Wemazi

Topic 8.10

Measure the radius from that line

Builds on locating the shifted rotation axis. 6 minutes.

The line rises while the radius shrinks

The teal region lies between the line \(y = 1 + x\) and the dashed line \(y = 3\), for \(0 \le x \le 2\). It turns about the dashed line, the axis of rotation, and the slider moves a slice of the region.

The orange segment from the axis down to the line \(y = 1 + x\) is the radius of the teal disk that the slice sweeps.

At \(x = 0\) the line is at height 1 and the axis at height 3, so the radius is the gap between them, 2.

Move the slider to the right. The line rises toward the axis, so the radius shrinks, and at \(x = 2\) it is 0.

The radius is the gap between the axis and the curve

With the numbers gone, the axis is at height \(k\) and the curve at height \(f(x)\). The orange segment between them is the radius, written \(R\).

If the axis is above the curve, the axis is the larger height, so the gap is \(k - f(x)\).

Press "Put the curve above the line". The curve is now the larger height, so the gap is \(f(x) - k\).

In both cases the radius is the larger height minus the smaller, so it is never negative.

Notation

  1. For a horizontal axis \(y = k\) and a curve \(y = f(x)\) below it, the axis is the larger height. The radius is the axis height minus the curve height.

    \[\cy{R(x)} = k - f(x)\]

  2. If the curve is above the axis, the curve is the larger height, so you subtract the other way.

    \[\cy{R(x)} = f(x) - k\]

  3. For a vertical axis \(x = h\), the same rule uses x-values. If the curve \(x = g(y)\) is to the left of the axis, \(h\) is the larger x-value, so the radius is \(h - g(y)\).

    \[\cy{R(y)} = h - g(y)\]

  4. In every case the radius is the larger coordinate minus the smaller one, which needs the region to lie on one side of the axis.

When it fails

Press the button to turn the curve into a line that crosses the axis at \(x = 2\). For \(x\) less than 2 the line is below the axis, so \(k - f(x)\) is positive.

At the slice, right of the crossing, the line is above the axis, so the red value \(k - f(x) = -0.6\) is negative. The orange segment is still 0.6 long, because a radius is a distance. One formula does not fit both sides.

Example 1: a vertical axis

The region between the curve \(x = \dfrac{y^2}{2}\) and the line \(x = 8\), for \(0 \le y \le 4\), turns about \(x = 8\). Find the radius at \(y\), and its values at \(y = 2\) and \(y = 4\).

  1. The curve is to the left of the axis \(x = 8\), so 8 is the larger x-value. Subtract the x-value of the curve from it.

    \[\cy{R(y)} = 8 - \dfrac{y^2}{2}\]

  2. At \(y = 2\) the curve is at \(x = 2\), so the orange segment is 6 units long.

    \[8 - \dfrac{2^2}{2} = 6\]

  3. At \(y = 4\) the curve reaches the axis at \(x = 8\), so the radius is 0.

    \[8 - \dfrac{4^2}{2} = 0\]

Example 2: an axis below the region

The region between the line \(y = -2\) and the curve \(y = 7 - x^2\), for \(0 \le x \le 3\), turns about \(y = -2\). Find the radius at \(x\).

  1. The axis \(y = -2\) lies below the region, so the radius runs up from the axis to the curve.

  2. A tempting wrong move: The radius is the height of the curve.

    \[R = 7 - x^2\]

    At \(x = 1\) the curve is at height 6, but the orange segment starts at the dashed line at height \(-2\). It is 8 units long, so the red disk with radius 6 is too small.

  3. The curve is above the axis, so the curve is the larger height. Subtract the axis height from the curve height.

    \[\cy{R(x)} = (7 - x^2) - (-2) = 9 - x^2\]

  4. At \(x = 1\) the radius is 8. At \(x = 3\) it is 0, because the curve reaches the axis.

    \[9 - 1^2 = 8, \quad 9 - 3^2 = 0\]

Practice

The region lies between the line \(y = 5\) and the curve \(y = 1 + 2\sqrt{x}\), for \(0 \le x \le 4\). It turns about the line \(y = 5\). Find the radius at \(x\), and its value at \(x = 1\).

Move the slider. The orange segment runs from the dashed axis down to the curve.

Solution

  1. The curve is below the axis, so the axis is the larger height. Subtract the curve height from the axis height.

    \[\begin{aligned} \cy{R(x)} &= 5 - \left(1 + 2\sqrt{x}\right) \\ &= 4 - 2\sqrt{x} \end{aligned}\]

  2. At \(x = 1\) the curve is at height 3, two units below the axis, so the radius is 2.

    \[4 - 2\sqrt{1} = 2\]

Questions

The region between the curve \(y = x^2 + 2\) and the line \(y = 7\), for \(-2 \le x \le 2\), turns about the line \(y = 7\).

  1. What is the radius of the disk at \(x = 1\)?

    • 4
    • 3
    • 7
    • 5
    Answer

    4

  2. What is the greatest radius of any disk in this solid?

    • 5
    • 2
    • 7
    • 1
    Answer

    5

  3. The region between the curve \(x = 3y\) and the line \(x = 12\), for \(0 \le y \le 4\), turns about the line \(x = 12\). What is the radius at \(y = 3\)?

    • 3
    • 9
    • 12
    Answer

    3

Summary

The radius of a disk is the distance from the axis to the curve, so it is the larger coordinate minus the smaller one.

Here the axis \(y = 3\) is above the line \(y = 1 + x\), so the radius is \(3 - (1 + x) = 2 - x\).

When the axis is a shifted line, the height of the curve alone is not the radius, so you subtract the two heights.

This is one of four skills in topic 8.10 of Which & Why: AP Calculus.