Vertical Bounds Between Cylindrical Surfaces

What Vertical Bounds Describe

In cylindrical coordinates, a vertical line holds \(r\) and \(\theta\) fixed while \(z\) varies. For a solid lying between two surfaces, the vertical bounds describe the values of \(z\) encountered along that line: the line begins at the lower surface and ends at the upper surface.

The key task is therefore to express both surfaces as heights, compare those heights over the projected region, and put the smaller one first. The projection tells us which \(r\)- and \(\theta\)-values occur; the vertical bounds tell us the \(z\)-values above each such point.

Definition: Suppose a solid is projected onto a region \(D\) in the \(r\)-\(\theta\) plane. If its lower and upper surfaces are \(z=L(r,\theta)\) and \(z=U(r,\theta)\), respectively, and \(L(r,\theta)\le U(r,\theta)\) throughout \(D\), then the vertical description of the solid is \(L(r,\theta)\le z\le U(r,\theta)\), for \((r,\theta)\in D\).

The inequality \(L\le U\) is essential: it ensures that each vertical interval has nonnegative length. In the cylindrical order \(dz\,dr\,d\theta\), these are the inner limits of integration. The cylindrical volume element is \(r\,dz\,dr\,d\theta\); choosing the surfaces does not change that factor.

Writing Surfaces in Cylindrical Coordinates

The basic substitutions are \(x=r\cos\theta\), \(y=r\sin\theta\), and \(x^2+y^2=r^2\). A surface written as a height \(z=f(x,y)\) becomes \(z=f(r\cos\theta,r\sin\theta)\). For a surface depending only on distance from the vertical axis, its height depends on \(r\) alone.

For example, the paraboloid \(z=x^2+y^2+1\) becomes \(z=r^2+1\), the horizontal plane \(z=5\) remains \(z=5\), and the cone \(z=2\sqrt{x^2+y^2}\) becomes \(z=2r\). The cylindrical equations describe heights, so they can be compared directly at the same \(r\) and \(\theta\).

When two surfaces meet, their heights are equal. Solving that equality helps locate where one surface can bound the solid above the other. The projection of the solid must then be restricted to points where the lower height does not exceed the upper height.

Worked Example: A Paraboloid Below a Horizontal Plane

Find the vertical bounds for the solid above \(z=x^2+y^2+1\) and below \(z=5\), using the full projected disk.

First write the two surfaces in cylindrical form. The paraboloid is \(z=r^2+1\), and the plane is \(z=5\). To find their intersection, set their heights equal:

r^2+1=5 \quad\Longrightarrow\quad r^2=4 \quad\Longrightarrow\quad r=2,

where \(r\ge 0\). At the axis, \(r=0\), the paraboloid has height \(1\), below the plane’s height \(5\). For \(0\le r\le2\), the difference between the plane and paraboloid is

5-(r^2+1)=4-r^2\ge0.

Thus the paraboloid is below the plane throughout the disk \(0\le r\le2\). The full disk has \(0\le\theta\le2\pi\), so the vertical bounds are

r^2+1\le z\le5, \qquad 0\le r\le2, \qquad 0\le\theta\le2\pi.

As checks, at \(r=0\) the interval is \(1\le z\le5\); at \(r=1\) it is \(2\le z\le5\); and at \(r=2\) both surfaces have height \(5\). The bounds meet exactly at the edge of the projected disk.

Comparing Heights Before Choosing the Bounds

A surface’s name or appearance is not enough to determine whether it is the lower or upper boundary. The correct comparison is pointwise: evaluate both height functions at the same \((r,\theta)\). If the difference \(U-L\) is nonnegative throughout the relevant projection, then \(L\) is the lower surface and \(U\) is the upper surface there.

The equality \(U=L\) identifies locations where the vertical interval collapses to a single point. Those locations often form the edge of a projected region. If \(U-L\) changes sign within the projection, however, the surface ordering changes. A single pair of vertical limits then cannot describe the whole solid: the projection must be divided into pieces on which the ordering is consistent.

Rule: For vertical bounds, put the smaller height first and the larger height second. Check the sign of the difference between the proposed upper and lower heights over the entire projected region, not just at one sample point.

Worked Example: A Cone Below a Horizontal Plane

A solid lies above the cone \(z=2\sqrt{x^2+y^2}\) and below the plane \(z=6\). Find the vertical bounds over its projection.

Since \(\sqrt{x^2+y^2}=r\), the cone is \(z=2r\). The plane is \(z=6\). Their intersection satisfies

2r=6 \quad\Longrightarrow\quad r=3.

At \(r=0\), the cone has height \(0\), while the plane has height \(6\). For \(0\le r\le3\),

6-2r\ge0,

so the cone is the lower surface and the plane is the upper surface over the disk of radius \(3\). The vertical bounds are

2r\le z\le6, \qquad 0\le r\le3, \qquad 0\le\theta\le2\pi.

Check the heights at \(r=0\): the interval runs from \(0\) to \(6\). At \(r=2\), it runs from \(4\) to \(6\). At the boundary \(r=3\), both surfaces have height \(6\), so the vertical interval has length zero there, as expected.

