Calculus 1

The washer method

When a region is turned about a line it does not touch, the solid has a hole down its middle. How do you find the volume of a solid with a hole?

You need

The disk methodCalculus 1

Where you are 10 in Applications of integrals

Calculus8

Applications of integrals10

▴ 5 above Arc length Average value of a function The disk method Surface area of a solid of revolution

The washer method

When you reach for it

The The disk method turns a region that lies against the axis of revolution, so every slice across the axis is a filled circle. Many solids are not like that. A pipe fitting, a bushing or a bead has a hole down its center, because the region that sweeps it out stays some distance away from the axis. Each slice across the axis is then a circle with a smaller circle cut out of its middle, shaped like the flat metal washer under a bolt.

You reach for the washer method when the region lies between two boundaries, both on the same side of the axis, and at least part of it is separated from the axis by a gap. It is the disk method applied twice, once to the outer boundary and once to the inner one.

The procedure

Day 1: washers about a coordinate axis

The slice is a washer. Let f and g be continuous on [a, b] with 0 \le g(x) \le f(x), and turn the region between them about the x-axis. The plane perpendicular to the axis at x meets the region in the segment from (x, g(x)) to (x, f(x)). Turning that segment about the axis sweeps every point whose distance from the axis lies between g(x) and f(x): a ring with

\begin{aligned} R(x) &= f(x) \quad \text{(outer radius)}, \\ r(x) &= g(x) \quad \text{(inner radius)}. \end{aligned}

The area of a ring. The ring is the disk of radius R with the disk of radius r removed (Figure 1), so its area is

\begin{aligned} A &= \pi R^2 - \pi r^2 = \pi\left(R^2 - r^2\right) \\ &= \pi(R - r)(R + r). \end{aligned}

This is not \pi(R - r)^2. That expression is the area of a disk whose radius is the width of the ring, and it leaves out how far the ring sits from its center. With R = 3 and r = 1 the ring has area \pi(9 - 1) = 8\pi, while \pi(3 - 1)^2 = 4\pi. Squaring each radius first and then subtracting is what matches the geometry.

Figure 1A washer seen face on. Its area is the big disk minus the small one, \pi R^2 - \pi r^2.

The integral. A solid whose slice at x has area A(x) has volume \int_a^b A(x)\,dx (Volumes with known cross sections). With washer slices this is the washer method:

V = \int_a^b \pi\left([R(x)]^2 - [r(x)]^2\right)dx.

A thin slab over a piece of width \Delta x is nearly a flat washer of volume \pi\left([R(x_i^*)]^2 - [r(x_i^*)]^2\right)\Delta x, and the integral is the limit of the sum of these, as for the disk method. Splitting the integral gives

V = \pi\int_a^b [R(x)]^2\,dx - \pi\int_a^b [r(x)]^2\,dx,

the solid swept by the outer boundary minus the solid swept by the inner one, which is the idea of the disk method’s Exercise 18. When r(x) = 0 throughout, the region reaches the axis and the formula is the disk method.

About the y-axis. If the region lies to the right of the y-axis between the curves x = q(y) and x = p(y), with 0 \le q(y) \le p(y) for c \le y \le d, the slices are horizontal washers with R(y) = p(y) and r(y) = q(y), and

V = \int_c^d \pi\left([R(y)]^2 - [r(y)]^2\right)dy.

The variable is the one running along the axis, and boundaries given as y = f(x) are solved for x first.

Day 2: other axes and changing boundaries

Radii are distances from the axis. For any axis, R is the distance from the axis to the farther boundary of the slice and r the distance to the nearer one, both measured along the slice. For a horizontal axis y = k below the region, with lower boundary g and upper boundary f,

R(x) = f(x) - k, \qquad r(x) = g(x) - k;

for an axis above the region the roles swap, R(x) = k - g(x) and r(x) = k - f(x). Which curve is outer depends on which side the axis is on. A vertical axis x = h works the same way with horizontal slices and the variable y.

