Calculus 2

The shell method

A region turned about a vertical line can be cut into thin vertical strips, and each strip sweeps out a thin tube. How do those tubes add up to a volume, and when is that easier than slicing across the axis?

Where you are 10 in Applications of integrals

Calculus8

Applications of integrals10

The shell method

Volumes with known cross sections Work as an integral Accumulation and net change3 Area between curves ▾ Show all 10 (5 more)

When you reach for it

The The disk method and the The washer method cut a solid of revolution across the axis, so that every slice is a circle or a ring. That needs the radius of each slice as a function of the variable along the axis. Turn a region about the y-axis, and the slices are horizontal: their radii have to be read off as functions of y, which means solving the boundary y = f(x) for x. For y = x(x - 3)^2 that is a cubic equation with two relevant roots at each height.

The shell method cuts the other way: into thin strips parallel to the axis. A vertical strip turned about the y-axis sweeps a thin tube, a cylindrical shell, and its size is read straight from f(x). You reach for shells when the boundary is given as y = f(x) and the axis is vertical, or as x = g(y) and the axis is horizontal, and solving for the other variable is hard or impossible.

The procedure

Day 1: shells about the y-axis

One shell. A cylindrical shell is the solid between two cylinders with the same axis and the same height h, of radii r_1 < r_2. Its volume is the outer cylinder minus the inner one, as in Volume of prisms and cylinders:

\begin{aligned} \pi r_2^2 h - \pi r_1^2 h &= \pi (r_2 + r_1)(r_2 - r_1)\,h \\ &= 2\pi \left(\frac{r_1 + r_2}{2}\right) h\,(r_2 - r_1). \end{aligned}

With \displaystyle r = \frac{r_1 + r_2}{2} the average radius and \Delta r = r_2 - r_1 the thickness, this is exactly

V_{\text{shell}} = 2\pi r\,h\,\Delta r.

The picture behind the formula: cut the shell along one side and unroll it (Figure 1). It flattens to nearly a slab whose length is the circumference 2\pi r, whose height is h and whose thickness is \Delta r, and length times height times thickness is 2\pi r h \Delta r. The unrolled slab is only an aid to memory: the outer side of the shell is slightly longer than the inner side. The algebra above is exact when r is the average radius.

Figure 1A thin shell cut along one side and flattened is nearly a slab of length 2\pi r, height h and thickness \Delta r.

Adding the shells. Let f be continuous and nonnegative on [a, b] with 0 \le a < b, and turn the region under y = f(x) and above the x-axis about the y-axis. Cut [a, b] into pieces of width \Delta x and let x_i^* be the midpoint of the i-th piece. The strip over that piece is nearly a rectangle of height f(x_i^*), and turned about the y-axis it sweeps nearly a shell of average radius x_i^*, height f(x_i^*) and thickness \Delta x (Figure 2). The shell volumes add to

\sum_{i=1}^{n} 2\pi x_i^* f(x_i^*)\,\Delta x,

a Riemann sum for 2\pi x f(x) (The definite integral as a limit of Riemann sums). As the strips get thinner it tends to the shell method:

V = \int_a^b 2\pi x\,f(x)\,dx.

Figure 2The strip at x turned about the y-axis sweeps a thin cylindrical shell of radius x and height f(x).

That this limit is the same volume the disk or washer method gives is stated here without a full proof. For a curve that can be inverted, Exercise 19 shows the two integrals agree by integration by parts; Examples 1, 3 and 5 check it on particular solids.

Radius and height in general. If the region lies between two curves, g(x) \le y \le f(x), each strip runs from the lower curve to the upper, so the shell’s height is h(x) = f(x) - g(x). Its radius is the distance from the axis to the strip, r(x) = x for the y-axis. In every case

V = \int_a^b 2\pi\,r(x)\,h(x)\,dx,

with the limits running across the region, perpendicular to the axis. The variable is the one that runs across the axis: strips parallel to a vertical axis are vertical, so they are indexed by x.

Why shells win. The disk or washer route for the same solid would need, at each height y, the inner and outer edges of the region as functions of y. When f cannot be solved for x in a usable form (a cubic, or any curve that rises and then falls, so that one height has two edges), shells avoid the problem entirely: the height of a vertical strip is f(x) as given. A second advantage is the factor x in the integrand, which can turn an integrand with no elementary antiderivative into one that a substitution handles (Example 3).

