Build the correct slice-area formula
Washer or disk?
The teal region lies between the lines \(y = x\) and \(y = 3\), for \(0 \le x \le 3\). It turns about the dashed line \(y = 1\), which passes through the region.
A slice is a thin vertical strip of the region. Drag the orange dot to move it. The readouts show what the slice sweeps and its area.
Left of \(x = 1\), the slice crosses the line \(y = 1\). The part above reaches \(y = 3\), which is 2 from the line, so it sweeps a disk of radius 2 and area \(4\pi\). The shorter part below turns inside that disk.
Right of \(x = 1\), a gap separates the slice from the line, so it sweeps a washer, a flat disk with a round hole. At \(x = 2\), the outer radius is \(R = 2\) and the hole has radius \(r = 1\), so the area is \(\pi(2^2 - 1^2) = 3\pi\).
A hole, or no hole
The lower boundary of the wavy region rises across the dashed axis, so the axis passes through the left part of the region. Slide the slice.
Where the slice crosses the axis, it has a long part on one side and a short part, shown faint, on the other. The short part turns inside the disk of the long part, so there is no hole.
Where a gap separates the slice from the axis, the slice sweeps a washer. A disk is a washer whose hole has area 0.
Notation
A washer has a hole, and its area is the area of the outer disk minus the area of the hole.
\[\cb{A(x)} = \pi\big(\cy{R(x)}^2 - \cy{r(x)}^2\big)\]A disk has no hole, because the slice reaches or crosses the axis. Its radius \(\cy{R(x)}\) is the larger of the distances from the axis to the two ends of the slice.
\[\cb{A(x)} = \pi\,\cy{R(x)}^2\]The shape changes at \(c\), where the lower boundary meets the axis. Split the interval there.
\[\begin{array}{l} a \le x \le c: \\ \quad \cb{A(x)} = \pi\,\cy{R(x)}^2 \\ c \le x \le b: \\ \quad \cb{A(x)} = \pi\big(\cy{R(x)}^2 - \cy{r(x)}^2\big) \end{array}\]The outer boundary can also change where the boundaries cross, at a point \(m\). Write one area for each piece, and add the integrals.
\[\cb{\int_a^m A_1\,dx + \int_m^b A_2\,dx}\]
When it fails
For a slice that crosses the axis, the washer formula describes the red ring, with the short part as the radius of its hole.
The short part turns inside the disk of the long part, so the slice has no hole. The correct area is \(\pi R^2\), which is more.
Example 1: the axis along the top
The region lies between \(y = x^2\) and \(y = 4\), for \(-2 \le x \le 2\). It turns about the line \(y = 4\). Find the area of the slice at \(x\).
The line \(y = 4\) is the top of the region, so every slice has one end on the axis. There is no gap, so the slice sweeps a disk.
Its radius \(R\) is the distance from the axis down to the parabola, and its area is \(\pi R^2\).
\[\begin{gathered} \cy{R = 4 - x^2}, \\ \cb{A(x) = \pi\left(4 - x^2\right)^2} \end{gathered}\]About the line \(y = 5\), a gap of 1 would leave a hole of radius 1.
As a check, at \(x = 1\) the radius is \(4 - 1 = 3\), so the area is \(\pi \cdot 3^2 = 9\pi\).
Example 2: two pieces
The region lies between \(y = x^2\) and \(y = 6 - x\), for \(0 \le x \le 3\). It turns about the line \(y = -1\). Write the area of the slice.
The axis is below the region, so every slice is a washer. The distances from \(y = -1\) are \(7 - x\) to the line and \(x^2 + 1\) to the parabola.
\[(6 - x) + 1 = 7 - x, \qquad x^2 + 1\]A tempting wrong move: The line is outer at \(x = 1\), so one area fits the whole interval.
\[A = \pi\left((7 - x)^2 - (x^2 + 1)^2\right)\]At \(x = 3\), the distances are 4 and 10, so the formula gives \(\pi(16 - 100) = -84\pi\). An area cannot be negative, so the formula fails there.
The distances are equal where the curves cross. There \(x^2 = 6 - x\), so \((x + 3)(x - 2) = 0\) and \(x = 2\) in \([0, 3]\). The line is outer before \(x = 2\) and the parabola after it.
\[\begin{aligned} \cb{A_1} &\cb{{}= \pi\left((7 - x)^2 - (x^2 + 1)^2\right)} \\ &\qquad \text{on } [0, 2] \\ \cb{A_2} &\cb{{}= \pi\left((x^2 + 1)^2 - (7 - x)^2\right)} \\ &\qquad \text{on } [2, 3] \end{aligned}\]Add the integrals of the two pieces.
\[\cb{\int_0^2 A_1\,dx + \int_2^3 A_2\,dx}\]
Practice
The region between the y-axis and \(x = 2 + y\), for \(0 \le y \le 2\), turns about the line \(x = 1\). Build \(A(y)\).
Slide the slice up and down. Which end of the slice is farther from the line \(x = 1\)?
Solution
The axis is vertical, so the slices are horizontal. The line \(x = 1\) cuts every slice, which runs from \(x = 0\) to \(x = 2 + y\), so every slice sweeps a disk.
The right end is \((2 + y) - 1 = 1 + y\) from the line, and the left end is 1 from it. Because \(y \ge 0\), \(1 + y \ge 1\), so the right end gives the radius.
\[\cy{R(y)} = 1 + y\]The slice has no hole, so its area is that of a disk of radius \(1 + y\).
\[\cb{A(y) = \pi(1 + y)^2}\]As a check, at \(y = 1\) the disk has radius 2 and area \(4\pi\).
Questions
The graph shows the region between \(y = x^3\) and \(y = 8\), for \(0 \le x \le 2\). It turns about the line \(y = 8\).
What does the slice at \(x = 1\) sweep?
- A disk of radius 7
- A washer with \(R = 7\) and \(r = 1\)
- A washer with \(R = 8\) and \(r = 1\)
Answer
A disk of radius 7
What is the area of the slice at \(x = 1\)?
- \(49\pi\)
- \(7\pi\)
- \(48\pi\)
- \(63\pi\)
Answer
\(49\pi\)
The region between \(y = 2x\) and \(y = 8 - 2x\) turns about the x-axis. The lines cross at \(x = 2\). What is the area of the slice for \(2 \le x \le 3\)?
- \(\pi(32x - 64)\)
- \(\pi(64 - 32x)\)
- \(\pi(4x - 8)^2\)
- \(4\pi x^2\)
Answer
\(\pi(32x - 64)\)
Summary
You write the disk area \(\pi R^2\) when the slice reaches or crosses the axis, and the washer area \(\pi(R^2 - r^2)\) when a gap separates them.
This works because the radius of the hole is the gap between the slice and the axis, so no gap means no hole.
You split the interval where the shape or the outer boundary changes, and write one area for each piece.
This is one of four skills in topic 8.12 of Which & Why: AP Calculus.