The axis fixes the variable. Washers are slices perpendicular to the axis, so a horizontal axis means integrating in x and a vertical axis means integrating in y. Slices parallel to the axis are not washers; building a solid from those is the The shell method of a later course.

Boundaries that change. If the farther or the nearer boundary is one curve on part of the interval and another curve on the rest, R or r has a different formula on each part. Split the interval where the boundary changes and add one washer integral for each piece, exactly as an area is split when the top curve changes (Area between curves).

Worked examples

Example 1

Day 1: a hole of constant radius

Find the volume when the region between y = 5 - x^2 and the line y = 1 is revolved about the x-axis.

Solution

The boundaries meet where 5 - x^2 = 1, at x = \pm 2, and between these 5 - x^2 \ge 1. The region lies between heights 1 and 5 - x^2, so it never touches the axis and the slice at x is a washer with R(x) = 5 - x^2 and r(x) = 1. Then [R]^2 - [r]^2 = 24 - 10x^2 + x^4, an even function, so

\begin{aligned} V &= 2\pi\int_0^2 \left(24 - 10x^2 + x^4\right)dx \\ &= 2\pi\left(48 - \frac{80}{3} + \frac{32}{5}\right) \\ &= \frac{832\pi}{15} \approx 174.3. \end{aligned}

A check by subtraction: the hole is a cylinder of radius 1 and length 4, volume \displaystyle 4\pi = \frac{60\pi}{15}, and the solid swept by the outer curve alone has volume \displaystyle \pi\int_{-2}^{2} (5 - x^2)^2\,dx = \frac{892\pi}{15}. The difference is \displaystyle \frac{832\pi}{15}.

Example 2

Day 1: Thamer checks a washer volume by subtraction

Thamer revolves the region between \displaystyle y = \sqrt{2x} and y = x about the x-axis. Find the volume, and check it as a solid minus a cone.

Solution

The curves meet where \displaystyle \sqrt{2x} = x. Squaring, 2x = x^2, so x = 0 or x = 2, and both satisfy the original equation. At x = 1, \displaystyle \sqrt{2} > 1, so the curve is on top. The slice at x is a washer with \displaystyle R(x) = \sqrt{2x} and r(x) = x (Figure 2), and [R(x)]^2 = 2x needs no root:

\begin{aligned} V &= \pi\int_0^2 \left(2x - x^2\right)dx = \pi\left(4 - \frac{8}{3}\right) \\ &= \frac{4\pi}{3} \approx 4.19. \end{aligned}

Figure 2The region of Example 2 and its reflection across the x-axis. The slice at x is a washer whose outer radius R reaches the curve and whose inner radius r reaches the line.

Thamer’s check: the region under the curve alone sweeps a solid of volume \pi\int_0^2 2x\,dx = 4\pi, and the triangle under y = x sweeps a cone of radius 2 and height 2, volume \displaystyle \frac{1}{3}\pi(2)^2(2) = \frac{8\pi}{3}. Removing the cone leaves \displaystyle 4\pi - \frac{8\pi}{3} = \frac{4\pi}{3}. The region touches the axis only at the origin, where both radii are 0; one point changes no integral.

Example 3

Day 1: washers in y

The region between \displaystyle y = \frac{x^2}{2} and y = 2x is revolved about the y-axis. Find the volume.

Solution

The curves meet where \displaystyle \frac{x^2}{2} = 2x, at x = 0 and x = 4, so at (0, 0) and (4, 8). The axis is vertical, so the slices are horizontal and 0 \le y \le 8. Solving each boundary for x (with x \ge 0): the line is \displaystyle x = \frac{y}{2} and the parabola is \displaystyle x = \sqrt{2y}. At y = 2 these give 1 and 2, so the parabola is farther from the y-axis:

R(y) = \sqrt{2y}, \qquad r(y) = \frac{y}{2}.

\begin{aligned} V &= \pi\int_0^8 \left(2y - \frac{y^2}{4}\right)dy = \pi\left(64 - \frac{128}{3}\right) \\ &= \frac{64\pi}{3} \approx 67.0. \end{aligned}

About the x-axis the roles would be the other way: there the line, which is on top, would be the outer boundary. Outer and inner are read from the axis, not from the picture’s top and bottom.

