Wemazi

Topic 8.10

Calculate the volume

Builds on setting up the disk integral. 7 minutes.

The disks add up to a volume

The teal region lies between the curve \(y = e^x\) and the dashed line \(y = e\), for \(0 \le x \le 1\). It turns about the dashed line.

The radius of the disk at \(x\) is the gap between the axis and the curve, which is \(e - e^x\).

The slider sets the number of thin disks. The volume of one disk is its area times its thickness.

With more disks, the sum of the volumes comes closer to 3.90. The integral gives this limit.

Expand, integrate, substitute

The volume is the integral of the disk area from one end of the region to the other.

The integrand is \(\pi\) times the squared radius. The radius is a difference, such as \(k - f(x)\), so its square has three terms.

To integrate, expand the square first. Then integrate each term, and substitute the two bounds.

A volume is positive, because every disk has a positive area. The answer is a number times \(\pi\), in cubic units.

Notation

  1. Start from the integral that you set up.

    \[V = \int_a^b \pi\,\cy{R(x)}^2\,dx\]

  2. Expand the square of the radius. The square of a difference has a middle term.

    \[(p - q)^2 = p^2 - 2pq + q^2\]

  3. Integrate each term, and substitute the bounds. Keep the factor \(\pi\).

  4. Write the answer in cubic units, such as cm³. Give an exact value, and add a decimal only as a check.

When it fails

The lower bound must be the smaller value. Press the button to swap the bounds.

The integral now runs from the larger bound to the smaller one, so it gives a negative number, shown in red. A volume cannot be negative, so this integral is not the volume.

Example 1: a cone

The region lies between the line \(y = 2x\), the line \(y = 4\) and the y-axis. It turns about the line \(y = 4\). Find the volume.

  1. The axis is horizontal, so the variable is \(x\). The radius is the axis height minus the line height, and the line reaches the axis height 4 at \(x = 2\).

    \[\cy{R(x)} = 4 - 2x, \quad 0 \le x \le 2\]

  2. Write the integral. Expand the square.

    \[\begin{aligned} V &= \pi\int_0^2 (4 - 2x)^2\,dx \\ &= \pi\int_0^2 \left(16 - 16x + 4x^2\right) dx \end{aligned}\]

  3. Find an antiderivative, and substitute the bounds.

    \[\begin{aligned} &\pi\left[16x - 8x^2 + \dfrac{4x^3}{3}\right]_0^2 \\ &\quad = \pi\left(32 - 32 + \dfrac{32}{3}\right) \end{aligned}\]

  4. The volume is \(\frac{32\pi}{3}\) cubic units. The solid is a cone with radius 4 and height 2, and the cone formula gives the same value.

    \[\dfrac{1}{3}\pi \cdot 4^2 \cdot 2 = \dfrac{32\pi}{3}\]

Example 2: a parabola

The region lies between the curve \(y = x^2\), the line \(y = 9\) and the y-axis. It turns about the line \(y = 9\). Find the volume.

  1. The curve meets the axis at \(x = 3\), so the bounds are 0 and 3. The radius is the gap between the axis and the curve.

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

  2. A tempting wrong move: Square each term of the radius.

    \[(9 - x^2)^2 = 81 - x^4\]

    At \(x = 2\) the orange radius is 5, so the disk has area \(25\pi\). The formula gives \(81 - 16 = 65\) for the squared radius, so the red disk has area \(65\pi\), far too large. The middle term is missing.

  3. Expand the square with its middle term.

    \[(9 - x^2)^2 = 81 - 18x^2 + x^4\]

  4. Integrate each term, and substitute the bounds.

    \[\begin{aligned} V &= \pi\left[81x - 6x^3 + \dfrac{x^5}{5}\right]_0^3 \\ &= \pi\left(243 - 162 + \dfrac{243}{5}\right) \end{aligned}\]

  5. The volume is \(\frac{648\pi}{5}\) cubic units. As a check, this is about 407.2.

    \[\pi\left(81 + \dfrac{243}{5}\right) = \cb{\dfrac{648\pi}{5}}\]

Practice

The region lies between the curve \(y = x^3\), the line \(y = 8\) and the y-axis. It turns about the line \(y = 8\). The lengths are in centimeters. Find the volume.

Move the slider to add disks. The radius is the gap \(8 - x^3\), and the curve meets the axis at \(x = 2\). Expand the square before you integrate.

Solution

  1. The radius is the gap between the axis and the curve. The bounds are 0 and 2.

    \[\cy{R(x)} = 8 - x^3, \quad 0 \le x \le 2\]

  2. Expand the square, and find an antiderivative.

    \[\begin{aligned} V &= \pi\int_0^2 \left(64 - 16x^3 + x^6\right) dx \\ &= \pi\left[64x - 4x^4 + \dfrac{x^7}{7}\right]_0^2 \end{aligned}\]

  3. Substitute the bounds. The volume is \(\frac{576\pi}{7}\) cubic centimeters, about 258.5 cm³.

    \[\pi\left(128 - 64 + \dfrac{128}{7}\right) = \cb{\dfrac{576\pi}{7}}\]

Questions

The region between the line \(y = x\), the line \(y = 3\) and the y-axis turns about the line \(y = 3\).

  1. At which value of \(x\) is the integrand \(\pi[R(x)]^2\) equal to 0?

    • 3
    • 0
    • 1.5
    • 6
    Answer

    3

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

    • \(4\pi\)
    • \(\pi\)
    • \(2\pi\)
    • 4
    Answer

    \(4\pi\)

  3. The region lies between the curve \(y = \sqrt{x}\), the line \(y = 2\) and the y-axis. It turns about the line \(y = 2\). What is the volume?

    • \(\dfrac{8\pi}{3}\)
    • \(8\pi\)
    • \(\dfrac{136\pi}{3}\)
    Answer

    \(\dfrac{8\pi}{3}\)

Summary

You expand the squared radius, integrate each term, and substitute the bounds to find the volume.

This works because the integral adds the volumes of the thin disks, and a volume is never negative.

Use it to find the volume of a disk solid about a shifted axis, in cubic units.

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