bound is \(0\), so only \(r=0\) remains. For angles with \(\cos\theta<0\), even \(r=0\) fails the height-overlap condition. These checks agree with the original inequalities.

Boundary Inclusion Must Follow the Constraints

Finding an algebraic boundary does not automatically tell you whether that boundary belongs to the projection. Preserve the strict or non-strict inequalities in the original solid. In particular, do not add boundary points merely because they are limits of points inside the projected region, and do not demand a height at boundary values that the constraints exclude.

For example, consider \(0<r<1\), \(0<z<1\), with all angles allowed. Every radius strictly between \(0\) and \(1\) has permissible heights, so the projection is \(0<r<1\). The limiting radii \(r=0\) and \(r=1\) are not included: the original radial constraints explicitly exclude them, and there is no point of the solid over either radius.

Common mistake: Replacing \(0<r<1\) by \(0\le r\le1\) because the endpoints appear when taking a closure changes the solid’s projection. The correction is to carry the original strict inequalities through the projection test: here the result remains \(0<r<1\), not \(0\le r\le1\).

More generally, if the overlap calculation produces an endpoint, substitute it into the original constraints and preserve the boundary’s inclusion or exclusion there. A boundary point belongs to the projection only if the original constraints admit at least one corresponding height at that point.

A Reliable Projection Procedure

For a solid described by cylindrical constraints, the following sequence keeps the projection test explicit:

1
Record the non-height restrictions.
Keep the stated conditions on \(r\) and \(\theta\), including whether their endpoints are included.
2
Identify the height constraints.
At a fixed \((r,\theta)\), distinguish lower bounds on \(z\) from upper bounds on \(z\).
3
Test for overlap.
For one lower and one upper bound, require the lower bound to be no greater than the upper bound. With several bounds, compare the largest lower bound to the smallest upper bound.
4
Describe the resulting parameter region.
Combine the overlap condition with the original \(r\)- and \(\theta\)-restrictions to obtain \(P\).
5
Check limiting values.
Substitute each proposed limit into the original constraints and preserve whether that boundary is included or excluded.

The result is the outer region for the cylindrical integral in the order \(dz\,dr\,d\theta\). If a single set of radial limits does not describe the projection across the entire angular interval, the region may need to be split into pieces before writing the iterated integral.

Check Your Understanding

For each question, focus on whether at least one permissible height exists at the proposed pair \((r,\theta)\).

  1. For \(r^2\le z\le 16-r^2\), what inequality determines the projection’s radial extent?
  2. Why does the condition \(L(r,\theta)\le U(r,\theta)\) characterize where a solid between two height surfaces exists?
  3. If \(2\le r\le4\) and the height-overlap condition gives \(r\le3\), what radial interval remains in the projection?
  4. For \(0\le z\le 9\cos\theta-r^2\), what must be true of \(\cos\theta\) for any radius to be possible?
  5. Why is it incorrect to include \(r=0\) in the projection of a solid constrained by \(0<r<1\)?

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