Example 4

Day 2: Nawal turns one region about two lines

Nawal takes the region between y = 4 - x^2 and y = 4 - 2x and revolves it (a) about the line y = 5 and (b) about the line y = -1. Find both volumes.

Solution

The curves meet where x^2 = 2x, at x = 0 and x = 2. For 0 < x < 2, x^2 < 2x, so 4 - x^2 is on top, and the region lies between heights 0 and 4. Both axes are horizontal, so both integrals are in x from 0 to 2.

(a) The line y = 5 is above the region, so the lower boundary is farther from it (Figure 3):

\begin{aligned} R(x) &= 5 - (4 - 2x) = 1 + 2x, \\ r(x) &= 5 - (4 - x^2) = 1 + x^2. \end{aligned}

Then [R]^2 - [r]^2 = 4x + 2x^2 - x^4, and

\begin{aligned} V_a &= \pi\int_0^2 \left(4x + 2x^2 - x^4\right)dx \\ &= \pi\left(8 + \frac{16}{3} - \frac{32}{5}\right) = \frac{104\pi}{15} \approx 21.8. \end{aligned}

Figure 3The region of Example 4 turned about the line y = 5. The axis is above the region, so the lower boundary is the farther one and gives the outer radius R; the parabola gives the inner radius r.

(b) The line y = -1 is below the region, so now the upper boundary is farther: R(x) = (4 - x^2) + 1 = 5 - x^2 and r(x) = (4 - 2x) + 1 = 5 - 2x. Then [R]^2 - [r]^2 = x^4 - 14x^2 + 20x, and

\begin{aligned} V_b &= \pi\int_0^2 \left(x^4 - 14x^2 + 20x\right)dx \\ &= \pi\left(\frac{32}{5} - \frac{112}{3} + 40\right) = \frac{136\pi}{15} \approx 28.5. \end{aligned}

The same region gives different solids. A small piece of the region far from the axis sweeps a wide ring, and this region sits nearer to y = 5 than to y = -1, so the second solid is larger. Exercise 20 makes exact how the volume grows as the axis moves away.

Example 5

Day 2: about the line x = -2

The region bounded by \displaystyle y = \sqrt{x}, the line x = 4 and the x-axis is revolved about the line x = -2. Find the volume.

Solution

The axis is vertical, so the variable is y. The curve meets x = 4 at y = 2, so 0 \le y \le 2. Solving \displaystyle y = \sqrt{x} for x gives x = y^2, and at height y the region runs from x = y^2 to x = 4. Both are to the right of x = -2, so the distances from the axis are

\begin{aligned} R(y) &= 4 - (-2) = 6, \\ r(y) &= y^2 - (-2) = y^2 + 2. \end{aligned}

Then [R]^2 - [r]^2 = 36 - (y^4 + 4y^2 + 4) = 32 - 4y^2 - y^4, and

\begin{aligned} V &= \pi\int_0^2 \left(32 - 4y^2 - y^4\right)dy \\ &= \pi\left(64 - \frac{32}{3} - \frac{32}{5}\right) = \frac{704\pi}{15} \approx 147.4. \end{aligned}

The outer radius is constant, so the outside of the solid is a cylinder of radius 6; the hole widens from radius 2 at the bottom to radius 6 at the top, where it meets the outside.

Example 6

Day 2: an outer radius that changes formula

The region under both y = 4x^2 and y = 5 - x and above the x-axis, for 0 \le x \le 5, is revolved about the line y = -2. Find the volume.

