Wemazi

Topic 8.7

Calculate and interpret volume

Builds on setting up the volume integral. 7 minutes.

Cubic meters of a solid

A solid has square cross sections perpendicular to the x-axis. Its base is the region between the x-axis and the line \(y = 2x + 1\), from \(x = 0\) to \(x = 3\). All lengths are in meters.

At \(x = 0\), the side is 1 m, so the face has area 1 m². At \(x = 3\), the side is 7 m, so the face has area 49 m².

Drag the end of the stack to the right. The readout shows the volume so far, and at x = 3 it reads 57.

Each slab has a face in square meters and a thickness in meters, so its volume is in cubic meters. The solid has volume 57 cubic meters.

The area under the area graph

The graph shows the cross-section area A against the position x. It stays above the x-axis because an area is positive.

Drag the end of the stack. The teal region under the graph, from a to x, is the volume of the solid so far.

A thin strip of the region has height A in square units and a small width in units. Its area is the volume of one slab, so the strips add up to the volume in cubic units.

Notation

  1. The volume is the integral of the cross-section area \(\cb{A(x)}\), from where the base starts, \(x = a\), to where it ends, \(x = b\).

    \[V = \int_a^b \cb{A(x)}\,dx\]

  2. To evaluate the integral, find an antiderivative \(F\) of \(\cb{A}\), so that \(F'(x) = A(x)\). Then \(V\) is \(F(b) - F(a)\).

    \[V = F(b) - F(a)\]

  3. For the opening's solid, raise the power of \((2x + 1)^2\) to 3. Then divide by the new power 3 and by the slope 2 of the inside. The division by 2 undoes the chain rule.

    \[\begin{aligned} &\int_0^3 (2x + 1)^2\,dx \\ &\quad = \left[\frac{(2x + 1)^3}{3 \cdot 2}\right]_0^3 \\ &\quad = \frac{343 - 1}{6} = 57 \end{aligned}\]

  4. \(\cb{A(x)}\) is in square units and \(dx\) is in units, so \(V\) is in cubic units.

  5. The result is positive. It is smaller than the largest face area times the length of the solid, because the solid fits inside that box.

Example 1: rectangles of height 3 meters

The base of a solid is the region between \(y = x\) and \(y = 2\sqrt{x}\), in meters. Its cross sections perpendicular to the x-axis are rectangles of height 3 m. Find the volume.

  1. The curves meet where \(x = 2\sqrt{x}\), at \(x = 0\) and \(x = 4\). At \(x = 1\) the curve is at height 2 and the line at height 1, so the curve is the upper boundary. Each face is 3 m tall on the base \(w(x)\).

    \[\begin{aligned} \cy{w(x)} &= 2\sqrt{x} - x \\ \cb{A(x)} &= 3\left(2\sqrt{x} - x\right) \end{aligned}\]

  2. Find an antiderivative of \(A(x) = 6x^{1/2} - 3x\). Raise each power by one and divide by the new power.

    \[F(x) = 4x^{3/2} - \tfrac{3}{2}x^2\]

  3. Evaluate at 4 and at 0, and subtract.

    \[\begin{aligned} V &= F(4) - F(0) \\ &= (32 - 24) - 0 = 8 \text{ m}^3 \end{aligned}\]

  4. As a check, the largest base is 1 m, at \(x = 1\), so the largest face has area 3 m². The solid is 4 m long, so the box has volume \(3 \cdot 4 = 12\) m³, and 8 is smaller.

Example 2: squares on a decreasing base

The base of a solid is the region between the x-axis and \(y = \frac{1}{\sqrt{x}}\), for \(1 \le x \le 4\), in meters. Its cross sections perpendicular to the x-axis are squares. Find the volume.

  1. The side of each square is the height of the curve, so the area is the side squared.

    \[\cy{w(x)} = x^{-1/2}, \quad \cb{A(x)} = \frac{1}{x}\]

  2. A tempting wrong move: Integrate the side from 1 to 4.

    \[\cr{\int_1^4 x^{-1/2}\,dx = 2}\]

    The integral of the side is the area of the base region, 2 square meters, not a volume. A volume is the integral of the cross-section area, in cubic meters.

  3. Integrate the area of the cross section. An antiderivative of \(\frac{1}{x}\) is \(\ln x\), and \(\ln 1 = 0\).

    \[\begin{aligned} V &= \int_1^4 \frac{1}{x}\,dx \\ &= \ln 4 - \ln 1 = \ln 4 \text{ m}^3 \end{aligned}\]

  4. As a check, \(\ln 4\) is about 1.386. The faces have areas between \(\frac{1}{4}\) and 1, and the solid is 3 m long, so the volume lies between 0.75 and 3.

Practice

The base of a solid is the region between the y-axis and \(x = y^3\), for \(0 \le y \le 2\), in centimeters. Its cross sections perpendicular to the y-axis are squares. Find the volume.

The slices are perpendicular to the y-axis, so integrate with respect to \(y\).

Solution

  1. Each base is horizontal and runs from \(x = 0\) to \(x = y^3\), so the side is \(y^3\) and the area of the square is \(y^6\).

    \[w(y) = y^3, \quad A(y) = y^6\]

  2. Integrate from \(y = 0\) to \(y = 2\).

    \[V = \int_0^2 y^6\,dy = \frac{2^7}{7} = \frac{128}{7} \text{ cm}^3\]

  3. As a check, the value is about 18.29. The largest face has area 64 cm² and the solid is 2 cm long, so the box has volume 128 cm³, and 18.29 is smaller.

Questions

The base of a solid is the region between the x-axis and \(y = 2x\), for \(0 \le x \le 3\). Lengths are in centimeters. The cross sections are squares perpendicular to the x-axis.

  1. What is the unit of the volume of this solid?

    • Cubic centimeters
    • Square centimeters
    • Centimeters
    Answer

    Cubic centimeters

  2. A box has the length of the solid and the largest face of the solid as its cross section. Which could be the volume of the solid?

    • 36 cm³
    • 108 cm³
    • 216 cm³
    Answer

    36 cm³

  3. A base is the region between the x-axis and \(y = 3x^2\), for \(0 \le x \le 2\), in meters. The cross sections are rectangles of height 2 m, perpendicular to the x-axis. What is the volume?

    • 16 m³
    • 8 m³
    • 48 m³
    • 16 m²
    Answer

    16 m³

Summary

You integrate the cross-section area between the ends of the base, and you write the answer in cubic units.

This works because the integral of the area adds the volumes of all the thin slabs.

You use it for any solid whose cross sections you can describe, and you test the result against the box around the solid.

This is one of four skills in topic 8.7 of Which & Why: AP Calculus.