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Work and Geometric Applications · Tutorial 1345 of 1495

Composite Solids From Slice Integrals

Build composite volumes from slice integrals by choosing clear partitions, using the right area on each interval, and accounting for shared volume only once.

Advanced 11 min read

What You'll Learn

  • Identify when a solid should be divided into regions with different slice formulas.
  • Set up a piecewise cross-sectional area function over the full interval.
  • Integrate the areas on separate intervals and add the resulting volumes.
  • Distinguish adjoining components from overlapping components.
  • Use cross-sectional unions or subtraction to prevent double-counting.

Why Composite Solids Need Careful Slicing

A solid does not always have the same kind of cross section from one end to the other. One portion might have triangular slices while another has semicircular slices. Or two solid components might occupy some of the same range along the slicing direction. In either situation, the volume calculation must represent the actual solid exactly once.

The central task is to describe the area of each slice as a function of position. Where the slice geometry changes, split the interval. Where components overlap, find the area of their union rather than simply adding their areas. Once the correct area function is in place, integrate it along the solid.

Key principle: If slices perpendicular to the \(x\)-axis have area \(A(x)\) for \(a\le x\le b\), then the volume is \(V=\int_a^b A(x)\,dx\), provided \(A\) is integrable. For a composite solid, define \(A(x)\) to be the area of the entire slice, counting every point of the solid only once.

Partitioning the Solid into Regions

Suppose a solid extends from \(x=a\) to \(x=b\), and its cross-sectional area formula changes at positions \(c_1,\ldots,c_n\). These positions partition the length into intervals. On each interval, use the formula that describes the slice there, then add the resulting volumes:

$$ V=\int_a^{c_1} A_1(x)\,dx +\int_{c_1}^{c_2} A_2(x)\,dx +\cdots +\int_{c_n}^{b} A_{n+1}(x)\,dx. $$

The split points must include every location where the slice formula changes. The area formulas should describe the actual slices on their respective intervals. Adjacent components can share an end face without sharing any volume; such a boundary has zero thickness and does not require a volume correction.

When two components occupy the same \(x\)-positions, however, their cross sections may overlap. If their slice regions at position \(x\) are \(S_1(x)\) and \(S_2(x)\), the slice of the combined solid is their union. For two regions with an overlap, its area is

$$ \operatorname{area}(S_1(x)\cup S_2(x)) =\operatorname{area}(S_1(x)) +\operatorname{area}(S_2(x)) -\operatorname{area}(S_1(x)\cap S_2(x)). $$

Thus, adding the component volumes directly is correct only if their interiors do not overlap. Otherwise the shared volume would be counted twice.

Why the Piecewise Integrals Add

The rule for adjoining regions follows from how a slice integral is formed. Divide the \(x\)-interval into many thin subintervals. Over a subinterval of width \(\Delta x\), the solid is approximated by a thin slab whose volume is approximately the slice area times the slab width. Adding the slab volumes gives a sum of the form \(\sum A(x_i^*)\Delta x\), where \(x_i^*\) is a sample position in the \(i\)th subinterval.

If the interval is split at each change of formula, the slabs on one part of the solid and the slabs on another part are separate groups. Their volumes add. Taking the limit as the maximum slab width tends to zero gives the corresponding integrals, so the total is the sum of the integrals on the pieces.

Derivation: For a finite partition \(a=c_0<c_1<\cdots<c_m=b\), the slice sums on the separate intervals combine into the slice sum over the whole interval. Taking the limit on each interval therefore gives \(V=\sum_{j=1}^{m}\int_{c_{j-1}}^{c_j} A_j(x)\,dx\). This argument requires the slice area to be integrable on each piece; piecewise continuous area formulas satisfy that condition.

This additivity concerns non-overlapping portions of the solid. If two named components overlap, their slice regions must first be combined as a union, or their shared volume must be subtracted. The partition rule itself does not make overlapping components disjoint.

Worked Example: A Cone Joined to a Cylinder

Worked Example: A Cone Joined to a Cylinder

A right circular cone occupies \(0\le x\le 2\), with its tip at \(x=0\) and radius \(r(x)=x\). It joins a circular cylinder occupying \(2\le x\le 5\), whose radius is \(2\). Find the volume of the combined solid.