Day 2: other axes, and choosing a method

A vertical line x = k. The strips are still vertical, and only the radius changes. The radius is the distance from the strip at x to the line:

r(x) = \lvert x - k \rvert,

which is k - x when the region lies to the left of the axis and x - k when it lies to the right. The height is still top minus bottom, unaffected by where the axis is.

A horizontal axis. Strips parallel to a horizontal axis are horizontal, so they are indexed by y. A strip at height y that runs from x = q(y) on the left to x = p(y) on the right has length h(y) = p(y) - q(y), and turned about the x-axis it sweeps a shell of radius y (for y \ge 0). About the line y = k the radius is \lvert y - k \rvert. Then

V = \int_c^d 2\pi\,r(y)\,h(y)\,dy.

Choosing. The two methods give the same volume, so the choice rests on which integral is easier to set up and evaluate. The table sums up what each one needs.

Axis	Washers need	Shells need
vertical	radii in y	height in x
horizontal	radii in x	length in y
integrand	\pi(R^2 - r^2)	2\pi r h

Look at how the boundary is given and at where the boundaries change formula. A region whose top is one curve and whose bottom is another, between two values of x, has a single shell integral about a vertical axis, while its washers may need several integrals if the left and right edges change formula along y. Example 5 computes one solid both ways.

Worked examples

Example 1

Day 1: the cone, by shells

Use shells to find the volume of a right circular cone of radius \rho and height H.

Solution

Put the cone’s apex at (0, H) on the y-axis and its base in the plane y = 0. Its edge in the first quadrant is the line from (0, H) to (\rho, 0), which is \displaystyle y = H\left(1 - \frac{x}{\rho}\right), and the cone is what the triangle under that line sweeps out about the y-axis. A strip at x has radius x and height \displaystyle H\left(1 - \frac{x}{\rho}\right), so

\begin{aligned} V &= \int_0^\rho 2\pi x\,H\left(1 - \frac{x}{\rho}\right)dx \\ &= 2\pi H\left(\frac{x^2}{2} - \frac{x^3}{3\rho}\right)\Big|_0^\rho \\ &= 2\pi H\left(\frac{\rho^2}{2} - \frac{\rho^2}{3}\right) \\ &= \frac{1}{3}\pi \rho^2 H, \end{aligned}

the formula from Geometry. The thin shells near the axis are tall but small around; the wide ones near the rim are short.

Example 2

Day 1: a cubic that cannot be turned around

The region under y = x(x - 3)^2 and above the x-axis for 0 \le x \le 3 is revolved about the y-axis. Find the volume.

Solution

The curve touches the x-axis at x = 0 and x = 3, is nonnegative between, and peaks at f(1) = 4. A vertical strip at x has radius x and height x(x - 3)^2 = x^3 - 6x^2 + 9x (Figure 2), so

\begin{aligned} V &= 2\pi\int_0^3 \left(x^4 - 6x^3 + 9x^2\right)dx \\ &= 2\pi\left(\frac{x^5}{5} - \frac{3x^4}{2} + 3x^3\right)\Big|_0^3 \\ &= 2\pi\left(\frac{243}{5} - \frac{243}{2} + 81\right) \\ &= 2\pi\cdot\frac{81}{10} \\ &= \frac{81\pi}{5} \approx 50.9. \end{aligned}

A check on size: the solid fits inside the cylinder of radius 3 and height 4, of volume 36\pi \approx 113.

Washers would need the region’s edges at each height. At y = 2 alone, the equation x^3 - 6x^2 + 9x - 2 = 0 factors as (x - 2)\left(x^2 - 4x + 1\right) = 0, so the washer runs from radius \displaystyle 2 - \sqrt{3} to radius 2. At a general height y the edges are roots of a cubic with no such factor; writing them out takes the cubic formula, and a washer integral built from those roots is not practical by hand. The shell integral never asks for them.

Example 3

Day 1: a bump whose area has no formula

The region under y = e^{-x^2} and above the x-axis for 0 \le x \le 1 is revolved about the y-axis. Find the volume.

Solution

Strips at x have radius x and height e^{-x^2}, so

V = \int_0^1 2\pi x\,e^{-x^2}\,dx.

