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Piecewise Radial Bounds for Solids With Steps

Learn to partition stepped solids into radial intervals and combine their cylindrical integrals correctly.

Intermediate 12 min read

What You'll Learn

  • Identify radial breakpoints where a wall or height changes.
  • Write cylindrical bounds separately on each radial interval.
  • Add integrals over disjoint radial pieces without gaps or overlap.
  • Include the cylindrical volume factor when setting up each integral.
  • Check piecewise bounds and computed volumes against the solid’s geometry.

Why Piecewise Radial Bounds Are Useful

A solid may have a circular footprint and still resist a single set of cylindrical bounds. Its roof might drop at a particular radius, its floor might rise, or its inner or outer wall might change. The right response is to divide the solid into radial bands where the bounding surfaces have one consistent description, set up an integral on each band, and add the results.

The key is to make the bands cover the solid exactly once. A missing radial interval leaves out part of the solid; overlapping intervals count part of it more than once. Careful attention to the radial breakpoints prevents both errors.

Central idea: When a solid’s cylindrical bounds change at a radius, split the radial range at that radius. Describe the solid on each resulting interval and add the corresponding triple integrals.

Cylindrical Bounds on a Radial Band

In cylindrical coordinates, \(r\) is the distance from the \(z\)-axis, \(\theta\) is the angle around that axis, and \(z\) is the vertical coordinate. The volume element is \(dV=r\,dz\,dr\,d\theta\). The factor \(r\) is part of the volume element and must appear in the integrand.

Suppose a solid covers angles \(\alpha\leq\theta\leq\beta\), and consider one radial band \(a\leq r\leq b\). If its lower and upper surfaces are \(z=L(r,\theta)\) and \(z=U(r,\theta)\), respectively, then the integral of a function \(f\) over this piece is

$$ \int_{\alpha}^{\beta}\int_a^b\int_{L(r,\theta)}^{U(r,\theta)} f(r,\theta,z)\,r\,dz\,dr\,d\theta. $$

This description requires \(a\leq b\) and \(L(r,\theta)\leq U(r,\theta)\) throughout the band and angular range. If the surfaces or walls change at \(r=c\), use separate descriptions on the intervals on either side of \(c\). At a breakpoint, the formulas on the two sides do not have to agree: a genuine step may make the height jump.

Definition: A piecewise radial description divides the radial range into intervals with disjoint interiors. On each interval, one set of cylindrical bounds describes the solid. The integral over the entire solid is the sum of the integrals over those pieces, provided the pieces cover the solid without interior overlap.

For a solid of revolution, the angular range is often \(0\leq\theta\leq2\pi\). The same radial partition can be used for each angle. In that case, if the integrand and bounds do not depend on \(\theta\), the angular integration contributes a factor of \(2\pi\).

Worked Example: A Roof That Drops at a Cylinder

Worked Example: A Roof That Drops at a Cylinder

Let a solid have full angular range \(0\leq\theta\leq2\pi\), base \(z=0\), and outer radius \(2\). Its roof is \(z=3\) for \(0\leq r\leq1\), then drops to \(z=1\) for \(1<r\leq2\). Find its volume.

The roof changes at \(r=1\), so the two radial bands are \(0\leq r\leq1\) and \(1\leq r\leq2\). Including their common boundary in both descriptions has no effect on volume. The volume integral is

$$ V= \int_0^{2\pi}\int_0^1\int_0^3 r\,dz\,dr\,d\theta + \int_0^{2\pi}\int_1^2\int_0^1 r\,dz\,dr\,d\theta. $$

Integrating with respect to \(z\) gives heights \(3\) and \(1\):

$$ V=2\pi\left(\int_0^1 3r\,dr+\int_1^2 r\,dr\right). $$

Evaluate the first integral:

$$ \int_0^1 3r\,dr =\left[\frac{3r^2}{2}\right]_0^1 =\frac32. $$

The derivative of \(3r^2/2\) is \(3r\), as required. For the second integral,

$$ \int_1^2 r\,dr =\left[\frac{r^2}{2}\right]_1^2 =2-\frac12 =\frac32, $$

and the derivative of \(r^2/2\) is \(r\). Therefore

$$ V=2\pi\left(\frac32+\frac32\right)=6\pi. $$

As a geometric check, the inner cylinder has radius \(1\) and height \(3\), so its volume is \(3\pi\). The outer annulus has area \(\pi(2^2-1^2)=3\pi\) and height \(1\), so its volume is \(3\pi\). The total is \(3\pi+3\pi=6\pi\), agreeing with the integral.

