Find the base width of a cross section
A solid with square cross sections
A solid stands on a flat floor. The floor is the region between the parabola \(y = x^2\) and the line \(y = 2x\), and this region is the base of the solid.
The y-axis runs back into the page, so the line is the back edge of the region.
Each slice is a plane perpendicular to the x-axis. The teal square is the cross section, the flat face where the plane cuts the solid.
Drag the slice. The orange segment is the part of the slice inside the region, so it is the base of the square. Its length is the base width \(w\).
At \(x = 1\), the line is at height 2 and the parabola is at height 1, so \(w = 1\). At \(x = 1.5\), the heights are 3 and 2.25, so \(w = 0.75\).
The upper curve minus the lower curve
At each \(x\), the upper curve is at height \(f(x)\) and the lower curve is at height \(g(x)\). The orange base runs between them.
Drag the slice. The base is a segment from the smaller height to the larger height, so its length is the larger height minus the smaller one.
Press Move the region down. Both curves move by the same amount, so the width at each x does not change. Only the gap between the curves matters, not their distance from the x-axis.
Notation
The upper curve, at the back of the region, is \(y = f(x)\), and the lower curve, at the front, is \(y = g(x)\). Here \(f(x) = 2x\) and \(g(x) = x^2\).
The base starts at the lower curve and ends at the upper curve, so its length is the upper height minus the lower height.
\[\cy{w(x)} = f(x) - g(x)\]This needs \(f(x) \ge g(x)\) on the whole interval, because a width cannot be negative.
The square stands on this base, so its upright orange side is also \(\cy{w(x)}\) long.
If the slices are perpendicular to the y-axis, each base runs from left to right. Write both curves as \(x\) in terms of \(y\), and subtract the left value from the right value.
\[\cy{w(y)} = x_{\text{right}}(y) - x_{\text{left}}(y)\]
When it fails
The direction of the slices decides which way the base is measured. Press the button to turn the slice so that it is perpendicular to the y-axis.
A red square stands on a base that runs from left to right, so its length is a difference of x-values. At \(y = 1\), the line is at \(x = 0.5\) and the parabola is at \(x = 1\), so the red base is 0.5 long. The orange base at \(x = 1\) is 1 long, so the two directions give different widths.
Example 1: slices perpendicular to the y-axis
The base of a solid is the region between \(x = y^2\) and \(x = 4\). Its cross sections perpendicular to the y-axis are squares. Find \(w(y)\).
The slices are perpendicular to the y-axis, so each base runs from left to right. The parabola is the left boundary, because at \(y = 0\) it is at \(x = 0\) and the line is at \(x = 4\).
Subtract the left x-value from the right x-value.
\[\cy{w(y)} = 4 - y^2\]At \(y = 1\), the base runs from \(x = 1\) to \(x = 4\), so it is 3 long, and the formula gives \(4 - 1 = 3\).
Example 2: a lower curve that is not the axis
The base of a solid is the region between \(y = x^2\) and \(y = x + 2\). Its cross sections perpendicular to the x-axis are squares. Find \(w(x)\).
The curves meet where \(x^2 = x + 2\), which gives \(x = -1\) and \(x = 2\). At \(x = 0\), the line is at height 2 and the parabola is at height 0, so the line is the upper curve.
\[x^2 - x - 2 = (x - 2)(x + 1) = 0\]A tempting wrong move: Use the height of the upper curve as the width.
\[\cr{w(x) = x + 2}\]At \(x = 1\) this gives 3, so the red square has side 3. The orange segment runs from height 1 to height 3, so it is only 2 long.
The base starts at the parabola, not at the x-axis, so subtract the lower curve.
\[\cy{w(x)} = (x + 2) - x^2\]At \(x = 1\), the formula gives \(3 - 1 = 2\), the length of the orange segment.
Practice
The base of a solid is the region between \(y = 3\) and \(y = 1 + x^2\). Its cross sections perpendicular to the x-axis are squares. Find \(w(x)\) and the values of \(x\) where the base begins and ends.
Drag the slice to each end of the region. The base begins and ends where the orange segment has length 0.
Solution
At \(x = 0\), the line is at height 3 and the parabola is at height 1, so the line is the upper curve. Subtract the lower height.
\[\cy{w(x)} = 3 - (1 + x^2) = 2 - x^2\]The width is 0 where the curves meet, so solve \(2 - x^2 = 0\).
\[x = -\sqrt{2} \quad\text{and}\quad x = \sqrt{2}\]
Questions
The base of a solid lies between the line \(y = x\) and the curve \(y = x^3\), for \(0 \le x \le 1\).
A slice stands at \(x = 0.5\), perpendicular to the x-axis. What is the base width of the slice?
- \(\frac{3}{8}\)
- \(\frac{1}{2}\)
- \(\frac{1}{8}\)
- \(\frac{5}{8}\)
Answer
\(\frac{3}{8}\)
A slice stands at height \(y\), perpendicular to the y-axis. What is its base width?
- \(\sqrt[3]{y} - y\)
- \(y - y^3\)
- \(y - \sqrt[3]{y}\)
Answer
\(\sqrt[3]{y} - y\)
A base lies between the lines \(y = x\) and \(y = 2x\). The cross sections are perpendicular to the y-axis. What is the base width at \(y = 4\)?
- 2
- 4
- 6
Answer
2
Summary
You find the base width as the larger boundary value minus the smaller one, measured along the slice.
This works because the base of a cross section is the part of the slice inside the region.
Use it first in every cross-section problem, because the area of each face depends on this width.
This is one of four skills in topic 8.7 of Which & Why: AP Calculus.