The factor x is what a substitution needs. With u = -x^2, du = -2x\,dx, and the limits 0 and 1 become 0 and -1:

\begin{aligned} V &= -\pi\int_0^{-1} e^{u}\,du = \pi\int_{-1}^{0} e^{u}\,du \\ &= \pi\left(1 - e^{-1}\right) \approx 1.99. \end{aligned}

The region’s area, \int_0^1 e^{-x^2}\,dx, and its volume about the x-axis, \pi\int_0^1 e^{-2x^2}\,dx, have integrands with no elementary antiderivative, so their values are found numerically. About the y-axis the volume is exact. Disks in y also reach it, but with more work: for 0 \le y \le e^{-1} the slice is a disk of radius 1, and above that its radius is \displaystyle \sqrt{-\ln y}, so

\begin{aligned} V &= \pi e^{-1} + \pi\int_{1/e}^{1} (-\ln y)\,dy \\ &= \frac{\pi}{e} + \pi\left(1 - \frac{2}{e}\right), \end{aligned}

the same number after a split and an integration by parts.

Example 4

Day 1: Ozoda’s bell

Ozoda, at a bell foundry, models a simple bell in centimeters. Its outer surface is \displaystyle y = 40 - \frac{x^2}{40} for 0 \le x \le 40 turned about the y-axis, its inner surface is \displaystyle y = 35 - \frac{x^2}{40} turned the same way, and its rim is cut off by the plane y = 0. Find the volume of bronze, and its mass if the bronze has a density of 8.7 grams per cubic centimeter.

Solution

The bronze is the solid swept by the region between the two curves and above the x-axis. A vertical strip at x runs from the inner curve up to the outer one while the inner curve is above the axis, that is for \displaystyle 35 - \frac{x^2}{40} \ge 0, or \displaystyle x \le \sqrt{1400} = 10\sqrt{14} \approx 37.4. There the height is the constant gap

h(x) = \left(40 - \frac{x^2}{40}\right) - \left(35 - \frac{x^2}{40}\right) = 5.

Beyond \displaystyle x = 10\sqrt{14} the strip runs from the axis up to the outer curve, so \displaystyle h(x) = 40 - \frac{x^2}{40}. The height changes formula, so the integral splits:

\begin{aligned} V &= 2\pi\int_0^{10\sqrt{14}} 5x\,dx \\ &\quad + 2\pi\int_{10\sqrt{14}}^{40} x\left(40 - \frac{x^2}{40}\right)dx. \end{aligned}

The first piece is \displaystyle 2\pi\cdot\frac{5}{2}\cdot 1400 = 7000\pi. In the second, \displaystyle u = 40 - \frac{x^2}{40} gives \displaystyle du = -\frac{x}{20}\,dx, so x\,dx = -20\,du, and the limits become u = 5 and u = 0:

2\pi\int_{0}^{5} 20u\,du = 2\pi\cdot 20\cdot\frac{25}{2} = 500\pi.

So V = 7500\pi \approx 23{,}562 cubic centimeters, about 23.6 liters, and the mass is 8.7 \times 7500\pi \approx 204{,}990 grams, about 205 kilograms.

A check: the bell is the solid dome under the outer surface minus the dome under the inner one. By shells, the outer dome is \displaystyle 2\pi\int_0^{40} x\left(40 - \frac{x^2}{40}\right)dx = 32{,}000\pi and the inner one is \displaystyle 2\pi\int_0^{10\sqrt{14}} x\left(35 - \frac{x^2}{40}\right)dx = 24{,}500\pi, and their difference is 7500\pi.

Example 5

Day 2: one solid, both ways

The region between \displaystyle y = \sqrt{x} and y = x^2 is revolved about the line x = 2. Find the volume with shells, then with washers.

Solution

The curves meet where \displaystyle \sqrt{x} = x^2, at x = 0 and x = 1, and \displaystyle \sqrt{x} \ge x^2 between. The whole region lies to the left of the axis x = 2 (Figure 3).