Finding the Right Breakpoints

A breakpoint belongs wherever the description of the solid changes as \(r\) increases. It may come from a cylindrical wall, a step in the roof or floor, or an intersection where two candidate surfaces exchange which one bounds the solid. For example, if the roof is the lower of two surfaces, find the radius where their heights are equal and split the interval there. Each piece must use the surface that actually bounds the solid on that interval.

After finding the breakpoints, check the intervals in order. Their union should be the entire radial range of the solid, and their interiors should not overlap. On every interval, also check that each proposed lower \(z\)-bound is no greater than its upper \(z\)-bound. A description that gives a negative height or puts the floor above the roof is not a valid bound for that piece.

1
Locate every change.
Find the radii at walls, steps, or intersections where the bounding description switches.
2
Order the radial endpoints.
Write them from the smallest to the largest, including the start and end of the solid’s radial range.
3
Describe each band.
For every interval, state the angular range and the lower and upper \(z\)-bounds that apply there.
4
Integrate and add.
Set up one cylindrical integral per band, including the factor \(r\), and sum them.

Why Adding the Pieces Works

The reason for splitting is not an approximation. It is an exact partition of the solid. Let its radial range be divided by ordered breakpoints \(a_0<a_1<\cdots<a_n\). The corresponding radial bands have disjoint interiors and together cover the full range. Thus every point of the solid, except possibly points on the dividing cylinders \(r=a_i\), belongs to exactly one piece.

Those dividing cylinders are boundaries, not extra interior pieces. They have zero volume, so assigning a boundary to either adjacent band does not change a volume integral. For an integrable function \(f\), additivity of the integral over the partition therefore gives

$$ \iiint_E f\,dV = \sum_{i=1}^{n}\iiint_{E_i}f\,dV, $$

where \(E\) is the whole solid and \(E_i\) is its part in the \(i\)th radial band. Writing each term in cylindrical coordinates yields the piecewise-bound setup. The formula requires that the pieces cover \(E\) and do not overlap in their interiors; additivity cannot correct a missing band or a region counted twice.

Worked Example: A Sloping Floor and a Step

Worked Example: A Sloping Floor and a Step

Consider a full solid with \(0\leq r\leq2\) and \(0\leq\theta\leq2\pi\). On \(0\leq r\leq1\), it lies between \(z=0\) and \(z=2-r^2\). On \(1<r\leq2\), it lies between \(z=r-1\) and \(z=1\). Find its volume.

The breakpoint \(r=1\) separates the two descriptions. Check the heights before integrating. On the inner band, \(2-r^2\geq1>0\). On the outer band, \(r-1\leq1\) for \(r\leq2\), so the lower surface does not exceed the upper surface.

$$ V= \int_0^{2\pi}\int_0^1\int_0^{2-r^2}r\,dz\,dr\,d\theta + \int_0^{2\pi}\int_1^2\int_{r-1}^1r\,dz\,dr\,d\theta. $$

Integrating with respect to \(z\) gives

$$ V=2\pi\left( \int_0^1(2-r^2)r\,dr+ \int_1^2\bigl(1-(r-1)\bigr)r\,dr \right) =2\pi\left( \int_0^1(2r-r^3)\,dr+ \int_1^2(2-r)r\,dr \right). $$

For the inner band,

$$ \int_0^1(2r-r^3)\,dr =\left[r^2-\frac{r^4}{4}\right]_0^1 =1-\frac14 =\frac34. $$

The derivative of \(r^2-r^4/4\) is \(2r-r^3\). For the outer band, expand \((2-r)r=2r-r^2\):

$$ \int_1^2(2r-r^2)\,dr =\left[r^2-\frac{r^3}{3}\right]_1^2 =\left(4-\frac83\right)-\left(1-\frac13\right) =\frac43-\frac23 =\frac23. $$

