Vertical Base Segments and Volume
A solid can be described by the shapes of its cross sections rather than by a familiar solid formula. When each cross section is built on a vertical base segment, the segment runs from a lower \(y\)-value to an upper \(y\)-value at a fixed \(x\). Moving from left to right through the solid gives the volume integral in terms of \(x\).
The essential geometric step is to find the length of that segment. If its top boundary is \(y=f(x)\) and its bottom boundary is \(y=g(x)\), its length is \(f(x)-g(x)\), provided \(f(x)\ge g(x)\) on the interval. The specified shape determines how this length becomes a cross-sectional area.
From a Vertical Segment to a Volume Integral
Suppose the region in the \(xy\)-plane extends from \(x=a\) to \(x=b\). At each \(x\), assume the vertical base segment runs between continuous boundary functions \(y=g(x)\) and \(y=f(x)\), with \(f(x)\ge g(x)\). Its length is
The cross-sectional area depends on the shape and on the measurement \(s(x)\). For example, if each cross section is a square, its area is \(s(x)^2\). If each is an equilateral triangle, its area is \(\frac{\sqrt{3}}{4}s(x)^2\). Once the area function \(A(x)\) is known, the volume is
Here \(a\) and \(b\) are the left and right endpoints of the region’s \(x\)-interval. The integration variable is \(x\) because the slices are indexed by their \(x\)-coordinates.
Why Integrating Cross-Sectional Area Gives Volume
Consider dividing \([a,b]\) into short intervals, each of width \(\Delta x\). Over a sufficiently short interval, the solid’s cross sections have areas close to the area at a sample value \(x_i^*\). A thin slice of the solid is therefore approximated by a thin slab with volume \(A(x_i^*)\Delta x\). Adding the slab volumes gives an approximation to the total volume.
As the interval widths shrink toward zero, these sums approach the definite integral of the cross-sectional area. For continuous area functions, this is the definition of the corresponding Riemann integral; for piecewise-continuous area functions, the integral can be taken piece by piece.
This reasoning explains both why the cross-sectional area belongs inside the integral and why the thickness is \(dx\). Each thin slice has width in the \(x\)-direction, while its cross section is formed on the vertical base segment.
Worked Example: Equilateral Triangles on Segments Between Two Curves
Worked Example: Equilateral Triangles on Segments Between Two Curves
The base region lies between \(y=x\) and \(y=x^2\) for \(0\le x\le 1\). Cross sections perpendicular to the \(x\)-axis are equilateral triangles whose side lengths are the vertical segments in the region. Find the volume.
On this interval, \(x\ge x^2\), so the top boundary is \(y=x\) and the bottom boundary is \(y=x^2\). The side length is
An equilateral triangle with side length \(s\) has area \(\frac{\sqrt{3}}{4}s^2\). Therefore, the cross-sectional area is
Integrate from the left endpoint \(0\) to the right endpoint \(1\):
Check the integration by differentiating the antiderivative inside the brackets: \(\frac{d}{dx}\left(\frac{x^3}{3}-\frac{x^4}{2}+\frac{x^5}{5}\right)=x^2-2x^3+x^4\), which is the expanded square of the side length. Thus the volume is \(\frac{\sqrt{3}}{120}\) cubic units.
Common Cross-Section Area Formulas
The same vertical-segment procedure works for different specified shapes. First find the segment length \(s(x)\); then use the appropriate area formula. For example, a semicircle with diameter \(s(x)\) has radius \(\frac{s(x)}{2}\), so its area is \(\frac{1}{2}\pi\left(\frac{s(x)}{2}\right)^2=\frac{\pi}{8}s(x)^2\).
| Cross-sectional shape | Area in terms of segment length \(s\) |
|---|---|
| Square | \(s^2\) |
| Equilateral triangle | \(\frac{\sqrt{3}}{4}s^2\) |
| Semicircle with diameter \(s\) | \(\frac{\pi}{8}s^2\) |
| Circle with diameter \(s\) | \(\frac{\pi}{4}s^2\) |
These formulas apply when the given vertical base segment is, respectively, the square’s side, the triangle’s side, or the stated circle’s diameter. Do not substitute a segment length into a formula that uses a different measurement without converting it first.
Worked Example: Semicircles with Vertical Diameters
Worked Example: Semicircles with Vertical Diameters
A base region is bounded by \(y=0\), \(y=\sqrt{x}\), \(x=0\), and \(x=4\). Each cross section perpendicular to the \(x\)-axis is a semicircle whose diameter is the vertical segment in the region. Find the volume.