Shells. A vertical strip at x is parallel to the axis. Its radius is the distance to the axis, r(x) = 2 - x, and its height is \displaystyle h(x) = \sqrt{x} - x^2:

\begin{aligned} V &= 2\pi\int_0^1 (2 - x)\left(x^{1/2} - x^2\right)dx \\ &= 2\pi\int_0^1 \left(2x^{1/2} - 2x^2 - x^{3/2} + x^3\right)dx \\ &= 2\pi\left(\frac{4}{3} - \frac{2}{3} - \frac{2}{5} + \frac{1}{4}\right) \\ &= 2\pi\cdot\frac{31}{60} = \frac{31\pi}{30}. \end{aligned}

Washers. The axis is vertical, so washers are horizontal and the variable is y, from 0 to 1. Solving, \displaystyle y = \sqrt{x} is x = y^2 and y = x^2 is \displaystyle x = \sqrt{y}; at height y the region runs from x = y^2 on the left to \displaystyle x = \sqrt{y} on the right. The far edge from the axis is the left one, so R(y) = 2 - y^2 and \displaystyle r(y) = 2 - \sqrt{y}:

\begin{aligned} V &= \pi\int_0^1 \left[\left(2 - y^2\right)^2 - \left(2 - \sqrt{y}\right)^2\right]dy \\ &= \pi\int_0^1 \left(-4y^2 + y^4 + 4\sqrt{y} - y\right)dy \\ &= \pi\left(-\frac{4}{3} + \frac{1}{5} + \frac{8}{3} - \frac{1}{2}\right) \\ &= \frac{31\pi}{30} \approx 3.25. \end{aligned}

Both routes work here because both curves invert easily. Shells used the curves as given; washers needed both inverted and a careful choice of which edge is outer.

Figure 3The region of Example 5. Each vertical strip is parallel to the axis x = 2 and sweeps a shell of radius 2 - x.

Example 6

Day 2: shells in y about the x-axis

The region between the curve x = 3y - y^2 and the y-axis is revolved about the x-axis. Find the volume.

Solution

The curve meets the y-axis where 3y - y^2 = 0, at y = 0 and y = 3, and 3y - y^2 > 0 between, so the region lies to the right of the y-axis for 0 \le y \le 3. The axis is horizontal, so strips parallel to it are horizontal. The strip at height y runs from x = 0 to x = 3y - y^2 and sits at distance y from the axis (Figure 4):

\begin{aligned} V &= 2\pi\int_0^3 y\left(3y - y^2\right)dy \\ &= 2\pi\left(y^3 - \frac{y^4}{4}\right)\Big|_0^3 \\ &= 2\pi\left(27 - \frac{81}{4}\right) = \frac{27\pi}{2} \approx 42.4. \end{aligned}

Washers here would be vertical, in x from 0 to \displaystyle \frac{9}{4}, and need the curve solved for y: \displaystyle y = \frac{3 \pm \sqrt{9 - 4x}}{2}, the larger root the outer radius and the smaller the inner. Then \displaystyle R^2 - r^2 = (R + r)(R - r) = 3\sqrt{9 - 4x}, and \displaystyle 3\pi\int_0^{9/4}\sqrt{9 - 4x}\,dx = 3\pi\cdot\frac{27}{6} = \frac{27\pi}{2}. The same answer, after the quadratic formula and a factoring step that shells never needed.

Figure 4The region of Example 6 turned about the x-axis. Horizontal strips are parallel to the axis, so the shells have radius y and length 3y - y^2.

Boundaries

The height h must be continuous on the closed interval, or continuous on pieces with the integral split at each place the top or bottom changes formula. An unbounded interval or a vertical asymptote turns the integral into an improper integral, which needs a limit.
The region must lie on one side of the axis. The radius is a distance, \lvert x - k \rvert, and if the region straddles the axis the two sides sweep overlapping solids. Writing 2\pi x with x negative on part of the interval subtracts volume rather than adding it (Exercise 20).
The strips are parallel to the axis, which fixes the variable: x for a vertical axis, y for a horizontal one. A strip perpendicular to the axis sweeps a disk or a washer, not a shell.
The height is top minus bottom (or right minus left) and does not depend on where the axis is; the radius is the distance to the axis and does.
The formula 2\pi r h \Delta r is exact for a shell when r is the average radius. The shell method’s integral rests on that, and the equality of its value with the disk or washer value is stated here, with a proof only for invertible boundaries (Exercise 19).

