Wemazi

Topic 8.8

Integrate the slice areas

Builds on building a semicircular cross-sectional area. 7 minutes.

Semicircles across a disk

The base of a solid is the disk \(x^2 + y^2 \le 36\), in centimeters. Its cross sections are semicircles perpendicular to the x-axis, with the diameter across the disk.

At \(x = 0\), the diameter is 12 and the face has area \(18\pi\) cm². At \(x = 6\), the diameter is 0 and the face is a point.

The stack of slices starts at the left end of the disk, \(x = -6\). Drag its end to the right and read the volume so far.

At \(x = 6\), the stack fills the solid and the volume is \(144\pi\) cm³, about 452.4 cm³.

From one end of the base to the other

Without numbers, the base lies between \(a\) and \(b\) on the slicing axis, the axis that the cross sections are perpendicular to.

Drag the end of the stack. The stack covers the whole solid only when it runs from \(a\) to \(b\).

Increase the number of slices. Each slice has volume equal to its face area \(A(x)\) times its thickness, so the sum of the slices is a Riemann sum. As the slices become thinner, the sum approaches the volume of the stack. For the whole solid, this limit is the integral of \(A\) from \(a\) to \(b\).

Notation

  1. Find the slicing variable from the problem. Cross sections perpendicular to the x-axis use \(x\), and cross sections perpendicular to the y-axis use \(y\).

  2. Write the area \(A\) of the cross section in that variable, as in the earlier skills.

    \[\cb{A(x)}\]

  3. The limits are the smallest and the largest values of that variable in the base, so the stack covers the whole solid. One of them can be negative.

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

  4. Find an antiderivative \(F\) of \(A\). Evaluate \(F\) at the limits, and subtract.

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

  5. If \(A(-x) = A(x)\), the two halves of a base from \(-b\) to \(b\) give equal volumes. Then you can integrate over one half and double the result.

    \[V = 2\int_0^b A(x)\,dx\]

  6. The result is a volume, so write it in cubic units.

Example 1: equilateral triangles

The base is the region between the x-axis and \(y = 6 - 3x\), in meters, for \(x \ge 0\). The cross sections are equilateral triangles perpendicular to the x-axis. Find the volume.

  1. The region lies between the x-axis and the line, so the side \(w\) of each triangle is the height of the line. The line meets the x-axis at \(x = 2\), so the limits are 0 and 2.

    \[\cy{w(x)} = 6 - 3x = 3(2 - x)\]

  2. The area of an equilateral triangle with side \(w\) is \(\tfrac{\sqrt{3}}{4}w^2\), and \(w^2 = 9(2 - x)^2\).

    \[\begin{aligned} \cb{A(x)} &= \frac{\sqrt{3}}{4}\cdot 9(2 - x)^2 \\ &= \frac{9\sqrt{3}}{4}(2 - x)^2 \end{aligned}\]

  3. Integrate from 0 to 2. An antiderivative of \((2 - x)^2\) is \(-\tfrac{(2 - x)^3}{3}\), which is 0 at \(x = 2\) and \(-\tfrac{8}{3}\) at \(x = 0\).

    \[\begin{aligned} V &= \frac{9\sqrt{3}}{4}\left[-\frac{(2 - x)^3}{3}\right]_0^2 \\ &= \frac{9\sqrt{3}}{4}\cdot\frac{8}{3} = 6\sqrt{3} \text{ m}^3 \end{aligned}\]

  4. As a check, the side shrinks linearly to a point, so the solid is a pyramid of height 2. Its base, the first face, has area \(9\sqrt{3}\) m², and one third of base area times height is \(\tfrac13\cdot 9\sqrt{3}\cdot 2 = 6\sqrt{3}\).

Example 2: both sides of the base

The base is the region between the x-axis and \(y = 1 - x^2\), in centimeters. The cross sections are isosceles right triangles perpendicular to the x-axis, with the hypotenuse on the base. Find the volume.

