Why a Volume Integral May Need Several Pieces
A solid can be sliced in one direction while its cross sections change from one part of the base region to another. A boundary curve may switch from being the upper boundary to the lower boundary, the left and right edges of a slice may exchange roles, or the prescribed cross-sectional shape may change. In each case, the area of a slice may follow one formula on part of the interval and another formula elsewhere.
The key is to split the volume integral wherever the area formula changes. On each resulting interval, use the formula that correctly describes the cross sections there. Then add the integrals. This is not a new volume principle: it is the slicing principle applied piece by piece.
Finding the Breakpoints
A breakpoint is a value of the slicing variable where the formula needed to describe a slice changes. It may occur where two boundary curves intersect, where the base region begins or ends a new piece, or where a problem specifies a new cross-sectional shape.
For vertical slices, first identify which boundary is above and which is below on each interval. The slice length is upper boundary minus lower boundary. For horizontal slices, identify which boundary is to the right and which is to the left; the slice length is right boundary minus left boundary. If the ordering changes, find where the boundaries are equal and split there.
The area formula must also match the specified shape. For example, a square with side length \(L\) has area \(L^2\), while a semicircle with diameter \(L\) has area \(\frac{\pi L^2}{8}\). If the shape changes, the area formula changes even if the base slice length does not.
Why Splitting Works
The slicing principle says that a volume is the integral of its cross-sectional area, provided the area function is integrable over the interval. Definite integrals are additive over adjacent intervals. Thus, for a breakpoint \(c\) between \(a\) and \(b\), the total volume can be computed by adding the two parts.
Here \(t\) stands for the slicing variable, either \(x\) or \(y\). With several breakpoints, apply the same property repeatedly. On each interval, use the area formula valid there.
Use the variable that matches the direction of the cross sections.
Find where boundaries switch order, the base interval changes, or the cross-sectional rule changes.
Find the slice length first, then apply the appropriate geometric area formula.
Evaluate each definite integral and sum the resulting volumes.
This reasoning is complete because the intervals meet only at their endpoints, and definite integrals are additive there. For a piecewise continuous area function, changing its value at finitely many breakpoints does not change the integral. The formula at the exact breakpoint therefore does not affect the volume.
Worked Example: Square Cross Sections When Boundaries Switch
Worked Example: Square Cross Sections When Boundaries Switch
A base region is bounded by \(y=x\), \(y=2-x\), and \(0\le x\le2\). Cross sections perpendicular to the \(x\)-axis are squares. Find the volume.
The boundaries intersect where \(x=2-x\), so \(2x=2\) and \(x=1\). On \(0\le x\le1\), \(2-x\) is above \(x\), giving slice length \(L(x)=(2-x)-x=2-2x\). On \(1\le x\le2\), \(x\) is above \(2-x\), giving \(L(x)=x-(2-x)=2x-2\). Since a square with side \(L\) has area \(L^2\),
Split at \(x=1\) and integrate each piece:
Check the antiderivatives: the derivative of \(4(x-x^2+x^3/3)\) is \(4(1-2x+x^2)=4(1-x)^2\), and the derivative of \(4(x-1)^3/3\) is \(4(x-1)^2\). At the breakpoint the slice length is zero from either side, as expected where the boundary curves meet. The volume is \(\frac83\) cubic units.
Worked Example: The Cross-Section Shape Changes
Worked Example: The Cross-Section Shape Changes
The base region is \(0\le x\le2\), \(0\le y\le x\). For \(0\le x\le1\), cross sections perpendicular to the \(x\)-axis are squares whose side length is the vertical slice length. For \(1\le x\le2\), they are semicircles whose diameter is that length. Find the volume.
