From a Base Region to a Slice Length
A solid described by cross-sections begins with a planar base region. At each position along the base, a slice cuts through that region. The part of the slice inside the region is a line segment, and its length gives a geometric measurement that can be used to describe the cross-section of the solid at that position.
The central task here is to read that segment from the boundary curves. A vertical slice has a fixed \(x\)-coordinate, so its endpoints are found by comparing \(y\)-values. A horizontal slice has a fixed \(y\)-coordinate, so its endpoints are found by comparing \(x\)-values. In either case, the length is the larger endpoint value minus the smaller one.
Vertical Slices: Upper Boundary Minus Lower Boundary
Suppose a vertical slice at \(x=c\) passes through a region bounded below by \(y=g(x)\) and above by \(y=f(x)\). At this fixed \(x\)-value, the chord runs from \((c,g(c))\) to \((c,f(c))\). Since the endpoints have the same \(x\)-coordinate, their distance is the difference between their \(y\)-coordinates.
This order assumes \(f(c)\geq g(c)\): \(f\) is the upper boundary and \(g\) is the lower boundary. If the curves exchange order, the upper and lower functions must be identified on each relevant interval. A length cannot be negative.
Worked Example: A Vertical Chord Between a Line and a Parabola
Consider the region between \(y=4\) and \(y=x^2\), for \(-2\leq x\leq 2\). Find the length of a vertical slice at \(x=c\), and identify the endpoints and length at \(c=1\).
First, compare the two boundary values on the given interval. Since \(x^2\leq 4\) for \(-2\leq x\leq 2\), the upper boundary is \(y=4\) and the lower boundary is \(y=x^2\). Therefore, at \(x=c\), the endpoint \(y\)-coordinates are \(4\) and \(c^2\).
At \(c=1\), the lower endpoint is \((1,1^2)=(1,1)\), and the upper endpoint is \((1,4)\). Their vertical distance is \(4-1=3\), so the slice length is \(3\).
Check the general expression at the interval ends: \(L(-2)=4-(-2)^2=0\) and \(L(2)=4-2^2=0\), as expected where the boundaries meet. At \(c=1\), the formula gives \(L(1)=4-1=3\), agreeing with the endpoint calculation.
If a solid is built from cross-sections perpendicular to the \(x\)-axis, this vertical chord is the base-region dimension available at position \(x=c\). The cross-section’s other dimensions, if any, come from the specified shape rule.
Horizontal Slices: Right Boundary Minus Left Boundary
For a horizontal slice at \(y=d\), both endpoints have the same \(y\)-coordinate. If the left boundary is \(x=p(y)\) and the right boundary is \(x=q(y)\), the chord length is the difference between their \(x\)-coordinates.
Here the boundary functions are expressed as \(x\)-values in terms of \(y\). This is why a horizontal slice is not generally found by subtracting two \(y=f(x)\) values: the endpoints are separated horizontally, not vertically.
Worked Example: Horizontal Chords in a Region Bounded by a Parabola and a Line
Consider the region bounded by \(x=y^2\) and \(x=4\), for \(-2\leq y\leq 2\). Find the horizontal chord length as a function of \(y\), then calculate the chord at \(y=-1\).
For every \(y\) in this interval, \(y^2\leq 4\). Thus \(x=y^2\) is the left boundary and \(x=4\) is the right boundary. A horizontal slice at \(y=d\) has endpoints \((d^2,d)\) and \((4,d)\), so its length is the difference of their \(x\)-coordinates:
At \(d=-1\), the endpoints are \(((-1)^2,-1)=(1,-1)\) and \((4,-1)\). The chord length is \(4-1=3\). The formula gives \(L(-1)=4-(-1)^2=3\), so it agrees with the distance between the endpoints.
At \(d=0\), the endpoints are \((0,0)\) and \((4,0)\), giving length \(4-0=4\), also equal to \(L(0)=4\). At \(d=2\), both boundaries meet at \(x=4\), and \(L(2)=4-4=0\). These checks confirm the endpoint order and the length formula.
For cross-sections perpendicular to the \(y\)-axis, this horizontal chord is the base-region dimension at the slice position \(y=d\).
Why the Subtraction Gives a Length
The rule follows directly from the coordinates of the endpoints. For a vertical slice, the endpoints have coordinates \((c,y_1)\) and \((c,y_2)\). Because their \(x\)-coordinates agree, the distance formula reduces to the absolute difference of their \(y\)-coordinates:
If the upper endpoint has value \(f(c)\) and the lower endpoint has value \(g(c)\), then \(f(c)\geq g(c)\), so the absolute difference is \(f(c)-g(c)\). The horizontal case is the same reasoning with the roles of the coordinates exchanged. Endpoints \((x_1,d)\) and \((x_2,d)\) have distance \(|x_2-x_1|\), which is \(q(d)-p(d)\) when \(q(d)\geq p(d)\).