For \(0\le x\le 2\), a slice of the cone is a disk of radius \(x\), so its area is \(A_1(x)=\pi x^2\). For \(2\le x\le 5\), a slice of the cylinder is a disk of radius \(2\), so \(A_2(x)=\pi(2)^2=4\pi\). The two pieces meet at \(x=2\); their cross sections there have the same radius, \(2\).

$$ V=\int_0^2 \pi x^2\,dx+\int_2^5 4\pi\,dx. $$

Evaluate the cone portion:

$$ \int_0^2 \pi x^2\,dx =\pi\left[\frac{x^3}{3}\right]_0^2 =\pi\left(\frac{8}{3}-0\right) =\frac{8\pi}{3}. $$

Evaluate the cylinder portion:

$$ \int_2^5 4\pi\,dx =4\pi[x]_2^5 =4\pi(5-2) =12\pi. $$

Therefore, the total volume is

$$ V=\frac{8\pi}{3}+12\pi =\frac{8\pi}{3}+\frac{36\pi}{3} =\frac{44\pi}{3}. $$

Check: The cone volume from its slice integral is \(\frac{8\pi}{3}\), matching \(\frac13\pi(2)^2(2)=\frac{8\pi}{3}\). The cylinder volume is \(4\pi(5-2)=12\pi\). Their sum is \(\frac{44\pi}{3}\). The components meet only at the joining face, so no positive volume is counted twice.

Worked Example: Triangular Slices Followed by Semicircular Slices

Worked Example: Triangular Slices Followed by Semicircular Slices

A solid extends from \(x=0\) to \(x=4\). For \(0\le x\le 2\), each slice is a triangle with base \(2-x\) and height \(2\). For \(2\le x\le 4\), each slice is a semicircle with diameter \(x-2\). Find the volume.

On the first interval, the area of a triangle is one-half its base times its height:

$$ A_1(x)=\frac12(2-x)(2)=2-x. $$

On the second interval, the radius of the semicircle is \(\frac{x-2}{2}\). A semicircle has half the area of a disk, so

$$ A_2(x)=\frac12\pi\left(\frac{x-2}{2}\right)^2 =\frac{\pi}{8}(x-2)^2. $$

Integrate each formula on its own interval:

$$ \begin{aligned} V &=\int_0^2(2-x)\,dx+\int_2^4\frac{\pi}{8}(x-2)^2\,dx\\ &=\left[2x-\frac{x^2}{2}\right]_0^2 +\frac{\pi}{8}\left[\frac{(x-2)^3}{3}\right]_2^4\\ &=\left(4-2\right)-0 +\frac{\pi}{8}\left(\frac{8}{3}-0\right)\\ &=2+\frac{\pi}{3}. \end{aligned} $$

Check: The first interval contributes \(2\) cubic units, since the antiderivative \(2x-\frac{x^2}{2}\) has values \(2\) at \(x=2\) and \(0\) at \(x=0\). The second contributes \(\frac{\pi}{3}\), since \(\frac{\pi}{8}\cdot\frac{8}{3}=\frac{\pi}{3}\). At the transition \(x=2\), both slice areas are zero, consistent with the specified shapes meeting at a point. The combined volume is \(2+\frac{\pi}{3}\).

When Components Overlap

For overlapping components, it can be clearer to find the area of the combined slice directly. Identify the regions in a typical slice, locate the overlap, and add the areas of the parts that make up the union. An equivalent method is to add the component areas and subtract the overlap area. The overlap can vary with \(x\), so its formula may itself change at a partition point.

Worked Example: Two Overlapping Rectangular Components

Use slices perpendicular to the \(x\)-axis. Component \(A\) extends from \(x=0\) to \(x=3\); at each position its cross section is the rectangle \(0\le y\le 1,\ 0\le z\le 1\). Component \(B\) extends from \(x=1\) to \(x=3\); its cross section is the rectangle \(\frac12\le y\le\frac32,\ 0\le z\le x-1\). Find the volume of their union.

