Determine radius order across the region
Which boundary is farther?
The teal region lies between the curves \(y = \sin x\) and \(y = \cos x\), for \(0 \le x \le \tfrac{\pi}{2}\). It turns about the dashed line \(y = -1\), below the region.
A thin vertical slice of the region sweeps a washer. The end of the slice farther from the axis gives the outer radius \(R\), and the nearer end gives the inner radius \(r\).
Drag the orange dot. Left of \(x = \tfrac{\pi}{4}\), the cosine curve is higher, so it is farther from the axis and \(R\) ends on it. Right of \(x = \tfrac{\pi}{4}\), the sine curve is higher, so \(R\) ends on it.
The curves cross at \(x = \tfrac{\pi}{4}\), where the two distances are equal, so the curve that gives \(R\) on one side gives \(r\) on the other.
Which side is the axis on?
Two boundaries cross once, and the region lies above the dashed axis. Drag the slice. The outer radius \(R\) ends on the boundary that is farther from the axis.
At the crossing, both boundaries are the same distance from the axis, so past it the other boundary is farther.
Move the region below the axis with the button. The axis is above the region now, so the lower boundary is farther from it.
While the axis stays on one side of the region, the order of the radii changes only where the boundaries cross.
Notation
At one slice, the larger distance from the axis to a boundary is the outer radius \(\cy{R}\), and the smaller is the inner radius \(\cy{r}\).
\[\cy{R} \ge \cy{r}\]If the axis is below the region, the higher boundary is outer. If the axis is above the region, the lower boundary is outer.
The two distances are equal where the boundaries cross, because the same number is subtracted from both heights. To find the crossing \(c\), solve \(f(c) = g(c)\).
Test one slice on each side of \(c\). The region runs from \(x = a\) to \(x = b\), and the axis \(y = k\) lies below it, so each radius is a height minus \(k\).
\[\begin{array}{l} a \le x \le c: \\ \quad \cy{R} = g(x) - k, \ \ \cy{r} = f(x) - k \\ c \le x \le b: \\ \quad \cy{R} = f(x) - k, \ \ \cy{r} = g(x) - k \end{array}\]
When it fails
These rules assume that the axis does not cut through the region. Tip the left end of the region down across the axis.
Near the left end, each slice crosses the axis, so it sweeps a full disk with no hole. Past the point where \(f\) meets the axis, the slice sweeps a washer again, so the interval splits there as well as at \(c\).
Example 1: an axis above the region
The region lies between \(y = \tfrac{4}{x}\) and \(y = 5 - x\), for \(1 \le x \le 4\). It turns about the line \(y = 6\). Which boundary is outer?
The line \(y = 6\) is above the region, so each distance is 6 minus a height.
\[\text{distance} = 6 - y\]At \(x = 2\), the curve \(y = \tfrac4x\) is at height 2, and the line \(y = 5 - x\) is at height 3. The distances are 4 and 3.
\[6 - 2 = 4, \qquad 6 - 3 = 3\]The axis is above the region, so the lower boundary is farther. The boundaries meet only at the ends, \(x = 1\) and \(x = 4\), so \(y = \tfrac4x\) is lower between them and gives the outer radius.
\[\cy{R(x) = 6 - \tfrac{4}{x}}, \qquad \cy{r(x) = 1 + x}\]
Example 2: the outer boundary changes
The region lies between \(y = x^2\) and \(y = 8 - x^2\), for \(0 \le x \le 3\). It turns about the line \(y = -2\). Find the outer radius.
The line \(y = -2\) is below the region, so each distance is a height plus 2. At \(x = 1\), the distances are 3 to \(y = x^2\) and 9 to \(y = 8 - x^2\).
\[1^2 + 2 = 3, \qquad (8 - 1^2) + 2 = 9\]A tempting wrong move: The curve \(y = 8 - x^2\) is outer at \(x = 1\), so it is outer on the whole interval.
\[R = 10 - x^2, \qquad r = x^2 + 2\]At \(x = 3\), the distances are 11 to \(y = x^2\) and 1 to \(y = 8 - x^2\). The claimed radius 1 draws the red disk, which is only the hole of the washer.
The order changes where the two distances are equal. Set them equal and solve. The root \(x = -2\) lies outside the interval.
\[\begin{aligned} x^2 + 2 &= 10 - x^2 \\ x^2 &= 4 \\ x &= 2 \end{aligned}\]Before \(x = 2\), the curve \(y = 8 - x^2\) is outer. After \(x = 2\), the curve \(y = x^2\) is outer.
\[\cy{R(x)} = \begin{cases} 10 - x^2, & 0 \le x \le 2 \\ x^2 + 2, & 2 \le x \le 3 \end{cases}\]
Practice
The region lies between \(y = x^2 - 4\) and \(y = 4\), for \(-2 \le x \le 2\). It turns about the x-axis. Find the two distances at \(x\), and decide which is larger.
Slide the slice to \(x = 0\). On which sides of the x-axis are the two ends of the slice?
Solution
At \(x = 0\), the ends of the slice are at \(y = 4\) and \(y = -4\), so the slice crosses the x-axis.
Above the axis, the distance to the line \(y = 4\) is 4. Below it, the distance down to the parabola is \(0 - (x^2 - 4)\).
\[\begin{aligned} \text{above:} \quad & 4 \\ \text{below:} \quad & 0 - (x^2 - 4) = 4 - x^2 \end{aligned}\]The distance above is larger, because \(4 - (4 - x^2) = x^2\) is not negative. The shorter side, the faint orange segment, turns inside the circle of the longer side, so the slice is a disk with no hole.
\[\cy{R = 4}\]
Questions
The graph shows the region between \(y = x + 3\) and \(y = 9 - x\), for \(0 \le x \le 5\). It turns about the x-axis.
At \(x = 5\), which boundary gives the outer radius?
- The line \(y = x + 3\)
- The line \(y = 9 - x\)
- Neither, because the lines meet
Answer
The line \(y = x + 3\)
Where does the outer radius change from one line to the other?
- \(x = 3\)
- \(x = 2.5\)
- \(x = 6\)
Answer
\(x = 3\)
The region between \(y = x^2\) and \(y = 2x + 8\), for \(0 \le x \le 5\), turns about the line \(y = -1\). Where does the outer radius change from one curve to the other?
- \(x = 4\)
- \(x = -2\)
- \(x = 5\)
Answer
\(x = 4\)
Summary
You compare the two distances from the axis at each slice, and the larger distance is the outer radius.
This works because the order changes only where the distances are equal, which is where the boundaries cross.
You use it before you write a washer area, to find which boundary is outer on each part of the interval.
This is one of four skills in topic 8.12 of Which & Why: AP Calculus.