Where this goes next

Shells and washers are two ways to slice one solid, and having both is what lets the integral fit the way the boundary is given. The same habit of building a total from thin pieces of known size already gave the length of a curve and the area of the band it sweeps out. It returns in Work as an integral, where a tank is pumped out slab by slab, and the factor “distance from the axis times a small piece” reappears in moments and the Centroid.

Exercises

Day 1: about the y-axis

The region under y = x^3 + 1 and above the x-axis for 0 \le x \le 1 is revolved about the y-axis. Find the volume.
The region under \displaystyle y = \sqrt{x} and above the x-axis for 0 \le x \le 9 is revolved about the y-axis. Find the volume.
The region between y = 2x - x^2 and the x-axis is revolved about the y-axis. Find the volume.
The region under y = \sin\left(x^2\right) and above the x-axis for \displaystyle 0 \le x \le \sqrt{\pi} is revolved about the y-axis. Find the volume.
The region under \displaystyle y = \frac{1}{x^2} and above the x-axis for 1 \le x \le 3 is revolved about the y-axis. Find the volume.
The region under y = \cos x and above the x-axis for \displaystyle 0 \le x \le \frac{\pi}{2} is revolved about the y-axis. Find the volume.
The region between y = x and y = x^2 is revolved about the y-axis. Find the volume.
The region in the first quadrant between y = 2x and y = x^3 is revolved about the y-axis. Find the volume.
Jahongir casts a hand bell whose outer surface is \displaystyle y = 9 - \frac{x^2}{9} for 0 \le x \le 9 and whose inner surface is \displaystyle y = 8 - \frac{x^2}{9}, both turned about the y-axis and cut off by the plane y = 0, in centimeters. Find the volume of bronze and its mass at 8.7 grams per cubic centimeter.

Day 2: other axes

The region under y = x^2 and above the x-axis for 0 \le x \le 1 is revolved about the line x = -1. Find the volume.
The region of Exercise 7 is revolved about the line x = 3. Find the volume.
The region between \displaystyle y = \sqrt{x} and \displaystyle y = \frac{x}{2} is revolved about the line x = -1. Find the volume.
The region under y = e^x and above the x-axis for 0 \le x \le 1 is revolved about the line x = 1. Find the volume.
The region in the first quadrant bounded by x = 2 - y^2 and the two axes is revolved about the x-axis. Find the volume with shells in y, and check it with disks in x.
The region bounded by y = \ln x, the x-axis and the line x = e is revolved about the line y = -1. Find the volume with shells in y, and check it with washers in x.
Behruz cuts a sand core for a mold by turning the triangle with vertices (0, 0), (2, 0) and (2, 1) about the line y = 2. Find the volume with shells, and check it with washers.

Day 2: choosing a method

The region bounded by \displaystyle y = \sqrt{x - 1}, the x-axis and the line x = 5 is revolved about the y-axis. Find the volume both with shells and with washers, and say which integral took less work.

Reasoning

Show that a shell of inner radius r_1, outer radius r_2 and height h has volume exactly 2\pi\bar{r}h\,\Delta r, where \bar{r} is the average radius and \Delta r = r_2 - r_1. Why is “circumference times height times thickness” only approximate if the inner or outer circumference is used instead?
Let f be differentiable and decreasing on [0, b] with f(b) = 0, and let x = g(y) be its inverse. Use integration by parts to show \int_0^b 2\pi x f(x)\,dx = -\pi\int_0^b x^2 f'(x)\,dx, and explain why the substitution y = f(x) turns the right side into the disk integral \pi\int_0^{f(0)} [g(y)]^2\,dy.
The region under y = 1 - x^2 and above the x-axis is revolved about the y-axis. Explain why \int_{-1}^{1} 2\pi x\left(1 - x^2\right)dx = 0 is not its volume, and find the volume.
A region lies between y = g(x) and y = f(x) for 0 \le a \le x \le b, with g \le f, and has area A. Let V_0 be its volume about the y-axis and V_k its volume about the line x = -k, with k > 0. Show that V_k = V_0 + 2\pi kA, and check it against Exercise 10.

