From a Planar Region to a Solid
A planar region can describe the base of a three-dimensional solid, but the base alone does not determine the solid's volume. We also need to know the shape of each cross-section. If sections perpendicular to a chosen axis are squares, triangles, or semicircles, their areas vary as we move along that axis. Adding those changing areas across the base gives the volume.
The central task is to make the geometry precise before integrating: choose the slice direction, find the interval of slices, and determine the length of the segment cut from the base region by each slice. That segment is the side length, base, or diameter of the specified cross-section.
Why the Area Integral Gives Volume
Divide the interval of slices into short pieces. Over a small interval of width \(\Delta x\), the solid is approximately a thin slab whose cross-sectional area is close to \(A(x_i^*)\) at some sample point \(x_i^*\). Its approximate volume is \(A(x_i^*)\Delta x\). Adding these slab volumes gives an approximation to the whole volume.
As the widths shrink, these sums approach the definite integral. This is the same limit of sums that defines integration, now applied to cross-sectional areas. The reasoning works when the areas vary continuously, and also piece by piece when the solid has finitely many changes in its description and the corresponding integrals are split at those changes.
The cross-sectional shape determines the formula for \(A\); the base region determines the length used in that formula. Keep these roles separate. For instance, the same chord length produces different areas for a square and a semicircle.
Finding the Slice Direction and Chord
A slice perpendicular to the \(x\)-axis is vertical when viewed in the base region. At a fixed \(x\), its intersection with the region is usually a segment running from a lower boundary curve to an upper boundary curve. Its length is upper \(y\)-value minus lower \(y\)-value. A slice perpendicular to the \(y\)-axis is horizontal; its chord length is right \(x\)-value minus left \(x\)-value.
The integration bounds are the endpoints of the region's projection onto the axis along which slices move. Thus vertical slices use the smallest and largest \(x\)-values in the region, while horizontal slices use its smallest and largest \(y\)-values. If one slice direction makes the boundary description or chord calculation awkward, consider the other direction—but use the direction specified by the solid's cross-section planes.
Area Formulas for Three Cross-Sections
Let \(s\) denote a square's side length, \(b\) the base of an equilateral triangle, and \(d\) the diameter of a semicircle. When the chord is specified as that side, base, or diameter, the cross-sectional areas are:
The triangle formula follows from its height. An equilateral triangle of side \(b\) has height \(\frac{\sqrt{3}}{2}b\), so its area is \(\frac{1}{2}b\left(\frac{\sqrt{3}}{2}b\right)=\frac{\sqrt{3}}{4}b^2\). For a semicircle, the radius is half its diameter, so its area is half the area of a circle with radius \(d/2\): \(\frac{1}{2}\pi(d/2)^2=\frac{\pi}{8}d^2\).
Worked Example: Squares on Vertical Chords
The base region is bounded by \(y=0\), \(y=2-x\), \(x=0\), and \(x=2\). Cross-sections perpendicular to the \(x\)-axis are squares. Find the volume.
1. Find the bounds. The region extends from \(x=0\) to \(x=2\), so integrate over \([0,2]\).
2. Find the chord length. A vertical slice runs from \(y=0\) to \(y=2-x\). Its length, and therefore the square's side, is \(s(x)=(2-x)-0=2-x\).
3. Write the area and integrate.
Check. Differentiating the antiderivative gives \((2-x)^2\), the cross-sectional area. The result is positive, as a volume must be. If lengths are measured in units, the volume is \(\frac{8}{3}\) cubic units.
Worked Example: Equilateral Triangles on Horizontal Chords
The base region is bounded by \(x=0\), \(x=y\), and \(y=2\). Cross-sections perpendicular to the \(y\)-axis are equilateral triangles whose bases are the horizontal chords of the region. Find the volume.
1. Find the bounds. The region extends vertically from \(y=0\) to \(y=2\), so use those bounds and integrate with respect to \(y\).
