Why a Volume Integral May Need Several Pieces
A solid’s cross-sectional area can vary as a slice moves through the base region. Sometimes the slice shape changes at a particular location. Sometimes the shape stays the same, but the boundaries of the base region that determine the slice dimensions change. Either way, an area formula that works on one interval may not work on the whole interval.
The remedy is to identify where the geometry changes, divide the interval there, and integrate the appropriate area formula on each piece. The resulting volume is the sum of those integrals.
This approach uses the cross-sectional-area interpretation of volume: a thin slice has approximately its cross-sectional area multiplied by its thickness, and adding the slices leads to an integral. The essential care in a piecewise problem is making sure that every formula matches the geometry on the interval where it is used.
Piecewise Cross-Sectional Area
Let \(x\) measure position along the direction in which slices are taken, and let \(A(x)\) be the area of the cross-section at position \(x\). If the area formula changes at positions \(c_1,\ldots,c_n\), divide the full interval at those positions and use a separate formula on each subinterval.
Here the variable need not be \(x\). For horizontal slices, position may be measured by \(y\), and the volume is integrated with respect to \(y\). The area must be nonnegative, and the piecewise area functions should be integrable on their intervals. In the common case of finitely many continuous formulas on bounded intervals, these conditions are satisfied.
A breakpoint is a position where the formula changes. It can arise because two base-region boundaries meet or exchange roles, because a slice changes shape, or because the stated cross-section rule itself changes. It is useful to distinguish these causes: a boundary change affects the slice dimensions, while a shape change affects how those dimensions are converted into area.
How to Find the Breakpoints and Set Up the Integral
Identify the variable that measures position along the solid and use slices perpendicular to that direction.
Locate where the base-region boundaries intersect or switch which one determines an endpoint, and note any position where the cross-section shape changes.
For each piece, determine the actual length or other dimension of the base-region chord.
Apply the correct cross-section area formula on that interval, then integrate it over the same interval.
Sum the integrals, check each antiderivative, and verify that the result is nonnegative and consistent with the slice areas.
The integral is split at each breakpoint because the area expression on one side may not describe the slices on the other. A single formula can still be used across multiple intervals if it genuinely describes all those slices; a change of boundary does not automatically require a new formula if the resulting area expression is unchanged.
Why the Piecewise Integral Gives the Volume
Suppose the cross-sectional area is given by an integrable formula \(A_i(x)\) on each interval \([c_{i-1},c_i]\), where \(c_0=a\) and \(c_{n+1}=b\). Divide each interval into thin subintervals of width \(\Delta x\). A thin piece of the solid has volume approximately equal to its cross-sectional area multiplied by its thickness. Adding these approximations over one interval and taking the limit gives the definite integral of the area there.
The solid is the union of the pieces between consecutive breakpoints. Those pieces meet only at their boundary slices, which have zero thickness and therefore contribute no volume. The volume of the whole solid is thus the sum of the volumes of the pieces.
The value assigned to the area formula at a single breakpoint does not affect the integral. What matters is that the formulas correctly describe the slices on the intervals between breakpoints.
Worked Examples
Worked Example: A Boundary Switch in a Triangular Base
The base region has vertices \((0,0)\), \((1,1)\), and \((2,0)\). Cross-sections perpendicular to the \(x\)-axis are squares, with one side equal to the vertical chord of the base. Find the volume.
For \(0\leq x\leq 1\), the upper boundary is \(y=x\), so the chord runs from \(y=0\) to \(y=x\). Its length is \(x\), and the square’s area is \(A_1(x)=x^2\).
For \(1\leq x\leq 2\), the upper boundary is \(y=2-x\). The chord length is \(2-x\), so the area is \(A_2(x)=(2-x)^2\). The boundary switch at \(x=1\) requires splitting the integral:
Evaluate each part:
Therefore \(V=\frac13+\frac13=\frac23\) cubic units. The antiderivatives check because \(\frac{d}{dx}(x^3/3)=x^2\) and \(\frac{d}{dx}(-(2-x)^3/3)=(2-x)^2\). The two area formulas also agree at the breakpoint: each gives area \(1\) at \(x=1\).