  1. The curve meets the x-axis at \(x = -1\) and \(x = 1\). The hypotenuse \(w\) is the height of the curve, and the triangle's height is half of its hypotenuse.

    \[\cb{A(x)} = \frac{1}{2}\,w\cdot\frac{w}{2} = \frac{(1 - x^2)^2}{4}\]

  2. A tempting wrong move: The base starts at the y-axis, so integrate from 0 to 1.

    \[\cr{V = \int_0^1 \frac{(1 - x^2)^2}{4}\,dx = \frac{2}{15}}\]

    The solid lies on both sides of the y-axis. The limits 0 to 1 count only the right half and leave the faint left half out, so the answer is half of the volume.

  3. The base reaches from \(x = -1\) to \(x = 1\). Because \(x\) appears only as \(x^2\), \(A(-x) = A(x)\), so the volume is twice the integral over the right half.

    \[\begin{aligned} V &= \int_{-1}^{1} \frac{(1 - x^2)^2}{4}\,dx \\ &= 2\cdot\frac{2}{15} = \frac{4}{15} \text{ cm}^3 \end{aligned}\]

Practice

The base is the region between \(y = x^2\) and \(y = 4\), in centimeters. The cross sections are semicircles perpendicular to the y-axis, with the diameter on the base. Find the volume.

The slices stand perpendicular to the y-axis, so the diameter is horizontal and the limits are y-values. Drag the slice up the y-axis.

Solution

  1. The slice at \(y\) crosses the region from \(x = -\sqrt{y}\) to \(x = \sqrt{y}\), where \(x^2 = y\). The diameter is the difference, and a semicircle with diameter \(w\) has area \(\tfrac{\pi}{8}w^2\).

    \[\begin{aligned} w(y) &= 2\sqrt{y} \\ \cb{A(y)} &= \frac{\pi}{8}\cdot 4y = \frac{\pi y}{2} \end{aligned}\]

  2. The base reaches from \(y = 0\), at the vertex of the parabola, to \(y = 4\), at the line.

    \[\begin{aligned} V &= \int_0^4 \frac{\pi y}{2}\,dy \\ &= \left[\frac{\pi y^2}{4}\right]_0^4 = 4\pi \text{ cm}^3 \end{aligned}\]

Questions

The base is the region between the y-axis and \(x = 2y\), for \(0 \le y \le 3\), in meters. The cross sections are semicircles perpendicular to the y-axis, with the diameter on the base.

  1. Which integral gives the volume?

    • \(\displaystyle\int_0^3 \frac{\pi y^2}{2}\,dy\)
    • \(\displaystyle\int_0^6 \frac{\pi y^2}{2}\,dy\)
    • \(\displaystyle\int_0^3 \frac{\pi x^2}{2}\,dx\)
    Answer

    \(\displaystyle\int_0^3 \frac{\pi y^2}{2}\,dy\)

  2. The stack stops at \(y = 2\). What does the integral from 0 to 2 give?

    • The volume of the part of the solid below \(y = 2\)
    • The volume of the whole solid
    • The area of the base region up to \(y = 2\)
    Answer

    The volume of the part of the solid below \(y = 2\)

  3. The base is the region between the x-axis and \(y = 4 - x\), for \(0 \le x \le 4\). The cross sections are isosceles right triangles with the hypotenuse on the base. What is the volume, in cubic units?

    • \(\tfrac{16}{3}\)
    • \(\tfrac{64}{3}\)
    • \(\tfrac{32}{3}\)
    • 8
    Answer

    \(\tfrac{16}{3}\)

Summary

You integrate the area of the cross section over the interval where the base lies, and write the answer in cubic units.

This works because the integral adds the volumes of the thin slices from one end of the solid to the other.

Use this for any triangular or semicircular cross section, after you write the area in the slicing variable.

This is one of four skills in topic 8.8 of Which & Why: AP Calculus.