Solution

The curves meet where 4x^2 + x - 5 = 0, that is (4x + 5)(x - 1) = 0, and in this interval only at x = 1. The top of the region is y = 4x^2 for 0 \le x \le 1 and y = 5 - x for 1 \le x \le 5; the bottom is the x-axis throughout. The axis y = -2 is below, so the inner radius is r = 0 - (-2) = 2 everywhere, while the outer radius has two formulas:

R(x) = \begin{cases} 4x^2 + 2, & 0 \le x \le 1, \\ 7 - x, & 1 \le x \le 5. \end{cases}

Split at x = 1:

\begin{aligned} V &= \pi\int_0^1 \left(16x^4 + 16x^2\right)dx \\ &\quad + \pi\int_1^5 \left((7 - x)^2 - 4\right)dx \\ &= \pi\left(\frac{16}{5} + \frac{16}{3}\right) + \pi\left(\frac{208}{3} - 16\right) \\ &= \frac{128\pi}{15} + \frac{800\pi}{15} = \frac{928\pi}{15} \approx 194.4. \end{aligned}

One formula for R across the whole interval would be wrong: at x = 3, 4x^2 + 2 = 38, far above the region. Slicing in y would avoid the split but give slices parallel to the axis, which are not washers.

Example 7

Day 2: Fahd’s brass pipe fitting

Fahd machines a brass pipe fitting 4 cm long. Along its center line, x cm from one end, its outside radius is

R(x) = 2 + \frac{(x - 2)^2}{4}\ \text{cm}, \qquad 0 \le x \le 4,

and a straight bore of radius 1 cm runs through it. Find the volume of brass and its mass at 8.5 grams per cubic centimeter.

Solution

The fitting is the solid swept by the region between y = 1 and y = R(x) turned about the center line, so each slice is a washer with outer radius R(x) and inner radius 1. The outside radius is 3 at the ends and 2 at the waist. Substituting u = x - 2 moves the waist to u = 0, with limits -2 and 2, and

\begin{aligned} [R]^2 - 1^2 &= \left(2 + \frac{u^2}{4}\right)^2 - 1 \\ &= 3 + u^2 + \frac{u^4}{16}, \end{aligned}

an even function, so

\begin{aligned} V &= 2\pi\int_0^2 \left(3 + u^2 + \frac{u^4}{16}\right)du \\ &= 2\pi\left(6 + \frac{8}{3} + \frac{2}{5}\right) \\ &= \frac{272\pi}{15} \approx 57.0\ \text{cm}^3. \end{aligned}

The mass is \displaystyle 8.5 \times \frac{272\pi}{15} \approx 484 grams. The bore itself is a cylinder of volume \pi(1)^2(4) = 4\pi \approx 12.6 cubic centimeters, the brass that a solid fitting of the same outline would have had in addition.

Boundaries

The region must lie on one side of the axis of revolution (it may touch it). If it straddles the axis, the two sides sweep overlapping solids and the washer formula does not apply as written.
At every slice, 0 \le r \le R, and the region must meet the slice in a single segment. If a slice crosses the region in two separate segments, the slice is two concentric rings and its area is the sum of two washer areas.
R and r must be continuous on the interval, or continuous on pieces with the integral split where a boundary changes.
The radii are distances from the axis of revolution, not heights above the x-axis, unless the axis is the x-axis.
The slices are perpendicular to the axis, which fixes the variable: x for a horizontal axis, y for a vertical one.

Where this goes next

Some solids are awkward in washers, for example a region between two curves turned about a vertical axis when the curves cannot be solved for x. Building the solid from thin cylindrical shells parallel to the axis, the The shell method, handles those. The bands swept by the outer and inner edges lead to the Surface area of a solid of revolution, and the same idea of adding thin pieces of known size returns in Work as an integral.