At each \(x\) between \(0\) and \(4\), the top boundary is \(\sqrt{x}\) and the bottom boundary is \(0\). Thus the diameter is \(s(x)=\sqrt{x}\). The semicircle area formula gives
Integrating over the stated \(x\)-bounds gives
Differentiate \(\frac{\pi x^2}{16}\), the antiderivative used in the calculation: its derivative is \(\frac{\pi x}{8}\), as required. The volume is \(\pi\) cubic units.
When the Top and Bottom Boundaries Switch
Two boundary graphs can cross within the interval. At a crossing, the vertical segment may shrink to length zero; on the other side, the graph that was lower may become the upper boundary. The segment length is always the nonnegative distance from bottom to top. If a single formula \(f(x)-g(x)\) becomes negative, the boundary roles have not been handled correctly on that part.
In such a situation, find the intersection \(x\)-value, split the integral there, and write the segment length using the correct top and bottom functions on each interval. The next example shows this explicitly for square cross sections.
Worked Example: Squares When the Boundary Graphs Cross
Worked Example: Squares When the Boundary Graphs Cross
For \(0\le x\le 2\), the base region lies between \(y=0\) and \(y=x^2-2\). Each cross section perpendicular to the \(x\)-axis is a square with the vertical segment as its side. Find the volume.
The graphs meet where \(x^2-2=0\), so \(x^2=2\), and the point in the interval is \(x=\sqrt{2}\). For \(0\le x\le\sqrt{2}\), \(x^2-2\le 0\): the top boundary is \(0\), the bottom boundary is \(x^2-2\), and the side length is \(2-x^2\). For \(\sqrt{2}\le x\le 2\), the top boundary is \(x^2-2\), the bottom boundary is \(0\), and the side length is \(x^2-2\).
Square area is the side length squared. The volume integral must therefore be split at \(\sqrt{2}\):
An antiderivative of \(4-4x^2+x^4\) is \(F(x)=4x-\frac{4}{3}x^3+\frac{x^5}{5}\), since
For the first integral, use \((\sqrt{2})^3=2\sqrt{2}\) and \((\sqrt{2})^5=4\sqrt{2}\):
For the second integral, evaluate both endpoints carefully. At \(x=2\),
At \(x=\sqrt{2}\), the earlier evaluation gives \(F(\sqrt{2})=\frac{32\sqrt{2}}{15}\). Thus
Adding the two pieces cancels the terms involving \(\sqrt{2}\):
The integrand on each piece is the square of the appropriate side length, and differentiating \(F\) gives the expanded integrand \(4-4x^2+x^4\). The volume is \(\frac{56}{15}\) cubic units.
A Pitfall: Keeping a Negative “Length”
A vertical segment’s length cannot be negative. For instance, on \(0\le x<\sqrt{2}\) in the previous example, \(x^2-2<0\). Using \(x^2-2\) as the square’s side length and then integrating that expression as though it were an area would give a negative contribution, which cannot represent volume. The correction is to identify the top and bottom boundaries: the segment length there is \(0-(x^2-2)=2-x^2\), and the square’s area is \((2-x^2)^2\). On the other side of the crossing, the length is \(x^2-2\). Correctly identifying the segment on each interval makes both cross-sectional areas nonnegative.
Check Your Setup
Before evaluating a volume integral based on vertical segments, confirm that the \(x\)-bounds match the region, the segment length is measured from the lower boundary to the upper boundary, and the area formula uses the specified geometric measurement. Then check that any boundary switch inside the interval has been handled.
Check Your Understanding
For each question, focus on how the vertical base segment determines the cross-sectional area.
- A region lies between \(y=3x\) and \(y=x\) for \(0\le x\le 1\). What is the vertical segment length on this interval?
- If that segment is the side of a square, what is the cross-sectional area function?
- A semicircle has a vertical base segment of length \(s(x)\) as its diameter. Why is its area \(\frac{\pi}{8}s(x)^2\), rather than \(\frac{\pi}{4}s(x)^2\)?
- For the region between \(y=0\) and \(y=x^2-9\) on \(0\le x\le 4\), at what \(x\)-value do the top and bottom boundaries switch?
- Why does a volume integral based on vertical base segments use \(dx\) rather than \(dy\)?