The derivative of \(r^2-r^3/3\) is \(2r-r^2\). Hence

$$ V=2\pi\left(\frac34+\frac23\right) =2\pi\left(\frac{9}{12}+\frac{8}{12}\right) =\frac{17\pi}{6}. $$

As an independent check, the inner piece has height \(2-r^2\), which decreases from \(2\) to \(1\) as \(r\) increases to \(1\). The outer piece has height \(2-r\), which decreases from \(1\) to \(0\) by \(r=2\). Both pieces have nonnegative height over their stated intervals, and their computed contributions, \(3\pi/2\) and \(4\pi/3\), sum to \(17\pi/6\).

Worked Example: A Cone-Like Roof Beside a Cylindrical Step

Worked Example: A Cone-Like Roof Beside a Cylindrical Step

A solid has full angular range \(0\leq\theta\leq2\pi\). For \(0\leq r\leq1\), it lies between \(z=0\) and \(z=1-r\). For \(1<r\leq2\), it lies between \(z=0\) and \(z=1\). Find the volume.

The roof changes at \(r=1\). On the inner band, \(1-r\) is nonnegative; on the outer band, the height is \(1\). Set up and evaluate the two contributions:

$$ V= 2\pi\left(\int_0^1(1-r)r\,dr+\int_1^2r\,dr\right). $$

First expand the inner integrand:

$$ \int_0^1(1-r)r\,dr =\int_0^1(r-r^2)\,dr =\left[\frac{r^2}{2}-\frac{r^3}{3}\right]_0^1 =\frac12-\frac13 =\frac16. $$

The derivative of \(r^2/2-r^3/3\) is \(r-r^2\). The outer integral is

$$ \int_1^2r\,dr =\left[\frac{r^2}{2}\right]_1^2 =2-\frac12 =\frac32. $$

The derivative of \(r^2/2\) is \(r\). Thus

$$ V=2\pi\left(\frac16+\frac32\right) =2\pi\left(\frac{1}{6}+\frac{9}{6}\right) =\frac{10\pi}{3}. $$

Check by geometry: the inner roof bounds a cone of base radius \(1\) and height \(1\), with volume \(\pi/3\). The outer piece is an annular cylinder with radii \(1\) and \(2\), and height \(1\), with volume \(\pi(2^2-1^2)=3\pi\). Their sum is \(\pi/3+3\pi=10\pi/3\), matching the integral.

Common Pitfall: Using One Height Across a Step

In the first example, the roof is \(3\) on the inner disk but only \(1\) on the outer annulus. A tempting but incorrect setup is to use \(z=0\) to \(z=3\) for every radius from \(0\) to \(2\). That calculation gives

$$ 2\pi\int_0^2 3r\,dr =2\pi\left[\frac{3r^2}{2}\right]_0^2 =12\pi, $$

which counts the outer annulus as though its height were \(3\), not \(1\). The correct outer contribution uses its actual height: \(2\pi\int_1^2r\,dr=3\pi\), rather than \(2\pi\int_1^2 3r\,dr=9\pi\). Replacing that incorrect contribution gives \(3\pi+3\pi=6\pi\), the correct volume.

Another useful check is to inspect every interval before integrating: does it start and end at the intended radii, and do its vertical bounds describe the solid throughout? A correct antiderivative cannot repair an incorrect partition.

Check Your Understanding

For each question, identify the radial pieces before carrying out any integration.

  1. A solid has radius \(3\), full angular range, and roof \(z=4\) for \(0\leq r\leq2\) and \(z=2\) for \(2<r\leq3\), with floor \(z=0\). What two radial intervals should be used?
  2. For the solid in question 1, write the cylindrical volume integral as a sum of two integrals.
  3. Suppose the lower surface is \(z=r\) and the upper surface is \(z=2\) on a radial band \(1\leq r\leq3\). Where does their order switch, and how should the band be handled if the solid requires the lower surface to lie below the upper one?
  4. Why does counting the shared boundary \(r=c\) in both neighboring radial intervals not change the volume?
  5. In a piecewise cylindrical volume integral, what factor must multiply the integrand, and where does it come from?

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