Exercises

Day 1: about the x-axis

The region between y = x^2 and the line y = 4 is revolved about the x-axis. Find the volume.
The region between y = x + 3 and y = 1 for 0 \le x \le 2 is revolved about the x-axis. Find the volume, and check it as a frustum of a cone minus a cylinder.
The region between \displaystyle y = \sqrt{x} and y = x^3 is revolved about the x-axis. Find the volume.
The region between y = e^x and y = 1 for 0 \le x \le 1 is revolved about the x-axis. Find the volume.
The region between y = \cos x and y = \sin x for \displaystyle 0 \le x \le \frac{\pi}{4} is revolved about the x-axis. Find the volume.
The region between \displaystyle y = \frac{4}{x} and y = 5 - x is revolved about the x-axis. Find the volume.

Day 1: about the y-axis

The region between x = y^2 + 1 and the line x = 2 is revolved about the y-axis. Find the volume.
The region to the right of the line x = 1 bounded by y = x^2, the line y = 4 and the line x = 1 is revolved about the y-axis. Find the volume.

Day 2: other lines

The region of Exercise 1 is revolved about the line y = -1. Find the volume.
The region of Exercise 3 is revolved about the line y = 1. Find the volume.
The region of Example 3 is revolved about the line x = 5. Find the volume.
The region of Example 3 is revolved about the line x = -3. Find the volume.
The region of Exercise 5 is revolved about the line y = -1. Find the volume.

Day 2: boundaries that change

The region bounded by \displaystyle y = \sqrt{x}, the line y = 3 - 2x and the x-axis is revolved (a) about the line y = -1, splitting the integral, and (b) about the y-axis, where no split is needed. Find both volumes.
The triangle bounded by y = 2x, y = x and the line x = 1 is revolved about the line x = 3. Find the volume. (The inner radius changes formula.)

Day 2: parts with a hole

Nawal makes a steel bushing 4 cm long with outside radius 2 cm. Its bore is tapered: x cm from one end its radius is \displaystyle 1 + \frac{x}{8} cm, for 0 \le x \le 4. Find the volume of steel and its mass at 7.9 grams per cubic centimeter.
Thamer drills a hole of radius 0.6 cm straight through the center of a glass ball of radius 1 cm to make a spacer. Find the volume of glass left and its mass at 2.5 grams per cubic centimeter.

Reasoning

The rectangle 0 \le x \le 3, 1 \le y \le 2 is revolved about the x-axis. Find the volume of the tube it sweeps, and use it to show that \pi\int_a^b \left(R - r\right)^2 dx does not give the volume of a solid with a hole. What does \pi(R - r)^2 measure?
Explain why the washer volume equals the volume swept by the outer boundary minus the volume swept by the inner boundary, and why the washer method becomes the disk method when the region reaches the axis.
A region lies between y = g(x) and y = f(x) with 0 \le g(x) \le f(x) on [a, b], and has area A. Let V_0 be its volume about the x-axis and V_k its volume about the line y = -k, with k > 0. Show that V_k = V_0 + 2\pi k A, and check it against Example 4(b).
A hole is drilled straight through the center of a ball of radius a, leaving a ring of height h (measured along the hole). Show that the volume left is \displaystyle \frac{\pi h^3}{6}, whatever the radius a, and check it against Exercise 17.

