Wemazi

Topic 8.9

Set up and evaluate the volume

Builds on building the disk area. 7 minutes.

Add the volumes of thin disks

The teal region lies under the curve \(y = \tfrac{1}{x}\) from \(x = 1\) to \(x = 4\). It turns about the x-axis. The slider sets how many thin disks fill the solid.

Each disk has a volume, which is the area of its face times its thickness. The radius of each disk is the height of the curve at the middle of the disk.

With 3 disks, the sum of the volumes is about 2.16. With 6 disks it is about 2.30, and with 24 disks it is about 2.35.

As the disks become thinner, the sum comes closer to a limit, about 2.356. The volume of the solid is this limit, and the integral computes it.

The integral adds the disks

The region runs from a to b along the axis, and a stack of thin disks fills the solid.

One disk has the face area \(\cb{A}\) and the thickness \(\cb{dx}\), so its volume is \(A\,dx\), the area times the thickness.

Add the volumes of all the disks from a to b. Whatever the shape of the curve, the sum comes closer to the volume of the solid as the disks become thinner. The integral is this sum for disks that are infinitely thin.

Notation

  1. The area of the disk at x is pi times the radius squared.

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

  2. The volume is the integral of the disk area from a to b, the two ends of the region. The disk area is the integrand, the function under the integral sign.

    \[V = \int_{\cy{a}}^{\cy{b}} \pi\,[\cy{r(x)}]^2\,dx\]

  3. The numbers \(a\) and \(b\) are the bounds of the integral: the smallest and the largest value of the variable along the axis. For the y-axis, replace \(x\) by \(y\), and \(a\) and \(b\) by \(c\) and \(d\).

  4. To evaluate the integral, find an antiderivative, and substitute the bounds. The volume is in cubic units, such as cubic centimeters (cm³).

When it fails

The integral adds disks only where the bounds reach. Press the button to stop the stack of disks early.

The disks now stop at \(x = 3\), and the part of the solid beyond that has no disks, so the sum is too small. The upper bound must be \(b\), the last value of the variable on the region.

Example 1: a volume about the x-axis

The region lies under \(y = x^3\) from \(x = 0\) to \(x = 1\). It turns about the x-axis. Find the volume.

  1. The axis is the x-axis, so the variable is \(x\). The region runs from \(x = 0\) to \(x = 1\). The radius is the height of the curve.

    \[\cy{r(x)} = x^3\]

  2. Write the integral of the disk area, with the radius squared.

    \[V = \int_0^1 \pi\,(x^3)^2\,dx = \pi\int_0^1 x^6\,dx\]

  3. Find an antiderivative of \(x^6\), and substitute the bounds.

    \[\pi\left[\frac{x^7}{7}\right]_0^1 = \pi \cdot \frac{1}{7}\]

  4. The volume is \(\tfrac{\pi}{7}\) cubic units.

    As a check, this is about 0.449.

Example 2: a volume about the y-axis

The region lies between the y-axis, the curve \(y = x^4\) and the line \(y = 16\). It turns about the y-axis. Find the volume.

  1. The axis is the y-axis, so the variable is \(y\). The radius is the x-value of the curve. Solve \(y = x^4\) for \(x\), which gives \(x = y^{1/4}\).

    \[\cy{r(y)} = y^{1/4}\]

  2. The area of the disk is pi times the radius squared.

    \[\cb{A(y)} = \pi\,(y^{1/4})^2 = \pi\sqrt{y}\]

  3. A tempting wrong move: The region reaches \(x = 2\), so the bounds are 0 and 2.

    \[V = \int_0^2 \pi\sqrt{y}\,dy\]

    The variable is \(y\), so the bounds are heights. The region reaches up to \(y = 16\), but the disks stop at \(y = 2\), so the stack is far too short.

  4. The region runs from \(y = 0\) to \(y = 16\). Use these as the bounds, and find an antiderivative.

    \[V = \pi\int_0^{16} y^{1/2}\,dy = \pi\left[\frac{2}{3}y^{3/2}\right]_0^{16}\]

  5. Substitute the bounds. Since \(16^{3/2} = 4^3 = 64\), the volume is \(\tfrac{128\pi}{3}\) cubic units.

    \[\pi \cdot \frac{2}{3} \cdot 64 = \frac{128\pi}{3}\]

Practice

A funnel is the solid formed when the region under \(y = 6 - 2x\) from \(x = 0\) to \(x = 3\) turns about the x-axis. The lengths are in centimeters. Find the volume.

The radius is the height of the line, \(6 - 2x\), and the disks run from \(x = 0\) to \(x = 3\). Expand the square before you integrate.

Solution

  1. The axis is the x-axis. The radius is the height of the line, and the bounds are 0 and 3.

    \[V = \int_0^3 \pi\,(6 - 2x)^2\,dx\]

  2. Expand the square, and find an antiderivative.

    \[\begin{aligned} &\pi\int_0^3 \left(36 - 24x + 4x^2\right)dx \\ &= \pi\left[36x - 12x^2 + \frac{4x^3}{3}\right]_0^3 \end{aligned}\]

  3. Substitute the bounds. The volume is \(36\pi\) cubic centimeters, which is about 113.1 cm³.

    \[\pi(108 - 108 + 36) = \cb{36\pi}\]

Questions

The figure shows the region between the x-axis and the curve \(y = 4 - x^2\), from \(x = -2\) to \(x = 2\). It turns about the x-axis.

  1. Which bounds does the integral use?

    • \(x = -2\) to \(x = 2\)
    • \(x = 0\) to \(x = 4\)
    • \(x = 0\) to \(x = 2\)
    Answer

    \(x = -2\) to \(x = 2\)

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

    • \(x = -2\) and \(x = 2\)
    • \(x = 0\)
    • \(x = 4\)
    Answer

    \(x = -2\) and \(x = 2\)

  3. The region under \(y = e^x\) from \(x = 0\) to \(x = 1\) turns about the x-axis. What is the volume?

    • \(\dfrac{\pi(e^2 - 1)}{2}\)
    • \(\pi(e - 1)^2\)
    • \(\pi(e^2 - 1)\)
    • \(\dfrac{\pi e^2}{2}\)
    Answer

    \(\dfrac{\pi(e^2 - 1)}{2}\)

Summary

You write the integral of pi times the squared radius over the whole range of the variable along the axis, and you evaluate it.

This works because the integral adds the volumes of thin disks, and each volume is the area of the face times the thickness.

Use it when the slices perpendicular to the axis are disks, and give the answer in cubic units.

This is one of four skills in topic 8.9 of Which & Why: AP Calculus.