Build a semicircular cross-sectional area
Semicircles on a sloping floor
A solid stands on the region between the x-axis and the line \(y = 12 - 2x\). Its cross sections are semicircles perpendicular to the x-axis, and the diameter of each semicircle is the orange segment on the floor.
At \(x = 2\) the diameter is 8, so the radius, the upright orange segment, is 4.
A semicircle is half of a circle with the same radius. That circle has area \(\pi r^2 = 16\pi\), so the semicircle has area \(8\pi\).
Drag the slice to \(x = 3\). The diameter is 6, the radius is 3, and the area is \(\tfrac{9\pi}{2}\). The area changes with \(x\), because the diameter changes.
Half a circle on the diameter
The orange segment \(w\) is the diameter, and the radius is \(\tfrac{w}{2}\).
A circle on this diameter has area \(\pi\left(\tfrac{w}{2}\right)^2\), and the semicircle is half of it. Drag the slice. The diameter changes, but the area divided by \(w^2\) stays at \(\tfrac{\pi}{8}\), about 0.39.
The area is a constant times \(w^2\), so a diameter twice as long gives an area four times as large.
Notation
The orange segment on the floor is the diameter, the top boundary \(f(x)\) minus the bottom boundary \(g(x)\).
\[\cy{d} = w(x) = f(x) - g(x)\]The radius is half of the diameter.
\[\cy{r} = \frac{w}{2}\]The area of a circle is \(\pi r^2\), and a semicircle is half of a circle.
\[\cb{A} = \frac{1}{2}\pi r^2\]Replace \(r\) by \(\tfrac{w}{2}\), so the area needs only the diameter.
\[\cb{A(x)} = \frac{1}{2}\pi\left(\frac{w}{2}\right)^2 = \frac{\pi}{8}\,w(x)^2\]If the problem gives the radius, skip the second step and use \(\tfrac12\pi r^2\) with that radius.
Example 1: semicircles between a parabola and a line
The base is the region between \(y = x^2\) and \(y = 2x + 3\). The cross sections are semicircles perpendicular to the x-axis, with the diameter on the base. Find \(A(x)\).
At \(x = 0\), the line is at height 3 and the parabola is at height 0, so the line is the upper curve. The diameter is the difference.
\[\cy{w(x)} = 2x + 3 - x^2\]The radius is half of the diameter.
\[\cy{r} = \frac{2x + 3 - x^2}{2}\]Use half of the area of a circle.
\[\begin{aligned} \cb{A(x)} &= \frac{1}{2}\pi r^2 \\ &= \frac{\pi}{8}\left(2x + 3 - x^2\right)^2 \end{aligned}\]As a check, take \(x = 1\). The diameter is 4 and the radius is 2, so the area is \(\tfrac12\pi\cdot 4 = 2\pi\). The formula gives \(\tfrac{\pi}{8}\cdot 16 = 2\pi\).
Example 2: semicircles under a sine curve
The base is the region between the x-axis and \(y = \sin x\), for \(0 \le x \le \pi\). The cross sections are semicircles perpendicular to the x-axis, with the diameter on the base. Find \(A(x)\).
The lower boundary is the x-axis, so the diameter is the height of the curve.
\[\cy{w(x)} = \sin x\]A tempting wrong move: Use the height of the curve as the radius.
\[\cr{A(x) = \frac{1}{2}\pi\sin^2 x}\]At \(x = \tfrac{\pi}{2}\) the diameter is 1. A radius of 1 gives the red semicircle, 2 wide, which reaches past both ends of the orange segment. This radius is twice the correct one, so the area is four times too large.
The orange segment is the diameter, so the radius is half of it.
\[\cy{r} = \frac{\sin x}{2}\]Use half of the area of a circle.
\[\cb{A(x)} = \frac{1}{2}\pi\left(\frac{\sin x}{2}\right)^2 = \frac{\pi}{8}\sin^2 x\]
Practice
The base is the region between \(y = x^3\) and \(y = 8\), for \(0 \le x \le 2\). The cross sections are semicircles perpendicular to the x-axis, with the diameter on the base. Find \(A(x)\).
The orange segment on the floor is the diameter, not the radius. Halve it before you square.
Solution
The line is above the curve, because \(x^3 \le 8\) when \(x \le 2\). The diameter is the difference, and the radius is half of it.
\[w(x) = 8 - x^3, \quad r = \frac{8 - x^3}{2}\]Use half of the area of a circle.
\[\begin{aligned} \cb{A(x)} &= \frac{1}{2}\pi\left(\frac{8 - x^3}{2}\right)^2 \\ &= \frac{\pi}{8}\left(8 - x^3\right)^2 \end{aligned}\]As a check, take \(x = 0\). The diameter is 8 and the radius is 4, so the area is \(\tfrac12\pi\cdot 16 = 8\pi\). The formula gives \(\tfrac{\pi}{8}\cdot 64 = 8\pi\).
Questions
The base lies between the x-axis and \(y = x^2 + 2\). The cross sections are semicircles perpendicular to the x-axis, with the diameter on the base.
What is the area of the cross section at \(x = 2\)?
- \(\tfrac{9\pi}{2}\)
- \(9\pi\)
- \(18\pi\)
- \(3\pi\)
Answer
\(\tfrac{9\pi}{2}\)
The slice moves from \(x = 1\) to \(x = 2\), and the diameter doubles from 3 to 6. By what factor does the area change?
- 4
- 2
- \(\tfrac12\)
Answer
4
The diameter of a semicircular cross section is \(4x + 2\). What is its area?
- \(\tfrac{\pi}{2}(2x + 1)^2\)
- \(\tfrac{\pi}{2}(4x + 2)^2\)
- \(\tfrac{\pi}{8}(16x^2 + 4)\)
- \(\tfrac{\pi}{2}(2x + 1)\)
Answer
\(\tfrac{\pi}{2}(2x + 1)^2\)
Summary
You read the orange segment as the diameter, halve it to find the radius, and use half of the area of a circle.
This works because the cross section is half of a circle, and the area of a circle is \(\pi\) times the radius squared.
Use this for every semicircular cross section before you set up the volume integral.
This is one of four skills in topic 8.8 of Which & Why: AP Calculus.