2. Find the chord length. At a fixed \(y\), the horizontal slice runs from \(x=0\) to \(x=y\). Thus \(b(y)=y-0=y\).
3. Use the triangle area formula and integrate.
Check. Differentiating \(\frac{\sqrt{3}}{12}y^3\) gives \(\frac{\sqrt{3}}{4}y^2\), the stated area. The volume is positive and has cubic units when the base coordinates have units of length.
Worked Example: Semicircles on Vertical Chords
The base region lies between \(y=x^2\) and \(y=2\). Cross-sections perpendicular to the \(x\)-axis are semicircles whose diameters are the vertical chords of the region. Find the volume.
1. Find the bounds. The curves meet where \(x^2=2\), so \(x=-\sqrt{2}\) and \(x=\sqrt{2}\). These give the full \(x\)-interval.
2. Find the diameter. The upper boundary is \(y=2\) and the lower boundary is \(y=x^2\). Therefore \(d(x)=2-x^2\).
3. Form the area and evaluate. A semicircle with diameter \(d\) has area \(\frac{\pi}{8}d^2\).
The second equality uses symmetry: the integrand is even, so the integral over the symmetric interval is twice the integral from \(0\) to \(\sqrt{2}\). Continue by integrating each term:
Check. Differentiating the antiderivative inside the brackets gives \(4-4x^2+x^4=(2-x^2)^2\), the squared diameter. The volume is positive, and the cross-sectional area vanishes at both endpoints where the base curves meet.
A Common Chord-Length Mistake
Suppose a region lies between \(y=x\) and \(y=2x\) for \(0\le x\le1\), with square cross-sections perpendicular to the \(x\)-axis. The vertical chord has length \(2x-x=x\), so its square area is \(x^2\). A common error is to use the upper boundary \(2x\) alone as the side length. That would give the incorrect volume \(\int_0^1(2x)^2\,dx=\frac{4}{3}\).
The boundary curve's height is not generally the chord length. The chord is the distance between the two boundaries. Here the correct volume is \(\int_0^1 x^2\,dx=\frac{1}{3}\). Indeed, an antiderivative is \(x^3/3\), whose derivative is \(x^2\), and evaluating at \(0\) and \(1\) gives \(1/3\). This check also makes the geometric error visible: at each \(x>0\), the actual chord is shorter than \(2x\).
Putting the Setup Together
Before evaluating a cross-section volume integral, make the geometry explicit. A short written record of the slice direction, bounds, and chord length helps prevent errors in both the integral and its interpretation.
Identify the axis perpendicular to the slices and the shape built on each chord.
Project the base region onto the axis along which the slices move.
Subtract lower from upper for vertical slices, or left from right for horizontal slices.
Use the chord as the side, base, or diameter required by the specified shape, then integrate that area across the bounds.
Differentiate the antiderivative, inspect the sign and endpoint behavior, and confirm that the units are cubic.
This method matters because it separates the geometric model from the integration. If a volume seems implausible, you can check whether the bounds cover the whole base, whether the chord is a difference of boundary values, and whether the correct shape formula was used. A correct integral evaluates the solid described by the setup; a mistaken chord or shape can produce a perfectly computed answer to a different solid.
Check Your Understanding
For each question, identify the geometry before carrying out any calculations.
- A base region is \(0\le x\le 3\), \(0\le y\le 3-x\). What is the vertical chord length at \(x\), and what are the bounds for cross-sections perpendicular to the \(x\)-axis?
- A horizontal slice of a region runs from \(x=1\) to \(x=4-y\). What is its chord length?
- Write the area of an equilateral-triangle cross-section whose base chord has length \(3x\).
- A semicircle has diameter \(d\). Explain why its area is \(\frac{\pi}{8}d^2\), rather than \(\frac{\pi}{4}d^2\).
- For vertical slices, the region lies between \(y=x^2\) and \(y=5\). Find the \(x\)-bounds by locating the intersections of the boundary curves.