From a Chord to a Triangular Cross-Section
A solid can be described by the shapes of its slices. In this tutorial, each slice is a triangle, and one dimension of that triangle comes from a chord of a planar base region. The chord length is not necessarily the triangle’s full perimeter or a side visible in a three-dimensional drawing: it is the segment cut out of the base region by the plane of the slice.
The central task is to connect that chord length to the triangle’s area. Once the cross-sectional area is written as a function of the slice position, integrating it gives the volume. The geometric condition on the triangles matters: “right triangle” alone may not determine an area from a chord, whereas a specified altitude-to-base ratio does.
Finding the Chord Length
For vertical slices, a chord runs from the lower boundary of the base region to its upper boundary. If those boundaries are \(y=g(x)\) and \(y=f(x)\), with \(f(x)\geq g(x)\) on the interval, then
For horizontal slices, the chord runs from the left boundary to the right boundary. If those boundaries are \(x=u(y)\) and \(x=v(y)\), with \(v(y)\geq u(y)\), then
These lengths are determined by the planar base region. The triangle’s altitude is a separate piece of information, supplied by the cross-section’s shape condition. Keep the two roles distinct: the chord provides the triangle’s base (or, in some right-triangle specifications, its hypotenuse), while the shape condition determines how the other dimensions relate to it.
Area Formulas for Triangular Slices
The area of any triangle is one-half its base times its perpendicular altitude. Therefore, if the chord is the chosen base and the altitude is \(h(x)\), then \(A(x)=\frac12 b(x)h(x)\). A ratio condition often makes this especially simple.
A common right-triangle condition says that the chord is one leg and the perpendicular leg has a specified length or ratio. In that case the perpendicular leg is the altitude to the chord, so the triangle area formula applies directly. For example, if both legs have length \(b(x)\), then \(h(x)=b(x)\) and \(A(x)=\frac12(b(x))^2\).
A different condition may make the chord the hypotenuse. Knowing only that a triangle is right-angled and that its hypotenuse has length \(b(x)\) does not determine its area: the other two sides can vary. An additional condition is needed, such as a specified altitude to the hypotenuse or a specified relationship between the legs. For an isosceles right triangle with chord as hypotenuse, the two legs are equal, and the area can be determined from the chord; this case is worked below.
Why Integrating the Areas Gives Volume
Suppose vertical slices run across \(a\leq x\leq b\), and \(A(x)\) is the area of the triangular cross-section at position \(x\). Divide the interval into short widths \(\Delta x\). Over a sufficiently short interval, a thin part of the solid is approximately a prism with volume \(A(x_i^*)\Delta x\), where \(x_i^*\) is a sample point in that interval. Adding the thin volumes gives an approximation to the whole volume:
As the widths become smaller, these sums approach the definite integral. Thus, when the cross-sectional area is integrable, the volume is
For horizontal slices over \(c\leq y\leq d\), the corresponding formula is \(V=\int_c^d A(y)\,dy\). The integration variable describes the direction in which the slices are stacked. Since area has units of length squared and \(dx\) or \(dy\) has units of length, the integral has units of volume.
Worked Examples
Worked Example: Right Isosceles Triangles With the Chord as a Leg
The base region is bounded by \(y=0\) and \(y=2-x\) for \(0\leq x\leq 2\). Perpendicular to the \(x\)-axis, the cross-sections are right isosceles triangles. The chord is one leg, and the other leg has the same length. Find the volume.
At a fixed \(x\), the vertical chord has length
Because the two perpendicular legs are equal, the altitude to the chord is \(h(x)=2-x\). The area is therefore
Integrate over the stated interval:
Check the area formula: a right triangle with legs \(2-x\) and \(2-x\) has area \(\frac12(2-x)(2-x)\), as used. Check the integration by differentiating the antiderivative: \(\frac{d}{dx}\left(-\frac{(2-x)^3}{6}\right)=\frac12(2-x)^2\). The chord shrinks to zero at \(x=2\), so the cross-sectional area also shrinks to zero there. The volume is \(\frac43\) cubic units.
Worked Example: A Prescribed Altitude-to-Base Ratio
The base region lies between \(y=0\) and \(y=x\) for \(0\leq x\leq 2\). Each cross-section perpendicular to the \(x\)-axis is a triangle whose altitude is \(\frac32\) times its chord length. Find the volume.