Worked Example: A Change in Cross-Section Shape
The base is the rectangle \(0\leq x\leq 2\), \(0\leq y\leq 1\). Slices are perpendicular to the \(x\)-axis. For \(0\leq x\leq 1\), each cross-section is a square whose side is the vertical chord of the base. For \(1\leq x\leq 2\), each is a semicircle whose diameter is that chord. Find the volume.
Every vertical chord has length \(1-0=1\). On the first interval, the square area is \(1^2=1\). On the second, the semicircle has radius \(1/2\), so its area is \(\frac12\pi(1/2)^2=\pi/8\). The shape changes at \(x=1\), giving
The volume is \(1+\pi/8\) cubic units. Each antiderivative checks: the derivative of \(x\) is \(1\), and the derivative of \(\pi x/8\) is \(\pi/8\). The result also matches the interval lengths: one unit of length with area \(1\), followed by one unit with area \(\pi/8\).
Worked Example: Horizontal Slices with a Shape Change
The base region is bounded below by \(y=x^2\) and above by \(y=2\). Slices perpendicular to the \(y\)-axis are squares for \(0\leq y\leq 1\), then equilateral triangles for \(1\leq y\leq 2\). In each case, one side of the cross-section is the horizontal chord of the base. Find the volume.
At height \(y\), the inequality \(x^2\leq y\) gives \(-\sqrt y\leq x\leq\sqrt y\). The chord length is therefore \(2\sqrt y\). For \(0\leq y\leq1\), the square’s area is \((2\sqrt y)^2=4y\). For \(1\leq y\leq2\), an equilateral triangle of side \(s\) has area \(\sqrt3s^2/4\), so the area is \(\frac{\sqrt3}{4}(2\sqrt y)^2=\sqrt3y\). Thus
The volume is \(2+\frac{3\sqrt3}{2}\) cubic units. The antiderivatives check because the derivatives of \(2y^2\) and \(\sqrt3y^2/2\) are \(4y\) and \(\sqrt3y\), respectively. At \(y=1\), the chord length is \(2\); the area formulas give \(4\) for the square and \(\sqrt3\) for the triangle, as expected for two different cross-section shapes.
A Common Mistake: Keeping the First Boundary Formula
In the triangular-base example, it is incorrect to use \(A(x)=x^2\) throughout \(0\leq x\leq2\). That formula comes from the boundary \(y=x\), which determines the top of the vertical chord only up to \(x=1\). Using it on the full interval gives
This is not the volume: for \(x>1\), the actual chord length is \(2-x\), not \(x\). The correction is to split at the boundary intersection and use the appropriate area on each side:
Checking the dimensions of the base-region chord before applying a cross-section formula helps prevent this error. In particular, a boundary expression should not be extended beyond the interval where it actually forms the boundary of the slice.
Why Careful Splitting Matters
Piecewise setup makes the geometry visible in the mathematics. Each integral records both the location of a group of slices and the formula for their areas. Correct breakpoints are especially important when the base region has changing boundaries: using the wrong boundary changes the chord, which changes every area and therefore the total volume.
Before evaluating, check that the intervals cover the full solid without gaps or overlaps, that each area expression is nonnegative on its interval, and that the variable of integration matches the slice direction. After evaluating, check the antiderivatives and consider whether the magnitude is plausible given the cross-sectional areas and interval lengths.
Check Your Understanding
For each question, identify the interval or geometry change that determines how the volume integral should be set up.
- A base-region boundary switches at \(x=3\). What should be checked before deciding whether to split the volume integral there?
- A square cross-section has side length \(2-x\) on \(1\leq x\leq2\). What area formula should be integrated on that interval?
- Every chord in a base region has length \(4\). Cross-sections are semicircles on \(0\leq x\leq1\) and squares on \(1\leq x\leq3\). Write the piecewise volume integral.
- For horizontal slices, what geometric information determines the endpoints of the chord at a fixed \(y\)?
- Why does changing the value assigned to the area function at one breakpoint not change the volume integral?