At each \(x\), the vertical slice runs from \(y=0\) to \(y=x\), so \(L(x)=x\). The cross-sectional shape changes at \(x=1\). For \(0\le x\le1\), the area is \(x^2\). For \(1\le x\le2\), the semicircle has radius \(x/2\), so its area is \(\frac12\pi(x/2)^2=\frac{\pi x^2}{8}\). Therefore,
Evaluate each part:
The checks are \(\frac{d}{dx}(x^3/3)=x^2\) and \(\frac{d}{dx}(\pi x^3/24)=\pi x^2/8\). The two volumes are \(\frac13\) and \(\frac{7\pi}{24}\), respectively, so their sum is \(\frac13+\frac{7\pi}{24}\) cubic units. The cross-sectional area can jump at \(x=1\) because the shape changes there; a value at that single point does not affect the volume integral.
Worked Example: Horizontal Slices with a Boundary Switch
Worked Example: Horizontal Slices with a Boundary Switch
Consider the base consisting of points between \(x=y^2\) and \(x=2-y\), for \(-1\le y\le2\). Cross sections perpendicular to the \(y\)-axis are rectangles with height \(2\) and base equal to the horizontal slice length. Find the volume.
To locate the switch, solve \(y^2=2-y\): \(y^2+y-2=0\), so \((y+2)(y-1)=0\). The switch in the specified interval is \(y=1\). For \(-1\le y\le1\), \(y^2\le2-y\), so the slice length is \(L(y)=2-y-y^2\). For \(1\le y\le2\), \(2-y\le y^2\), so \(L(y)=y^2+y-2\). Since rectangle area is height times base, \(A(y)=2L(y)\).
For the first integral, an antiderivative of \(2-y-y^2\) is \(2y-y^2/2-y^3/3\). Thus,
For the second integral, an antiderivative of \(y^2+y-2\) is \(y^3/3+y^2/2-2y\). Therefore,
Combining the pieces gives \(V=2(10/3+11/6)=2(31/6)=31/3\) cubic units. Differentiating the two antiderivatives gives \(2-y-y^2\) and \(y^2+y-2\), respectively. At \(y=1\), both formulas give a slice length of zero, confirming that the boundary curves meet there.
Common Mistake: Keeping a Signed Slice Length
A slice length is never negative. If two boundary curves switch order, subtracting the same named curve from the other across the entire interval can create a negative expression. That is not a negative area; it is a signal that the boundaries must be reordered.
For example, use the base between \(y=x\) and \(y=2-x\) on \(0\le x\le2\), but make the cross sections rectangles of height \(2\). A mistaken setup keeps \(2-x\) as the upper boundary throughout:
This is wrong: on \(1<x\le2\), the expression \((2-x)-x\) is negative even though a rectangle cannot have negative base length. The correction is to split at the intersection \(x=1\) and use the positive length on each interval:
As a check, the derivatives of \(2x-x^2\) and \(x^2-2x\) are \(2-2x\) and \(2x-2\). The corrected volume is \(4\) cubic units. The error came from using a signed difference instead of the nonnegative slice length.
A Reliable Setup Check
Before integrating a piecewise area function, check each interval separately. Confirm that the interval covers the base exactly once, the boundary order is correct there, and the shape formula uses the stated dimension correctly. In particular, a semicircle whose diameter is \(L\) has radius \(L/2\), not \(L\).
It is not necessary to split merely because an algebraic expression could be rewritten in another form. Split when the geometric description or the resulting area formula changes. At a switch, solving for the intersection gives the candidate breakpoint; then check the ordering on either side. If several changes occur, sort all their \(x\)- or \(y\)-values and integrate over every resulting interval.
Check Your Understanding
For each question, identify where the area formula changes before attempting the integral.
- Two boundary curves intersect at \(x=3\) within the base interval. What must you check on either side of \(x=3\) before writing a slice length?
- A square cross section has side length \(L(x)\) on one interval. What is its area formula?
- A semicircle has diameter \(L(y)\). What is its area formula in terms of \(L(y)\)?
- Why can the area function have different formulas on adjacent intervals and still give a valid volume integral?
- For horizontal slices, the boundary curves are \(x=y^2\) and \(x=2-y\). At what \(y\)-values do they intersect?
