Why a Volume Model May Need More Than One Formula
A solid can have cross sections whose dimensions change as we move along its base. Sometimes the base region itself changes width: different curves form its boundary on different intervals. In other cases, the rule for the cross-sectional shape changes partway along the base. Either change can make one area formula inadequate for the whole solid.
The method is to locate the points where the rule changes, divide the base interval there, and write the cross-sectional area on each resulting interval. The total volume is the sum of the volumes from those intervals.
This is not a different principle of volume from the one used for solids with a single slice rule. It is the same principle applied carefully to several intervals, each with its own valid area expression.
Piecewise Cross-Section Areas
Suppose slices perpendicular to the \(x\)-axis run from \(x=a\) to \(x=b\). At each \(x\), let \(A(x)\) denote the area of the cross section. When the formula for \(A(x)\) changes at points \(c_1,\ldots,c_{n-1}\), use those points to partition the interval. On each part, use the area rule that is valid there.
The area formula may change because the slice width changes according to a different boundary expression, because the cross-sectional shape changes, or because both happen. A breakpoint is a location where one of the rules changes. It is not necessary to split at every point where a formula happens to have a different-looking algebraic form; split where the geometric rule or the expression valid for it changes.
For a slice whose base width is \(w(x)\), the cross-sectional shape determines how that width is used. For example, a square with side \(w(x)\) has area \(w(x)^2\). A semicircle with diameter \(w(x)\) has radius \(w(x)/2\), so its area is \(\frac{\pi}{8}w(x)^2\). The area expressions must match both the slice width and the stated shape.
Why the Interval Volumes Add
The volume integral adds the areas of thin slices across the base. If the base is divided into adjacent intervals, the slices from those intervals together make up the whole solid. Their volumes therefore add. The integral notation records that addition exactly.
To justify the formula, take a Riemann sum for the area function over the full interval. Choose a partition that includes each breakpoint \(c_i\). The sum then separates into groups of terms, one group for each interval \([c_{i-1},c_i]\). As the subdivisions become finer, each group approaches the integral on its interval. The limit of the full sum is the sum of these finitely many limits. Hence the integral over the full interval equals the sum of the interval integrals. A cross section exactly at a breakpoint has no thickness, so it contributes no separate volume.
The conditions matter: the area must describe the actual cross section on each interval, and the integrals must exist. Piecewise continuous area formulas on a finite interval meet the integrability condition used here.
Worked Example: A Square Slice Rule with a Changing Base Width
Worked Example: A Square Slice Rule with a Changing Base Width
A base region lies above \(y=0\) and below \(y=x\) for \(0\le x\le1\), then below \(y=2-x\) for \(1\le x\le2\). Perpendicular cross sections are squares whose sides equal the vertical width of the base. Find the volume.
Step 1: Find the slice width on each interval. On \(0\le x\le1\), the vertical width is \(x-0=x\). On \(1\le x\le2\), it is \((2-x)-0=2-x\). The boundary rule changes at \(x=1\), so split the volume there.
Step 2: Convert width to area. A square of side \(s\) has area \(s^2\). Therefore the slice area is \(x^2\) on the first interval and \((2-x)^2\) on the second.
Step 3: Integrate each area expression.
Check: Each interval has length \(1\), and the square side rises from \(0\) to \(1\) on the first interval and falls from \(1\) to \(0\) on the second. Each interval contributes \(\frac{1}{3}\) cubic units by its integral, so the total \(\frac{2}{3}\) is consistent with the two matching contributions. The two area formulas also agree at the breakpoint: \(1^2=(2-1)^2=1\).
Worked Example: Changing the Cross-Section Shape
Worked Example: Changing the Cross-Section Shape
Let the base be the rectangle \(0\le x\le1\), \(0\le y\le2\), so every vertical slice has width \(2\). For \(0\le x\le\frac12\), the cross sections are squares with side \(2\). For \(\frac12\le x\le1\), they are semicircles with diameter \(2\). Find the volume.