Answers

\displaystyle \frac{7\pi}{5}: \displaystyle 2\pi\int_0^1 \left(x^4 + x\right)dx = 2\pi\left(\frac{1}{5} + \frac{1}{2}\right).
\displaystyle \frac{972\pi}{5} \approx 610.7: \displaystyle 2\pi\int_0^9 x^{3/2}\,dx = 2\pi\cdot\frac{2}{5}\cdot 9^{5/2} = 2\pi\cdot\frac{486}{5}.
\displaystyle \frac{8\pi}{3}: the curve meets the axis at x = 0 and x = 2, and \displaystyle 2\pi\int_0^2 \left(2x^2 - x^3\right)dx = 2\pi\left(\frac{16}{3} - 4\right).
2\pi: \sin\left(x^2\right) \ge 0 for 0 \le x^2 \le \pi. With u = x^2, du = 2x\,dx, the integral \displaystyle 2\pi\int_0^{\sqrt{\pi}} x\sin\left(x^2\right)dx becomes \pi\int_0^{\pi}\sin u\,du, which is \pi(1 + 1).
2\pi\ln 3 \approx 6.90: the radius cancels one power of x, so \displaystyle 2\pi\int_1^3 x\cdot\frac{1}{x^2}\,dx = 2\pi\int_1^3 \frac{1}{x}\,dx = 2\pi\left(\ln 3 - \ln 1\right).
\pi^2 - 2\pi \approx 3.59: by parts, \int x\cos x\,dx = x\sin x + \cos x, so \displaystyle 2\pi\left(x\sin x + \cos x\right)\Big|_0^{\pi/2} = 2\pi\left(\frac{\pi}{2} - 1\right).
\displaystyle \frac{\pi}{6}: they meet at x = 0 and x = 1 with x \ge x^2 between, and \displaystyle 2\pi\int_0^1 \left(x^2 - x^3\right)dx = 2\pi\left(\frac{1}{3} - \frac{1}{4}\right).
\displaystyle \frac{16\sqrt{2}\,\pi}{15} \approx 4.74: 2x = x^3 at x = 0 and \displaystyle x = \sqrt{2} in the first quadrant, and \displaystyle 2\pi\int_0^{\sqrt{2}} \left(2x^2 - x^4\right)dx = 2\pi\left(\frac{4\sqrt{2}}{3} - \frac{4\sqrt{2}}{5}\right).
\displaystyle \frac{153\pi}{2} \approx 240.3 cubic centimeters, mass \displaystyle 8.7 \cdot \frac{153\pi}{2} \approx 2091 grams: the inner curve meets y = 0 at \displaystyle x = \sqrt{72} = 6\sqrt{2}. The height is 1 for \displaystyle 0 \le x \le 6\sqrt{2}, giving \displaystyle 2\pi\cdot\frac{72}{2} = 72\pi, and \displaystyle 9 - \frac{x^2}{9} beyond, where \displaystyle u = 9 - \frac{x^2}{9}, \displaystyle x\,dx = -\frac{9}{2}\,du gives \displaystyle 2\pi\int_0^1 \frac{9}{2}u\,du = \frac{9\pi}{2}.
\displaystyle \frac{7\pi}{6}: r(x) = x + 1, and \displaystyle 2\pi\int_0^1 (x + 1)x^2\,dx = 2\pi\left(\frac{1}{4} + \frac{1}{3}\right).
\displaystyle \frac{5\pi}{6}: r(x) = 3 - x, and 2\pi\int_0^1 (3 - x)\left(x - x^2\right)dx is 2\pi\int_0^1 \left(3x - 4x^2 + x^3\right)dx, which is \displaystyle 2\pi\cdot\frac{5}{12}.
\displaystyle \frac{104\pi}{15} \approx 21.8: \displaystyle \sqrt{x} = \frac{x}{2} at x = 0 and x = 4, with \displaystyle \sqrt{x} on top. \displaystyle 2\pi\int_0^4 (x + 1)\left(\sqrt{x} - \frac{x}{2}\right)dx is \displaystyle 2\pi\left(\frac{64}{5} + \frac{16}{3} - \frac{32}{3} - 4\right), which is \displaystyle 2\pi\cdot\frac{52}{15}.
2\pi(e - 2) \approx 4.51: r(x) = 1 - x, and by parts \int (1 - x)e^x\,dx = (2 - x)e^x, so 2\pi\left[(2 - x)e^x\right]_0^1 = 2\pi(e - 2).
2\pi: the curve meets the x-axis at x = 2 and the y-axis at \displaystyle y = \sqrt{2}. Shells: \displaystyle 2\pi\int_0^{\sqrt{2}} y\left(2 - y^2\right)dy = 2\pi(2 - 1). Disks: \displaystyle y = \sqrt{2 - x}, so \pi\int_0^2 (2 - x)\,dx = 2\pi.
