Identify the outer and inner radii
A slice with a hole
The teal region lies between the line \(y = 5\) and the curve \(y = x^2 + 1\), for \(-2 \le x \le 2\). It does not reach the x-axis, so when it turns about that axis the solid has a hollow center.
A slice is a thin vertical strip of the region. As the region turns, the slice sweeps the teal washer, a disk with a round hole. Drag the orange dot to move the slice.
Two orange segments start on the x-axis. The long one, to the outer edge of the washer, is the outer radius \(R\), the distance to the farther end of the slice. The short one, to the edge of the hole, is the inner radius \(r\), the distance to the nearer end.
At \(x = 1\), the line is at height 5 and the curve is at height \(1^2 + 1 = 2\), so \(R = 5\) and \(r = 2\).
Farther and nearer
A slice crosses the region perpendicular to the axis, and its ends lie on the two wavy boundaries.
The end farther from the axis gives the outer radius \(R\), and the nearer end gives the inner radius \(r\).
Fold the region below the axis. The lower curve is now the farther one, so \(R\) ends on it. Which curve gives \(R\) depends on distance from the axis, not on which curve is on top.
Notation
Draw a slice across the region, perpendicular to the axis.
The outer radius is the distance from the axis to the farther end of the slice. The inner radius is the distance to the nearer end.
\[\begin{aligned} \cy{R(x)} &= \text{farther distance} \\ \cy{r(x)} &= \text{nearer distance} \end{aligned}\]Call the upper boundary \(f\) and the lower boundary \(g\). Above the x-axis, the upper curve \(f\) is farther, and the distance to each curve is its height.
\[\cy{R(x)} = f(x), \qquad \cy{r(x)} = g(x)\]Below the x-axis, the lower curve \(g\) is farther. Both heights are negative, so each radius is minus a height.
\[\cy{R(x)} = -g(x), \qquad \cy{r(x)} = -f(x)\]About the y-axis, the slice is horizontal, so each distance is an \(x\)-value and the radii are \(\cy{R(y)}\) and \(\cy{r(y)}\).
When it fails
These formulas need the same boundary to be farther on the whole interval. Raise the right end of \(g\) until it crosses \(f\).
Past the crossing, \(g\) is farther, so \(\cy{R(x)} = g(x)\) there. Split the interval where the boundaries cross.
Example 1: turning about the y-axis
The region between \(x = 5y\) and \(x = y^2 + 4\), for \(1 \le y \le 4\), turns about the y-axis. Find the outer and inner radii at height \(y\).
The axis is the y-axis, so the slice is horizontal and each distance from the axis is an \(x\)-value.
At \(y = 2\), the line is at \(x = 10\) and the parabola is at \(x = 8\).
\[5(2) = 10, \qquad 2^2 + 4 = 8\]The line is farther from the axis, so it gives the outer radius, and the nearer parabola gives the inner radius.
\[\cy{R(y) = 5y}, \qquad \cy{r(y) = y^2 + 4}\]The line stays farther on the whole interval, because \(5y - (y^2 + 4) = -(y - 1)(y - 4)\) is positive for \(1 < y < 4\).
Example 2: a region below the axis
The region between \(y = -1\) and \(y = x^2 - 5\), for \(-2 \le x \le 2\), turns about the x-axis. Find the outer and inner radii.
At \(x = 0\), the line is at height \(-1\) and the parabola is at height \(-5\). The whole region lies below the axis.
A tempting wrong move: The line \(y = -1\) is the higher curve, so it gives the outer radius: \(R = 1\), its distance from the axis.
The red disk of radius 1 is what the line sweeps, and it is only the hole of the washer. The parabola is farther from the axis, so it gives the outer radius.
The region is below the axis, so each radius is minus a height. The parabola is \(-(x^2 - 5) = 5 - x^2\) from the axis, and the line is \(-(-1) = 1\).
\[\cy{R(x) = 5 - x^2}, \qquad \cy{r(x) = 1}\]The parabola stays farther, because \((5 - x^2) - 1 = 4 - x^2\) is not negative for \(-2 \le x \le 2\).
Practice
The region between \(y = 12 - 2x\) and \(y = x^2 + 4\), for \(0 \le x \le 2\), turns about the x-axis. Find \(R(x)\) and \(r(x)\).
Drag the slice to \(x = 1\). Which end of the slice is farther from the x-axis?
Solution
The axis is the x-axis, so each distance is a height. At \(x = 1\), the line has height 10 and the parabola has height 5.
\[12 - 2(1) = 10, \qquad 1^2 + 4 = 5\]The line is farther from the axis, so it gives the outer radius.
\[\begin{aligned} \cy{R(x)} &= 12 - 2x \\ \cy{r(x)} &= x^2 + 4 \end{aligned}\]The line stays farther on the whole interval, because both factors of the difference are positive for \(0 \le x < 2\).
\[\begin{aligned} &(12 - 2x) - (x^2 + 4) \\ &\quad = (2 - x)(x + 4) \end{aligned}\]
Questions
The region between \(y = 8 - x^2\) and \(y = 4\), for \(-2 \le x \le 2\), turns about the x-axis.
What is the outer radius at \(x = 1\)?
- 7
- 4
- 3
Answer
7
The slice moves from \(x = 0\) to \(x = 2\). How do the two radii change?
- \(R\) falls from 8 to 4, and \(r\) stays at 4.
- \(R\) and \(r\) both fall to 4.
- \(R\) stays at 8, and \(r\) falls to 4.
Answer
\(R\) falls from 8 to 4, and \(r\) stays at 4.
The region between \(x = y^2\) and \(x = 2y + 3\) turns about the y-axis. What are \(R\) and \(r\) at \(y = 1\)?
- \(R = 5\) and \(r = 1\)
- \(R = 1\) and \(r = 5\)
- \(R = 1\) and \(r = 1\)
Answer
\(R = 5\) and \(r = 1\)
Summary
You draw a slice across the region, perpendicular to the axis, and measure its two ends from the axis.
The farther end gives the outer radius and the nearer end gives the inner radius, because the washer is a disk with a hole.
You use this when a region that does not reach the axis turns about it, because the gap becomes the hole.
This is one of four skills in topic 8.11 of Which & Why: AP Calculus.