Measure both radii from a shifted axis
An axis below the region
The teal region lies between the line \(y = 2x + 1\) and the parabola \(y = x^2 + 1\). It turns about the dashed line \(y = -1\), below the region.
When the region turns, a thin vertical slice of it sweeps a washer, a flat disk with a round hole. Drag the orange dot to move the slice.
A radius is a distance from the axis, here \(y = -1\) and not the x-axis, so the orange segments start on that line.
At \(x = 1\), the line is at height 3 and the parabola at height 2, so their distances from \(y = -1\) are 4 and 3. The larger distance is the outer radius \(R\), and the smaller is the inner radius \(r\).
Any horizontal axis
The axis is the dashed line \(y = k\), and the wavy region lies below it.
Drag the slice. Each radius is the distance from the axis to one boundary, so both radii start on the axis.
Press the first button to move the region above the axis. The radii now reach up, but each is still the distance from the axis to a boundary.
Notation
A shifted axis is a horizontal line \(y = k\) or a vertical line \(x = h\), not a coordinate axis.
A radius is the distance from the axis to a boundary, measured perpendicular to the axis.
Suppose the line \(y = k\) does not cut through the region. The distance to a boundary \(y = f(x)\) is \(f(x) - k\) if the region is above the line, and \(k - f(x)\) if it is below.
\[\begin{array}{ll} \text{region above:} & \cy{f(x) - k} \\ \text{region below:} & \cy{k - f(x)} \end{array}\]The same holds for a vertical line \(x = h\). The distance to a boundary \(x = f(y)\) is \(f(y) - h\) if the region is to the right, and \(h - f(y)\) if it is to the left.
\[\begin{array}{ll} \text{region on the right:} & \cy{f(y) - h} \\ \text{region on the left:} & \cy{h - f(y)} \end{array}\]
When it fails
Press the button to measure from the x-axis instead. The red ring uses the heights of the boundaries as radii, which at \(x = 1\) are 3 and 2.
The solid turns about \(y = -1\), which is 1 lower than the x-axis, so each radius is 1 longer than the height. The correct radii are 4 and 3.
Example 1: an axis above the region
The region lies between \(y = \sqrt{x}\) and \(y = \tfrac{x}{2}\), for \(0 \le x \le 4\). It turns about the line \(y = 3\). Find the outer and inner radii.
The line \(y = 3\) is above the region, so each distance is 3 minus a height.
\[\text{distance} = 3 - y\]At \(x = 1\), the curve \(y = \sqrt{x}\) is at height 1 and the line \(y = \tfrac{x}{2}\) at height \(\tfrac12\), so the distances are 2 and \(\tfrac52\).
\[3 - 1 = 2, \qquad 3 - \tfrac12 = \tfrac52\]The axis is above the region, so the lower boundary is farther and gives the outer radius.
\[\begin{aligned} \cy{R(x)} &= 3 - \tfrac{x}{2} \\ \cy{r(x)} &= 3 - \sqrt{x} \end{aligned}\]The line \(y = \tfrac{x}{2}\) stays the lower boundary, because \(\sqrt{x} \ge \tfrac{x}{2}\) for \(0 \le x \le 4\).
Example 2: a vertical axis on the left
The region lies between \(x = y\) and \(x = 3 - y\), for \(0 \le y \le \tfrac32\). It turns about the line \(x = -2\). Find the outer and inner radii.
The axis is vertical, so each slice is horizontal. At \(y = 1\), the region runs from \(x = 1\) to \(x = 2\), to the right of the line \(x = -2\).
A tempting wrong move: Take the \(x\)-value of each boundary as its radius, so \(R = 3 - y\) and \(r = y\).
\[R = 3 - y, \qquad r = y\]These are distances from the y-axis, not from \(x = -2\). At \(y = 1\) they are 2 and 1, so the red ring lies inside the hole of the washer, whose radii are 4 and 3.
The distance from the line \(x = -2\) to a boundary is its \(x\)-value minus \(-2\), which is the \(x\)-value plus 2. The boundary \(x = 3 - y\) is farther.
\[\cy{R(y) = 5 - y}, \qquad \cy{r(y) = y + 2}\]The difference \(R - r = 3 - 2y\) is not negative for \(y \le \tfrac32\), so \(x = 3 - y\) stays the farther boundary. At \(y = 1\), \(R = 4\) and \(r = 3\).
Practice
The region lies between \(x = y^2\) and \(x = y + 2\), for \(-1 \le y \le 2\). It turns about the line \(x = 5\). Find \(R(y)\) and \(r(y)\).
The region is to the left of the line \(x = 5\). Which subtraction gives a positive distance?
Solution
The region is to the left of the line \(x = 5\), so each distance is 5 minus an \(x\)-value. At \(y = 1\), the parabola is at \(x = 1\) and the line at \(x = 3\).
\[5 - 1 = 4, \qquad 5 - 3 = 2\]The parabola is farther from the line \(x = 5\), so it gives the outer radius.
\[\cy{R(y) = 5 - y^2}, \qquad \cy{r(y) = 3 - y}\]The difference \((5 - y^2) - (3 - y) = (2 - y)(1 + y)\) is not negative for \(-1 \le y \le 2\).
Questions
The graph shows the region between \(y = x + 1\) and \(y = 7 - x\), for \(0 \le x \le 3\). It turns about the line \(y = -2\).
What is the inner radius at \(x = 1\)?
- 4
- 2
- 8
- \(-4\)
Answer
4
Which boundary gives the outer radius?
- The upper boundary, \(y = 7 - x\)
- The lower boundary, \(y = x + 1\)
- Neither, because the two radii are equal
Answer
The upper boundary, \(y = 7 - x\)
The region between \(x = y^2\) and \(x = 4\) turns about the line \(x = 6\). What are \(R\) and \(r\) at \(y = 1\)?
- \(R = 5\) and \(r = 2\)
- \(R = 2\) and \(r = 5\)
- \(R = 4\) and \(r = 1\)
- \(R = 10\) and \(r = 7\)
Answer
\(R = 5\) and \(r = 2\)
Summary
You measure each radius from the stated line to a boundary, so here \(R = (2x + 1) - (-1) = 2x + 2\) and \(r = x^2 + 2\).
This works because a radius is a distance from the axis of the turn, the line \(y = -1\), not the x-axis.
You use it when a region turns about any line other than the x-axis or the y-axis.
This is one of four skills in topic 8.12 of Which & Why: AP Calculus.