\pi e \approx 8.54: for 0 \le y \le 1 the strip runs from x = e^y to x = e at radius y + 1, so \displaystyle 2\pi\int_0^1 (y + 1)\left(e - e^y\right)dy = 2\pi\left(\frac{3e}{2} - e\right), using \int_0^1 (y + 1)e^y\,dy = \left[ye^y\right]_0^1 = e. Washers: R = \ln x + 1, r = 1, and \pi\int_1^e \left(\ln^2 x + 2\ln x\right)dx = \pi\left((e - 2) + 2\right).
\displaystyle \frac{10\pi}{3}: the hypotenuse is x = 2y. Shells: the strip at height y runs from x = 2y to x = 2, at radius 2 - y, so \displaystyle 2\pi\int_0^1 (2 - y)(2 - 2y)\,dy = 2\pi\left(4 - 3 + \frac{2}{3}\right). Washers: R = 2, \displaystyle r = 2 - \frac{x}{2}, and \displaystyle \pi\int_0^2 \left[4 - \left(2 - \frac{x}{2}\right)^2\right]dx = \pi\left(8 - \frac{14}{3}\right).
\displaystyle \frac{544\pi}{15} \approx 113.9. Shells: \displaystyle 2\pi\int_1^5 x\sqrt{x - 1}\,dx with u = x - 1 is \displaystyle 2\pi\int_0^4 (u + 1)\sqrt{u}\,du = 2\pi\left(\frac{64}{5} + \frac{16}{3}\right). Washers: y runs from 0 to 2, R = 5, r = y^2 + 1, and \displaystyle \pi\int_0^2 \left[25 - \left(y^2 + 1\right)^2\right]dy = \pi\left(50 - \frac{206}{15}\right). The washer integral is a polynomial and needs no substitution, so it is a little less work here; both are short.
The volume is \pi r_2^2 h - \pi r_1^2 h, which factors as \pi(r_2 + r_1)(r_2 - r_1)h, and that is \displaystyle 2\pi\cdot\frac{r_1 + r_2}{2}\cdot h\,\Delta r with nothing dropped. Using r_1 gives 2\pi r_1 h\Delta r, short by \pi h(\Delta r)^2, and using r_2 is over by the same amount; that error is of second order in \Delta r, so it vanishes in the limit, but it is there for any actual shell.
With u = f(x) and dv = 2\pi x\,dx, v = \pi x^2, so \int_0^b 2\pi x f(x)\,dx equals \left[\pi x^2 f(x)\right]_0^b - \pi\int_0^b x^2 f'(x)\,dx. The bracket is 0 because f(b) = 0 and the lower end has x = 0. Now set y = f(x), dy = f'(x)\,dx, x = g(y): as x runs from 0 to b, y runs from f(0) down to 0, so -\pi\int_0^b x^2 f'(x)\,dx becomes -\pi\int_{f(0)}^{0} [g(y)]^2\,dy, which is \pi\int_0^{f(0)} [g(y)]^2\,dy, the disks of radius g(y).
\displaystyle \frac{\pi}{2}. The integrand is odd, so the integral is 0; on [-1, 0] the factor x is negative, while a radius is a distance |x|. The left half sweeps exactly the same solid as the right half, so the volume is that of the right half alone: \displaystyle 2\pi\int_0^1 x\left(1 - x^2\right)dx = 2\pi\left(\frac{1}{2} - \frac{1}{4}\right) = \frac{\pi}{2}.
About x = -k the radius is x + k, so V_k = 2\pi\int_a^b (x + k)\left(f - g\right)dx. Splitting the factor x + k, this is 2\pi\int_a^b x(f - g)\,dx + 2\pi k\int_a^b (f - g)\,dx, which is V_0 + 2\pi kA. For Exercise 10, \displaystyle V_0 = 2\pi\int_0^1 x^3\,dx = \frac{\pi}{2} and \displaystyle A = \frac{1}{3}, so \displaystyle V_1 = \frac{\pi}{2} + \frac{2\pi}{3} = \frac{7\pi}{6}.