Answers

\displaystyle \frac{256\pi}{5}: they meet at x = \pm 2, R = 4, r = x^2, and \displaystyle 2\pi\int_0^2 \left(16 - x^4\right)dx = 2\pi\left(32 - \frac{32}{5}\right).
\displaystyle \frac{92\pi}{3}: \pi\int_0^2 \left((x + 3)^2 - 1\right)dx is \displaystyle \pi\left(\frac{(x + 3)^3}{3} - x\right)\Big|_0^2, which is \displaystyle \pi\left(\frac{125}{3} - 2 - 9\right). The frustum has radii 3 and 5 and length 2, volume \displaystyle \frac{\pi\cdot 2}{3}(9 + 15 + 25) = \frac{98\pi}{3}, and the cylinder 2\pi.
\displaystyle \frac{5\pi}{14}: they meet at x = 0 and x = 1, with \displaystyle \sqrt{x} \ge x^3 between, and \displaystyle \pi\int_0^1 \left(x - x^6\right)dx = \pi\left(\frac{1}{2} - \frac{1}{7}\right).
\displaystyle \frac{\pi(e^2 - 3)}{2} \approx 6.89: e^x \ge 1 on [0, 1], and \displaystyle \pi\int_0^1 \left(e^{2x} - 1\right)dx = \pi\left(\frac{e^2 - 1}{2} - 1\right).
\displaystyle \frac{\pi}{2}: \cos x \ge \sin x \ge 0 there, and \cos^2 x - \sin^2 x = \cos 2x, so \displaystyle \pi\int_0^{\pi/4} \cos 2x\,dx = \frac{\pi}{2}\sin 2x\,\Big|_0^{\pi/4}.
9\pi: \displaystyle \frac{4}{x} = 5 - x gives x^2 - 5x + 4 = 0, so x = 1 or 4, with the line on top. \pi\int_1^4 (5 - x)^2\,dx = 21\pi and \displaystyle \pi\int_1^4 \frac{16}{x^2}\,dx = 12\pi.
\displaystyle \frac{64\pi}{15}: they meet at y = \pm 1, R = 2, r = y^2 + 1, and \displaystyle 2\pi\int_0^1 \left(3 - 2y^2 - y^4\right)dy = 2\pi\left(3 - \frac{2}{3} - \frac{1}{5}\right).
\displaystyle \frac{9\pi}{2}: for 1 \le y \le 4 the region runs from x = 1 to \displaystyle x = \sqrt{y}, so \pi\int_1^4 (y - 1)\,dy.
\displaystyle \frac{1088\pi}{15}: R = 4 + 1 = 5, r = x^2 + 1, and 2\pi\int_0^2 \left(24 - 2x^2 - x^4\right)dx is \displaystyle 2\pi\left(48 - \frac{16}{3} - \frac{32}{5}\right).
\displaystyle \frac{10\pi}{21}: the axis is above, so R = 1 - x^3 and \displaystyle r = 1 - \sqrt{x}, and \displaystyle \pi\int_0^1 \left(2\sqrt{x} - x - 2x^3 + x^6\right)dx is \displaystyle \pi\left(\frac{4}{3} - \frac{1}{2} - \frac{1}{2} + \frac{1}{7}\right).
32\pi: the left boundary \displaystyle x = \frac{y}{2} is farther from x = 5, so \displaystyle R = 5 - \frac{y}{2}, \displaystyle r = 5 - \sqrt{2y}, and \displaystyle \pi\int_0^8 \left(10\sqrt{2y} - 7y + \frac{y^2}{4}\right)dy is \displaystyle \pi\left(\frac{640}{3} - 224 + \frac{128}{3}\right).
\displaystyle \frac{160\pi}{3}: \displaystyle R = \sqrt{2y} + 3, \displaystyle r = \frac{y}{2} + 3, and \displaystyle \pi\int_0^8 \left(6\sqrt{2y} - y - \frac{y^2}{4}\right)dy is \displaystyle \pi\left(128 - 32 - \frac{128}{3}\right).
\displaystyle \pi\left(2\sqrt{2} - \frac{3}{2}\right) \approx 4.17: R = \cos x + 1, r = \sin x + 1, and [R]^2 - [r]^2 = \cos 2x + 2(\cos x - \sin x), which integrates to \displaystyle \frac{1}{2} + 2\left(\sqrt{2} - 1\right).