A Region Can Give Different Slice Formulas
The same region can be sliced in either direction, but the endpoints—and therefore the formulas—can differ. This is especially important when the boundary curves are more naturally written in one direction than the other. The following region gives a direct comparison.
Worked Example: Reading the Same Region in Both Directions
Let the region lie between \(y=x\) and \(y=x^2\) for \(0\leq x\leq 1\). Find its vertical chord length. Then rewrite the boundaries to find the horizontal chord length. Finally, state what these lengths mean for cross-sections perpendicular to each axis.
For \(0\leq x\leq 1\), \(x\geq x^2\), because \(x-x^2=x(1-x)\geq 0\). Thus the line \(y=x\) is above the parabola \(y=x^2\). The vertical chord length is
For the horizontal description, the region’s \(y\)-values range from \(0\) to \(1\). The line \(y=x\) gives \(x=y\). On this region the parabola gives \(x=\sqrt{y}\), using the nonnegative square root because \(x\geq 0\). For \(0\leq y\leq 1\), \(\sqrt{y}\geq y\), so \(x=y\) is the left boundary and \(x=\sqrt{y}\) is the right boundary. Therefore,
Check at \(x=\tfrac12\): the upper value is \(\tfrac12\), the lower value is \(\tfrac14\), and the vertical chord length is \(\tfrac12-\tfrac14=\tfrac14\). The formula gives \(L_{\mathrm{vertical}}(\tfrac12)=\tfrac12-(\tfrac12)^2=\tfrac14\).
Check the horizontal formula at \(y=\tfrac14\): the left endpoint has \(x=\tfrac14\), while the right endpoint has \(x=\sqrt{\tfrac14}=\tfrac12\). The length is \(\tfrac12-\tfrac14=\tfrac14\), and the formula gives \(L_{\mathrm{horizontal}}(\tfrac14)=\tfrac12-\tfrac14=\tfrac14\). Both formulas also give zero at their respective interval endpoints \(0\) and \(1\), where the boundary curves meet.
For slices perpendicular to the \(x\)-axis, \(L_{\mathrm{vertical}}(x)\) supplies the base-region dimension at \(x\). For slices perpendicular to the \(y\)-axis, \(L_{\mathrm{horizontal}}(y)\) supplies it at \(y\). These are two descriptions of chords in the same region, not interchangeable formulas for a single chosen slice direction.
Using the Chord as a Cross-Section Dimension
In a solid formed by cross-sections, the base-region chord at a given position becomes a specified dimension of the cross-section at that position. The slice direction tells you which chord to calculate: vertical base-region slices correspond to positions measured along the \(x\)-axis, while horizontal base-region slices correspond to positions measured along the \(y\)-axis.
The chord length alone does not determine every cross-section. The problem must also state what shape is used and how its other dimensions relate to the chord. Once that shape rule gives the cross-sectional area \(A\) as a function of position, the volume-by-cross-sections setup uses that area across the interval of positions. The key work in this tutorial is supplying the correct chord length as the relevant dimension, not guessing the rest of the cross-section.
A vertical slice fixes \(x\); a horizontal slice fixes \(y\).
For a vertical slice, find the two \(y\)-coordinates. For a horizontal slice, find the two \(x\)-coordinates.
Use upper minus lower or right minus left, so the chord length is nonnegative.
Apply the problem’s shape rule to determine any additional dimensions and the cross-sectional area.
Common Mistake: Subtracting in the Wrong Order
A frequent error is to subtract the lower boundary from the upper one in the wrong order, or to use a vertical-slice subtraction for a horizontal slice. For example, in the region between \(y=4\) and \(y=x^2\) at \(x=1\), the incorrect calculation \(1-4=-3\) is not a valid slice length. It results from subtracting the upper value from the lower value.
The correction is to identify the endpoint coordinates first. At \(x=1\), the lower value is \(1\) and the upper value is \(4\), so the length is \(4-1=3\). Likewise, for a horizontal slice, compare the left and right \(x\)-coordinates rather than the upper and lower \(y\)-values. Writing the endpoints before subtracting makes both the direction and the sign clear.
Check Your Understanding
For each question, identify the slice endpoints before calculating a length.
- A region has lower boundary \(y=x\) and upper boundary \(y=6\). What is the vertical chord length at \(x=2\)?
- A horizontal slice has left endpoint \(x=y^2\) and right endpoint \(x=5\). What is its length at \(y=1\)?
- In the region between \(y=x^2\) and \(y=2x\) for \(0\leq x\leq 2\), which curve is upper, and what is the vertical chord length?
- Why must a horizontal chord length be computed from a difference of \(x\)-coordinates?
- What additional information is needed to turn a base-region chord length into a cross-sectional area?