Step 1: Write the areas. A square of side \(2\) has area \(2^2=4\). For a semicircle with diameter \(2\), the radius is \(1\), so the area is \(\frac12\pi(1)^2=\frac{\pi}{2}\).
Step 2: Integrate on the intervals where each rule applies.
Check: The square-rule part has length \(\frac12\) and area \(4\), giving volume \(4\cdot\frac12=2\). The semicircle-rule part has the same length and area \(\frac{\pi}{2}\), giving volume \(\frac{\pi}{2}\cdot\frac12=\frac{\pi}{4}\). Their sum is \(2+\frac{\pi}{4}\) cubic units, agreeing with the integral.
Worked Example: Both the Width and the Shape Rule Change
Worked Example: Both the Width and the Shape Rule Change
A base region lies above \(y=0\). Its upper boundary is \(y=x\) for \(0\le x\le1\), and \(y=\frac{3-x}{2}\) for \(1\le x\le3\). For \(0\le x\le1\), the cross section is a square whose side is the vertical width of the base. For \(1\le x\le3\), it is a semicircle whose diameter is that width. Find the volume.
Step 1: Determine the width. The width is \(x\) on the first interval and \(\frac{3-x}{2}\) on the second. The base boundary changes at \(x=1\), and the cross-section shape also changes there.
Step 2: Find the area rules. On \([0,1]\), the square area is \(x^2\). On \([1,3]\), the semicircle diameter is \(w(x)=\frac{3-x}{2}\). Its area is \(\frac{\pi}{8}w(x)^2\), giving $$ A(x)=\frac{\pi}{8}\left(\frac{3-x}{2}\right)^2 =\frac{\pi(3-x)^2}{32}. $$
Step 3: Integrate and add.
Check: The square area is \(1\) at \(x=1\). The semicircle diameter there is also \(1\), so its area is \(\frac{\pi}{8}\); the cross-sectional areas need not match, because the shape changes. At \(x=3\), the diameter is \(0\), as required at the end of the base. The integral contributions are \(\frac13\) and \(\frac{\pi}{12}\), respectively, giving the stated total.
A Common Mistake: Using One Width Formula Everywhere
In the first example, the width is \(x\) only on \(0\le x\le1\). Extending that formula across the whole interval gives this incorrect calculation:
This model treats the square side as \(2\) at \(x=2\), even though the base width there is \(0\). It has used a formula outside the interval where it describes the region. The correction is to use \(x\) on \([0,1]\), use \(2-x\) on \([1,2]\), and add the two integrals, which gives \(\frac23\).
A Reliable Setup
For a piecewise cross-section problem, organize the work in this order:
Identify where boundary curves exchange roles or where the stated slice shape changes.
Subtract the lower boundary from the upper boundary for vertical slices, or use the corresponding horizontal distance for horizontal slices.
Use the width and the specified cross-sectional shape to obtain the area expression on that interval.
Integrate each area formula over exactly the interval where it applies, then sum the results.
Pay particular attention to what is changing. A width can change while the cross-section shape stays the same, as in the square example. The width can stay fixed while the shape changes, as in the rectangle example. Both can change together. In all three cases, the volume calculation follows the same rule: use the correct area expression on each part of the base and add the resulting volumes.
Check Your Understanding
For each question, identify the interval-specific area rules before evaluating any integrals.
- A base width is \(x\) on \([0,1]\) and \(2-x\) on \([1,2]\). Cross sections are squares. What area function applies on each interval, and where must the volume integral be split?
- A rectangular base has constant slice width \(3\) for \(0\le x\le2\). Slices are squares on \([0,1]\) and semicircles with diameter \(3\) on \([1,2]\). Write the volume as a sum of two integrals.
- Why does a cross section at a single breakpoint not add a separate volume term?
- For a semicircle with diameter \(w(x)\), explain why the area is \(\frac{\pi}{8}w(x)^2\), rather than \(\frac{\pi}{2}w(x)^2\).
- What geometric error is made by using a width formula on an interval where it no longer describes the base?
