Wemazi

Topic 8.11

Build a washer area

Builds on identifying the outer and inner radii. 6 minutes.

A slice of a pipe

The region between the lines \(y = 3\) and \(y = 5\), for \(0 \le x \le 4\), turns about the x-axis and makes a pipe. The teal slice, cut across the pipe, has outer radius 5 and inner radius 3.

At the start, the hole is filled, so the slice is a whole disk of radius 5 with area \(\pi \cdot 5^2 = 25\pi\).

Move the slider to remove the hole. The hole is a disk of radius 3, so its area is \(\pi \cdot 3^2 = 9\pi\). The ring that remains is the washer, with area \(25\pi - 9\pi = 16\pi\).

Any washer

The same pipe has no numbers. The outer disk has radius \(R\), and the hole has radius \(r\).

Fill the hole, and the washer becomes the whole outer disk. Take the hole out again to remove the disk of radius \(r\).

Change the inner radius. Whatever \(r\) is, the washer is the outer disk without the hole, so its area is the difference of the two disk areas. If \(r\) is 0, there is no hole and the washer is a full disk.

Notation

  1. The outer disk has radius \(\cy{R(x)}\), the distance to the farther boundary, so its area is \(\pi\cy{R(x)}^2\).

    \[\cb{\pi R^2}\]

  2. The hole is a disk with radius \(\cy{r(x)}\), the distance to the nearer boundary, so its area is \(\pi\cy{r(x)}^2\).

    \[\cb{\pi r^2}\]

  3. The washer is the outer disk without the hole, so subtract the area of the hole from the area of the outer disk. Both areas have the factor \(\pi\), so take it out in front. \(R\) is the larger radius, so the area is not negative.

    \[\begin{aligned} \cb{A} &= \pi\cy{R}^2 - \pi\cy{r}^2 \\ &= \pi\big(\cy{R}^2 - \cy{r}^2\big) \end{aligned}\]

    Square each radius first. Then subtract. Then multiply by \(\pi\).

  4. In a problem, \(R\) and \(r\) depend on the place \(x\) of the slice, so the area depends on \(x\) too.

    \[\cb{A(x)} = \pi\big(\cy{R(x)}^2 - \cy{r(x)}^2\big)\]

    This needs the whole region on one side of the axis.

When it fails

The hole comes from the gap between the region and the axis. Move the region down so that it crosses the axis.

The gap is gone, so each slice sweeps a full disk with no hole. The red ring is what \(\pi(R^2 - r^2)\) gives when \(r\) is still measured to the nearer boundary. It leaves out the middle of the disk, so the formula gives too little.

Example 1: two curved radii

A washer has outer radius \(R(x) = x^2 + 3\) and inner radius \(r(x) = x + 1\), for \(0 \le x \le 2\). Find its area \(A(x)\). Then find the area at \(x = 1\).

  1. The outer disk has area \(\pi\) times the square of the outer radius, so expand \((x^2 + 3)^2\).

    \[(x^2 + 3)^2 = x^4 + 6x^2 + 9\]

  2. The hole has area \(\pi\) times the square of the inner radius, so expand \((x + 1)^2\).

    \[(x + 1)^2 = x^2 + 2x + 1\]

  3. Subtract the inner square from the outer square, so every term of \(x^2 + 2x + 1\) changes sign. Then multiply by \(\pi\).

    \[\begin{aligned} \cb{A(x)} &= \pi\big(x^4 + 6x^2 + 9 \\ &\qquad - x^2 - 2x - 1\big) \\ &= \pi\big(x^4 + 5x^2 - 2x + 8\big) \end{aligned}\]

    At \(x = 1\), the radii are 4 and 2, so the area is \(\pi(16 - 4) = 12\pi\). The formula gives \(\pi(1 + 5 - 2 + 8) = 12\pi\).

Example 2: turning about the y-axis

The region between \(x = 2\) and \(x = 2 + \sqrt{y}\), for \(0 \le y \le 9\), turns about the y-axis. Find the area \(A(y)\) of a slice.

  1. The axis is the y-axis, so the radii are \(x\)-values. Because \(\sqrt{y} \ge 0\), the curve \(x = 2 + \sqrt{y}\) is farther from the axis than the line \(x = 2\).

    \[\cy{R(y) = 2 + \sqrt{y}}, \qquad \cy{r(y) = 2}\]

  2. A tempting wrong move: Subtract the radii. Then square the difference.

    \[\begin{aligned} A &= \pi(R - r)^2 \\ &= \pi\big(\sqrt{y}\big)^2 \\ &= \pi y \end{aligned}\]

    At \(y = 4\), the washer is the ring from radius 2 to radius 4. The formula gives \(4\pi\), the area of the red disk of radius 2, which only fills the hole.

  3. Square each radius first. Then subtract.

    \[\begin{aligned} \cb{A(y)} &= \pi\big((2 + \sqrt{y})^2 - 2^2\big) \\ &= \pi\big(4 + 4\sqrt{y} + y - 4\big) \\ &= \pi\big(4\sqrt{y} + y\big) \end{aligned}\]

    At \(y = 4\), this gives \(\pi(8 + 4) = 12\pi\), which is the area of the ring.

Practice

A washer has outer radius \(R(x) = 2x + 3\) and inner radius \(r(x) = x^2\), for \(0 \le x \le 3\). Find \(A(x)\). Then find the area at \(x = 2\).

Drag the slice to \(x = 2\). Read the two radii. Then square each one.

Solution

  1. At \(x = 2\), the radii are 7 and 4.

    \[2(2) + 3 = 7, \qquad 2^2 = 4\]

  2. Square each radius first. Then subtract and multiply by \(\pi\).

    \[\cb{A(2)} = \pi\big(7^2 - 4^2\big) = 33\pi\]

  3. For any \(x\), square the two radii and subtract. The square of \(x^2\) is \(x^4\).

    \[\begin{aligned} \cb{A(x)} &= \pi\big((2x + 3)^2 - x^4\big) \\ &= \pi\big(4x^2 + 12x + 9 - x^4\big) \end{aligned}\]

Questions

The region between \(y = 3\) and \(y = 7\), for \(0 \le x \le 4\), turns about the x-axis. A slice is a washer with outer radius 7 and inner radius 3.

  1. What is the area of the hole?

    • \(9\pi\)
    • \(3\pi\)
    • \(6\pi\)
    • \(49\pi\)
    Answer

    \(9\pi\)

  2. What is the area of the washer?

    • \(40\pi\)
    • \(16\pi\)
    • \(58\pi\)
    Answer

    \(40\pi\)

  3. A washer has \(R(x) = 2x + 1\) and \(r(x) = x\). What is its area at \(x = 2\)?

    • \(21\pi\)
    • \(9\pi\)
    • \(29\pi\)
    • \(3\pi\)
    Answer

    \(21\pi\)

Summary

You square each radius separately, subtract the inner square from the outer square, and multiply by \(\pi\).

The area is a difference because a washer is the outer disk without the hole.

You use this for the area of each slice of a solid that has a hole.

This is one of four skills in topic 8.11 of Which & Why: AP Calculus.