(a) \displaystyle \frac{5\pi}{2}, (b) \displaystyle \frac{83\pi}{60}. The curve and line meet at x = 1; the top is \displaystyle \sqrt{x} on [0, 1] and 3 - 2x on \displaystyle \left[1, \frac{3}{2}\right]. (a) r = 1; the pieces \displaystyle \pi\int_0^1 \left(x + 2\sqrt{x}\right)dx and \pi\int_1^{3/2} \left((4 - 2x)^2 - 1\right)dx give \displaystyle \frac{11\pi}{6} and \displaystyle \frac{2\pi}{3}. (b) For 0 \le y \le 1 the region runs from x = y^2 to \displaystyle x = \frac{3 - y}{2}, and \displaystyle \pi\int_0^1 \left(\frac{(3 - y)^2}{4} - y^4\right)dy = \pi\left(\frac{19}{12} - \frac{1}{5}\right).
\displaystyle \frac{7\pi}{3}: the vertices are (0, 0), (1, 1) and (1, 2). The outer radius is \displaystyle 3 - \frac{y}{2} throughout; the inner radius is 3 - y for 0 \le y \le 1 and 3 - 1 = 2 for 1 \le y \le 2. The pieces give \displaystyle \pi\cdot\frac{5}{4} and \displaystyle \pi\cdot\frac{13}{12}.
\displaystyle \frac{29\pi}{3} \approx 30.37 cubic centimeters, about 239.9 grams: \displaystyle \pi\int_0^4 \left(4 - \left(1 + \frac{x}{8}\right)^2\right)dx = \pi\left(16 - \frac{19}{3}\right), since \displaystyle \int_0^4 \left(1 + \frac{x}{8}\right)^2 dx = \frac{8}{3}\left(\frac{27}{8} - 1\right) = \frac{19}{3}.
\displaystyle \frac{256\pi}{375} \approx 2.14 cubic centimeters, about 5.36 grams: the circle \displaystyle y = \sqrt{1 - x^2} meets y = 0.6 at x = \pm 0.8, and \displaystyle \pi\int_{-0.8}^{0.8} \left(0.64 - x^2\right)dx = 2\pi\left(0.512 - \frac{0.512}{3}\right).
9\pi: \pi\int_0^3 (4 - 1)\,dx, the cylinder of radius 2 minus the cylinder of radius 1. The other integral gives \pi\int_0^3 1\,dx = 3\pi, only a third of it. \pi(R - r)^2 is the area of a disk whose radius is the ring’s width; the ring’s area is \pi(R - r)(R + r), which also grows with how far the ring is from its center.
The integral of a difference is the difference of the integrals, so \pi\int_a^b \left(R^2 - r^2\right) = \pi\int_a^b R^2 - \pi\int_a^b r^2: the disk solid of the outer boundary minus the disk solid of the inner one, which is exactly the hole. If the region reaches the axis, r = 0 and the second integral is 0.
About y = -k, R = f + k and r = g + k, so R^2 - r^2 = f^2 - g^2 + 2k(f - g). Integrating, V_k is \pi\int_a^b \left(f^2 - g^2\right)dx + 2\pi k\int_a^b (f - g)\,dx, which is V_0 + 2\pi kA. For Example 4, \displaystyle V_0 = \pi\int_0^2 \left(x^4 - 12x^2 + 16x\right)dx = \frac{32\pi}{5} and \displaystyle A = \int_0^2 \left(2x - x^2\right)dx = \frac{4}{3}, so \displaystyle V_1 = \frac{32\pi}{5} + \frac{8\pi}{3} = \frac{136\pi}{15}, as found.
Turn the region between \displaystyle y = \sqrt{a^2 - x^2} and y = c about the x-axis, where c is the hole’s radius. They meet at \displaystyle x = \pm\frac{h}{2}, so \displaystyle c^2 = a^2 - \frac{h^2}{4} and \displaystyle R^2 - r^2 = \frac{h^2}{4} - x^2, with no a left. Then \displaystyle V = \pi\int_{-h/2}^{h/2} \left(\frac{h^2}{4} - x^2\right)dx, which is \displaystyle \pi\left(\frac{h^3}{4} - \frac{h^3}{12}\right) = \frac{\pi h^3}{6}. In Exercise 17, h = 1.6, and \displaystyle \frac{\pi(1.6)^3}{6} = \frac{256\pi}{375}.