The vertical chord length is
The altitude-to-base condition gives \(h(x)=\frac32x\). Substituting both dimensions into the triangle area formula gives
Now integrate from \(0\) to \(2\):
Check the cross-sectional area: at \(x=2\), the chord is \(2\) units and the altitude is \(\frac32(2)=3\) units, so the area is \(\frac12(2)(3)=3\) square units. The formula gives \(A(2)=\frac34(4)=3\), agreeing. Differentiating the antiderivative \(\frac{x^3}{4}\) gives \(\frac34x^2\), the integrand. The volume is \(2\) cubic units.
Worked Example: An Isosceles Right Triangle With the Chord as Hypotenuse
The base region is bounded by \(x=0\) and \(x=y\) for \(0\leq y\leq 2\). The cross-sections perpendicular to the \(y\)-axis are isosceles right triangles, and the horizontal chord is their hypotenuse. Find the volume.
At height \(y\), the horizontal chord runs from \(x=0\) to \(x=y\), so
Let each equal leg have length \(s\). Since the triangle is right isosceles, its hypotenuse is \(s\sqrt{2}\). Setting that equal to the chord length gives \(s\sqrt{2}=y\), hence \(s=\frac{y}{\sqrt{2}}\). Its area is
Integrate with respect to \(y\) across the base region’s vertical extent:
Check the triangle geometry: when \(y=2\), the hypotenuse is \(2\), each leg is \(2/\sqrt{2}=\sqrt{2}\), and the area is \(\frac12(\sqrt{2})(\sqrt{2})=1\) square unit. The area formula gives \(A(2)=4/4=1\). Also, differentiating \(y^3/12\) gives \(y^2/4\), the integrand. The volume is \(\frac23\) cubic units.
A Reliable Setup
Before integrating, make each geometric choice explicit. A short setup sequence helps prevent a correct integration from answering the wrong geometric question.
Vertical slices are perpendicular to the \(x\)-axis and use \(dx\); horizontal slices are perpendicular to the \(y\)-axis and use \(dy\).
Subtract the lower boundary from the upper boundary for vertical slices, or the left boundary from the right boundary for horizontal slices.
Determine the altitude from the stated ratio or right-triangle geometry. Do not assume that the chord is a leg or hypotenuse unless the condition says so.
Use \(A=\frac12 bh\), then integrate the resulting area over the interval covered by the slices.
Confirm that the dimensions give area, that the integrand is nonnegative, and that the final units are cubic units.
Common Pitfall: Omitting the One-Half
A frequent error is to multiply the chord by the altitude but forget the factor \(\frac12\). For example, if the chord is \(b=2\) and the altitude is \(h=\frac32b=3\), the incorrect calculation \(A=bh=2(3)=6\) is not the area of the triangle. The correct calculation is
The same error changes the whole volume integral: it doubles every cross-sectional area and therefore doubles the volume. A quick check is to compare the proposed triangular area with the area \(bh\) of a rectangle with the same base and altitude. The triangle’s area must be half of that rectangle’s area.
Another important check concerns right triangles. If the chord is the hypotenuse, a right angle alone does not fix the altitude or the area. A definite area formula requires additional information, such as the isosceles condition in the third example or a specified altitude-to-hypotenuse ratio.
Why the Geometry Matters
The chord is determined by the base region, but the cross-sectional shape determines how much area is built above that chord. The same base region can therefore produce different volumes under different triangle conditions. For instance, changing the altitude-to-base ratio changes the coefficient in \(A(x)=\frac{r}{2}(b(x))^2\); changing which segment the chord represents can require a different geometric calculation altogether.
Keeping the base region, slice direction, chord, and triangle condition separate makes the method dependable. It also provides useful checks: a zero-length chord should give a zero-area cross-section, a positive chord and positive altitude should give positive area, and integrating area over length should produce volume.
Check Your Understanding
For each question, identify the chord and the geometric information needed before writing a volume integral.
- A vertical chord has length \(b(x)\), and the triangle’s altitude is twice its base. What is the cross-sectional area in terms of \(b(x)\)?
- A right triangle has a base-region chord as one leg, and the other leg is three times as long. What is its area in terms of the chord length?
- Why is knowing only that a right triangle has a given hypotenuse not enough to determine its area?
- A horizontal chord runs from \(x=u(y)\) to \(x=v(y)\). Write its length and the volume integral if the cross-sections have area \(A(y)\) for \(c\leq y\leq d\).
- If a proposed triangle area is \(bh\), what is the correction, and why?