Why a Volume Setup May Need More Than One Integral
A volume integral is built from the areas of slices. When those areas follow one formula across the entire interval, a single integral may be enough. But the geometry can change: the upper and lower boundaries of the base region can switch, or the prescribed cross-sectional shape can change. In either situation, using one formula without accounting for the change can produce an incorrect setup.
The key is to identify the intervals on which the geometry is consistent. Find where boundaries meet or where the slice shape changes, split the interval at those points, and use the appropriate area formula on each piece. The volume is then the sum of the resulting integrals.
How to Identify the Intervals
Suppose a base region is described by curves and vertical lines, and its cross sections are perpendicular to the \(x\)-axis. At a fixed \(x\), the slice through the base region is a vertical segment. Its endpoints are the relevant \(y\)-values on the region’s boundary. The slice length is the upper \(y\)-value minus the lower \(y\)-value, so that length is nonnegative.
If two boundary curves exchange which one is above the other, their intersection gives a point where the slice formula may change. Solve for the intersection, then check the order of the curves on each side of it. A change in cross-sectional shape, if one is specified, is another reason to split the interval. A boundary point alone is not automatically a split point: split where the formula for the area actually changes.
Use the stated limits of the base region or its boundary intersections to determine the starting and ending \(x\)-values.
Solve for intersections of relevant boundary curves and note any stated changes in cross-sectional shape.
Identify the upper and lower boundaries and calculate the nonnegative slice length.
Use the area formula for the specified cross section on each interval, then sum the definite integrals.
Why Adding the Slice Areas Gives Volume
Divide the interval from \(a\) to \(b\) into short subintervals. On a subinterval of width \(\Delta x\), the solid is approximately a thin slab whose volume is its cross-sectional area multiplied by its thickness. If a sample point in that subinterval is \(x_i^*\), the slab’s approximate volume is \(A(x_i^*)\Delta x\). Adding the slab estimates gives a Riemann sum.
As the widths of the subintervals approach zero, the sum approaches the definite integral of the cross-sectional area. If the area formula changes at finitely many points, split the interval there. The volume of the whole solid is the sum of the volumes on the resulting intervals; the cross sections at the dividing points do not add any separate volume.
Worked Example: Squares with Switching Boundaries
Worked Example: Squares with Switching Boundaries
The base region lies between \(y=x\) and \(y=2-x\) for \(0\le x\le 2\). Cross sections perpendicular to the \(x\)-axis are squares whose side lengths are the vertical lengths of the base region. Find the volume.
Step 1: Find where the boundaries meet. Set their \(y\)-values equal:
The curves meet at \(x=1\), where \(y=1\). For \(0\le x\le 1\), \(2-x\ge x\), so the vertical length is \((2-x)-x=2-2x\). For \(1\le x\le 2\), \(x\ge 2-x\), so the vertical length is \(x-(2-x)=2x-2\).
Step 2: Find each cross-sectional area. A square with side length \(s\) has area \(s^2\). Thus the area formulas on the two intervals are \((2-2x)^2\) and \((2x-2)^2\). The volume is
Step 3: Evaluate both integrals. On the first interval, \((2-2x)^2=4(1-x)^2\), so
On the second interval, \((2x-2)^2=4(x-1)^2\), so
Therefore, \(V=\frac43+\frac43=\frac83\) cubic units.
Check: The cross-sectional area is positive for \(x\in(0,1)\cup(1,2)\), and zero at \(x=1\), where the curves meet. At \(x=0\) and \(x=2\), the vertical segment has length \(2\), so the endpoint cross sections have area \(4\). The positive total volume is consistent with these cross sections. The two interval integrals are equal because the geometry on the second interval mirrors the first.
Worked Example: Semicircles on the Same Base Region
Worked Example: Semicircles on the Same Base Region
Use the same base region between \(y=x\) and \(y=2-x\) for \(0\le x\le 2\), but now let each cross section perpendicular to the \(x\)-axis be a semicircle whose diameter is the vertical segment in the base. Find the volume.
The intersection remains at \(x=1\), found from \(x=2-x\). The diameter is \(2-2x\) on \(0\le x\le1\) and \(2x-2\) on \(1\le x\le2\). A semicircle with diameter \(d\) has radius \(d/2\), so its area is
Therefore, the volume is
Each integral inside the parentheses equals \(\frac43\), as calculated in the previous example. Hence
Check: The semicircle area is \(\pi d^2/8\), not \(\pi d^2/4\), because the latter is the area of a full circle with diameter \(d\). The result is positive, and the cross-sectional area is zero only at \(x=1\).
Worked Example: A Change in Cross-Sectional Shape
Worked Example: A Change in Cross-Sectional Shape
The base is the rectangle \(0\le x\le2\), \(0\le y\le1\). For \(0\le x\le1\), cross sections perpendicular to the \(x\)-axis are equilateral triangles whose side is the vertical segment in the base. For \(1\le x\le2\), the cross sections are squares whose side is that segment. Find the volume.
Every vertical segment in the base has length \(1-0=1\). An equilateral triangle with side \(s\) has area \(\frac{\sqrt{3}}{4}s^2\), so the first interval has area \(\frac{\sqrt{3}}{4}\). A square with side \(1\) has area \(1\), so the second interval has area \(1\). Split at \(x=1\), where the prescribed shape changes:
Check: The first cross sections have constant area \(\frac{\sqrt{3}}4\) over a length of \(1\), and the second have constant area \(1\) over a length of \(1\). Adding those two slab volumes gives \(1+\frac{\sqrt{3}}4\) cubic units. The single point \(x=1\) has no effect on the integral.
A Common Setup Error
A frequent mistake is to use the vertical distance between two curves as though it were the cross-sectional area. In the square example, the distance between the curves on the first interval is \(2-2x\), but the cross section is a square, whose area is the square of that distance. Integrating the distance alone gives the wrong quantity.
The correction is to identify the nonnegative side length on each interval and then apply the square-area formula. The correct setup is
Even if a particular formula happens to conceal a boundary switch—for example, squaring a signed difference can produce the same result—finding the correct slice length on each interval is the reliable method. It keeps the geometric meaning, the area, and the units clear.
What to Check Before Evaluating
Before calculating a piecewise volume, verify that each integral represents a cross-sectional area over the interval shown. In particular, check the intersection values, the order of the boundaries between intersections, the specified cross-sectional shape, and whether the area should use a radius, diameter, side, or another dimension. After evaluation, check that the answer is nonnegative and has units of volume.
The central procedure is simple: identify each change in the geometry, write the appropriate area formula on every resulting interval, integrate those areas, and add the results. Careful splitting is not an extra step added to integration; it is how the integral accurately represents the solid.
Check Your Understanding
For each question, identify the relevant change in geometry before evaluating or interpreting the integral.
- For \(y=x\) and \(y=2-x\) on \(0\le x\le2\), at what \(x\)-value do the upper and lower boundaries switch?
- If square cross sections are built on vertical segments between those curves, what is the side length on \(0\le x\le1\)?
- What is the area of a semicircle whose diameter is \(d\), and why must the radius be \(d/2\)?
- Why should the volume integral be split at a point where the prescribed cross-sectional shape changes?
- What units should a volume have if the coordinates are measured in centimeters?