Component \(A\) has slice area \(1\), so its volume is

$$ V_A=\int_0^3 1\,dx=3. $$

Component \(B\) has a rectangle of width \(1\) in the \(y\)-direction and width \(x-1\) in the \(z\)-direction. Its area is \(x-1\), and its volume is

$$ V_B=\int_1^3(x-1)\,dx =\left[\frac{(x-1)^2}{2}\right]_1^3 =\frac{4}{2}-0 =2. $$

The components overlap only for \(1\le x\le 3\). Their \(y\)-ranges overlap by a width of \(\frac12\). In the \(z\)-direction, the overlap width is \(x-1\) for \(1\le x\le 2\), and \(1\) for \(2\le x\le 3\), because component \(A\) ends at \(z=1\). Thus the intersection area is \(\frac12(x-1)\) on the first interval and \(\frac12\) on the second. Its volume is

$$ \begin{aligned} V_{\mathrm{overlap}} &=\int_1^2\frac12(x-1)\,dx+\int_2^3\frac12\,dx\\ &=\frac12\left[\frac{(x-1)^2}{2}\right]_1^2 +\frac12[x]_2^3\\ &=\frac12\left(\frac12-0\right)+\frac12(3-2)\\ &=\frac14+\frac12=\frac34. \end{aligned} $$

Subtract the shared volume once:

$$ V_{\mathrm{union}} =V_A+V_B-V_{\mathrm{overlap}} =3+2-\frac34 =\frac{17}{4}. $$

Check by integrating the union slice areas: For \(0\le x\le 1\), only \(A\) is present, with area \(1\). For \(1\le x\le 2\), the union area is \(1+(x-1)-\frac12(x-1)=1+\frac12(x-1)\). For \(2\le x\le 3\), it is \(1+(x-1)-\frac12=x-\frac12\). Therefore

$$ \begin{aligned} V_{\mathrm{union}} &=\int_0^1 1\,dx +\int_1^2\left(1+\frac12(x-1)\right)\,dx +\int_2^3\left(x-\frac12\right)\,dx\\ &=1+\frac54+2 =\frac{17}{4}. \end{aligned} $$

Both methods give the same volume. Simply adding \(V_A+V_B=5\) would be wrong: it counts the intersection, of volume \(\frac34\), twice.

Common Pitfalls

A frequent error is to identify two solid components, calculate each volume correctly, and then add them without checking whether they occupy shared space. In the overlapping-rectangles example, the incorrect calculation is

$$ V_{\mathrm{wrong}}=V_A+V_B=3+2=5. $$

This is too large because each point in the intersection belongs to both components. The correction is to subtract that intersection once, giving \(5-\frac34=\frac{17}{4}\). Alternatively, form the area of the union at each slice position before integrating.

A second error is to use one area formula across a point where the cross-sectional shape or dimensions change. The cone-cylinder example requires one formula on \([0,2]\) and another on \([2,5]\); using the cone formula throughout would omit the cylindrical portion. Always mark the transition positions before setting up the integrals.

Finally, do not confuse an end face shared by adjoining pieces with a volume overlap. A single cross section has zero thickness along the \(x\)-direction, so that shared boundary contributes no volume. An overlap across an interval of \(x\)-values has positive thickness and must be accounted for.

A Reliable Setup

Before evaluating any integrals, make the geometric structure explicit. A useful sequence is:

1
Choose the slice direction.
Use the direction in which the solid can be described with clear cross sections and a manageable interval.
2
Mark every transition.
Partition wherever a component begins or ends, or the slice shape, dimensions, or overlap changes.
3
Find the full slice area.
On each interval, include the union of the component regions; subtract any overlap if adding component areas.
4
Integrate and combine.
Integrate each area formula over its interval, then add the interval volumes.
5
Check the model.
Confirm that the formulas apply on the stated intervals and that no part of the solid has been omitted or counted twice.

Composite slice integrals are useful whenever a single geometric formula cannot describe the whole solid. The integral does not change: the careful work is constructing an area function that matches the solid, handles each transition, and represents overlapping pieces only once.

Check Your Understanding

Answer these questions by identifying the slice areas and the intervals on which they apply.

  1. A solid has triangular slices for \(0\le x\le 1\) and square slices for \(1\le x\le 3\). What information is needed to write its total volume as a sum of integrals?
  2. Two components meet only along a face at \(x=2\). Does the shared face require a volume subtraction? Explain.
  3. Two components occupy the same \(x\)-interval and their cross sections overlap. What slice-area calculation prevents double-counting?
  4. If component volumes are \(8\) and \(5\), and their intersection has volume \(2\), what is the volume of their union?
  5. Why should a new partition point be included when the overlap width changes formula?

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