Why the Height Comparison Works

Fix any point \((r,\theta)\) in the projection. Along the vertical line through that point, the lower surface has height \(L(r,\theta)\) and the upper surface has height \(U(r,\theta)\). The points of the solid on that line are exactly the heights between those two values. This gives the interval \(L(r,\theta)\le z\le U(r,\theta)\), provided \(L(r,\theta)\le U(r,\theta)\).

The comparison can also be read from the interval’s length. Its length is \(U(r,\theta)-L(r,\theta)\). A valid vertical interval has nonnegative length, and its endpoints coincide precisely where that difference is zero. Thus solving \(U=L\) locates potential edges, while checking the sign of \(U-L\) determines which surface is above on each side.

Express both surfaces as heights.
Use the cylindrical substitutions to write \(z=L(r,\theta)\) and \(z=U(r,\theta)\) in terms of the projected coordinates.

Find where the heights agree.
Solve \(L(r,\theta)=U(r,\theta)\) to identify where the vertical interval can collapse and where the projection may end.

Compare the heights on the projection.
Check the sign of \(U-L\) throughout the proposed region. If its sign changes, divide the region where necessary.

Write the vertical interval.
Place the lower height as the \(z\)-lower limit and the upper height as the \(z\)-upper limit.

Worked Example: A Solid Between Two Paraboloids

Find the vertical bounds for the region above \(z=2+r^2\) and below \(z=8-r^2\), over the full projection where these surfaces enclose the region.

Set the heights equal to find their intersection:

2+r^2=8-r^2 \quad\Longrightarrow\quad 2r^2=6 \quad\Longrightarrow\quad r^2=3 \quad\Longrightarrow\quad r=\sqrt{3},

using \(r\ge0\). To determine which surface is above, subtract the proposed lower height from the proposed upper height:

$$ (8-r^2)-(2+r^2)=6-2r^2. $$

This difference is nonnegative when \(r^2\le3\), or \(0\le r\le\sqrt3\). Therefore the projection is the disk of radius \(\sqrt3\), and the lower and upper surfaces are respectively \(z=2+r^2\) and \(z=8-r^2\). The bounds are

2+r^2\le z\le8-r^2, \qquad 0\le r\le\sqrt3, \qquad 0\le\theta\le2\pi.

At \(r=0\), the surface heights are \(2\) and \(8\), in the stated order. At \(r=1\), they are \(3\) and \(7\). At \(r=\sqrt3\), they are both \(5\). These checks confirm both the ordering inside the disk and the meeting of the surfaces at its edge.

A Common Reversal Error

Suppose a solid is between the plane \(z=1\) and the downward-opening paraboloid \(z=4-r^2\). A common mistake is to write \(4-r^2\le z\le1\) simply because the paraboloid was mentioned first. At \(r=0\), this gives \(4\le z\le1\), an impossible interval: the purported lower limit exceeds the upper limit.

The correction is to compare the heights and find their intersection:

4-r^2=1 \quad\Longrightarrow\quad r^2=3 \quad\Longrightarrow\quad r=\sqrt3.

For \(0\le r\le\sqrt3\), the difference between the paraboloid and plane is

(4-r^2)-1=3-r^2\ge0.

Hence the plane is below and the paraboloid is above. The corrected vertical bounds are \(1\le z\le4-r^2\) over that disk. At \(r=0\), these read \(1\le z\le4\), and at \(r=\sqrt3\), both limits equal \(1\). This example illustrates why the order should be established by comparing heights rather than by the order in which surfaces appear in a description.

When One Pair of Bounds Is Not Enough

Two surfaces can cross more than once over a proposed projection, or their height difference can change sign as \(r\) or \(\theta\) varies. In that case, the same surface is not necessarily lower everywhere. A single pair of \(z\)-limits would then describe some vertical lines incorrectly, or produce reversed limits on part of the region.

The remedy is to identify where the surfaces meet and divide the projection into pieces on which the height difference has a consistent sign. On each piece, write the lower height first and the upper height second. This is the vertical-bound version of splitting a solid when one set of iterated bounds does not describe it throughout.

The method matters because correct vertical limits encode the solid itself. Once they are set, a triple integral in cylindrical coordinates can use those limits to include exactly the points between the bounding surfaces. Before proceeding to an integral, verify that the surfaces have been written as heights, that their order is correct throughout the projection, and that the bounds meet where the projected region ends.

Check Your Understanding

For each question, identify the heights and explain how their ordering is determined.

Rewrite the surface \(z=x^2+y^2+3\) as a height in cylindrical coordinates.
The solid lies above \(z=r^2\) and below \(z=9\). At what radius do the surfaces meet, and which surface is lower at \(r=1\)?
For surfaces \(z=1+r^2\) and \(z=7-r^2\), find their intersection radius and state the region where the first surface is lower.
Why is checking the sign of the upper height minus the lower height more reliable than guessing the surface order from a sketch?
What should be done if the difference between two surface heights changes sign within the projected region?