Volumes The Shell Method

The shell method slices a region into thin vertical rectangles and revolves each one about a vertical axis. Instead of a disk, each rectangle sweeps out a hollow cylindrical shell, and unrolling that shell into a flat slab shows that its volume is circumference times height times thickness: $d V = 2 π x f (x) d x$ .

Item	Statement
Shell volume element	$d V = \underset{circumference}{\underset{⏟}{2 π x}} \underset{height}{\underset{⏟}{f (x)}} \underset{thickness}{\underset{⏟}{d x}}$
Volume about the $y$ -axis	$V = \int_{a}^{b} 2 π x f (x) d x$
Region between two curves	$V = 2 π \int_{a}^{b} x [f (x) - g (x)] d x$
General form	$V = \int_{a}^{b} 2 π ρ d A$ , with $ρ$ the shell radius and $d A$ the area of the thin rectangle
Axis $x = c$ , region to the right	$ρ = x - c$
Axis $x = c$ , region to the left	$ρ = c - x$
Variable of integration	$x$ for a vertical axis, the same variable that measures the width of the region
Sphere of radius $a$	$\frac{4}{3} π a^{3}$
Ellipsoid from $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ about the $y$ -axis	$\frac{4}{3} π a^{2} b$
Torus, tube radius $r$ , centre at distance $a$	$2 π^{2} r^{2} a$
Pappus's theorem	$V = (area of the region) \times (distance travelled by its centroid)$
Prefer shells when	solving $y = f (x)$ for $x$ is painful, or washers would need two or more integrals

Building a Cylindrical Shell

Let $R$ be the region under the curve $y = f (x)$ between $x = a$ and $x = b$ , where $0 \leq a < b$ .

The region R shaded under the curve y = f(x) and above the x-axis, between the vertical lines x = a and x = b. — The region $R$ under the curve $y = f (x)$ between $x = a$ and $x = b$ .

In Section 8.2 we revolved such a region about the $x$ -axis and computed the volume with disks. Revolving the same region about the $y$ -axis produces a completely different solid, and disks are no longer the natural slice.

The solid swept out when the region R makes a full turn about the vertical y-axis, hollow along the axis. — The region $R$ revolved about the $y$ -axis.

To compute the volume of this new solid, consider a thin vertical rectangle of height $f (x)$ and width $d x$ , standing at distance $x$ from the $y$ -axis.

When this rectangle is revolved about the $y$ -axis, it generates a hollow, thin-walled shell of radius $x$ , height $f (x)$ and thickness $d x$ .

A thin vertical rectangle of height f(x) and width dx sweeping out a hollow cylindrical shell of radius x as it turns about the y-axis. — The revolved rectangle sweeps out a hollow shell of radius $x$ , height $f (x)$ and thickness $d x$ .

If the cylindrical shell is rolled out flat, like a thin sheet of tin cut down one vertical side, it becomes a thin slab of height $f (x)$ , thickness $d x$ , and length $2 π x$ , which is the circumference of the shell. This is the picture worth remembering:

The cylindrical shell cut down one side and unrolled into a flat slab of length 2 pi x, height f(x) and thickness dx. — Unrolled flat, the shell becomes a slab of length $2 π x$ , height $f (x)$ and thickness $d x$ .

The slab has volume length times height times thickness, so the element of volume is

\bbox[8px, #E6F0FA, border: 3px solid #0066CC]{dV=\underbrace{2\pi x}_{\text{circumference}}\underbrace{f(x)}_{\text{height}}\underbrace{dx}_{\text{thickness}}.}

The total volume is then obtained by adding up the nested infinitesimal shells, one for each $x$ between $a$ and $b$ :

V = \int_{a}^{b} d V = \int_{a}^{b} 2 π x f (x) d x .

In principle the volume of this solid can also be obtained with thin disks generated by revolving infinitesimally thin horizontal rectangles. In practice that is often harder, because (1) the equation $y = f (x)$ has to be solved for $x$ in terms of $y$ , and (2) the formula for the length of the horizontal rectangle may change partway up the region. In such cases we have to compute more than one integral.
Notice that the shell method integrates with respect to $x$ even though the axis of revolution is vertical. That is exactly the reverse of the washer method, and it is the whole reason the shell method exists.

Regions Between Two Curves

If the region lies between two curves $y = f (x)$ and $y = g (x)$ with $f (x) \geq g (x)$ , then the height of the vertical rectangle, which is also the height of the cylindrical shell, is $f (x) - g (x)$ .

A region bounded above by y = f(x) and below by y = g(x), with a vertical rectangle of height f(x) minus g(x) standing at distance x from the y-axis. — When the region lies between two curves, the shell height is $f (x) - g (x)$ .

Therefore, in this case

\bbox[8px, #E6F0FA, border: 3px solid #0066CC]{dV=2\pi x\left[f(x)-g(x)\right]dx}

and

V = 2 π \int_{a}^{b} x [f (x) - g (x)] d x .

In general we can write

\bbox[8px, #E6F0FA, border: 3px solid #0066CC]{V=\int_{a}^{b}2\pi(\text{shell radius})(\text{area of thin rectangle})=\int_{a}^{b}2\pi\rho\,dA}

where $ρ$ is the distance from the thin rectangle to the axis of revolution and $d A$ is the area of that rectangle. Writing the formula this way makes it easy to adapt when the axis is a vertical line other than the $y$ -axis: if the axis is $x = c$ and the region lies to the right of it, then $ρ = x - c$ ; if the region lies to the left of it, then $ρ = c - x$ . In both cases $ρ$ must be positive throughout the region, because it is a radius.

Shell or Washer? How to Choose

Both methods compute the same number, so the choice is purely a matter of which integral you would rather evaluate. The following comparison covers almost every case you will meet.

Question	Washer (Section 8.2)	Shell
Slice orientation	Perpendicular to the axis	Parallel to the axis
Axis vertical (a line $x = c$ )	Integrate in $y$	Integrate in $x$
Axis horizontal (a line $y = k$ )	Integrate in $x$	Integrate in $y$
What you must solve for	The boundary curves as functions of the integration variable	Usually nothing, the curves are already given in that form
Volume element	$d V = π (R^{2} - r^{2}) d y$	$d V = 2 π ρ d A$

Reach for shells when any of the following is true.

The region is described by $y = f (x)$ and solving for $x$ in terms of $y$ is painful or impossible. For example $y = x e^{x}$ or $y = x + \sin x$ cannot be inverted in closed form at all, so washers about the $y$ -axis are simply unavailable, while shells give $\int 2 π x f (x) d x$ immediately.
The region's horizontal cross sections change formula partway up, so the washer setup would have to be split into two or more integrals. A vertical strip, by contrast, runs from one single lower curve to one single upper curve for every $x$ .
The region touches the axis of revolution only at a point, or sits away from it, so that the washer picture would require you to track an inner radius that appears and disappears.

Reach for washers when any of the following is true.

The axis is vertical and the region is already described conveniently by $x = h (y)$ .
The shell integrand $2 π x f (x)$ is harder than the washer integrand. Multiplying by the radius $x$ raises the degree of a polynomial by one and can turn a friendly integral into an integration by parts.
The solid is a plain disk or washer of revolution and one integral in $y$ finishes the job.

A practical routine: sketch the region, draw the axis, then draw one rectangle parallel to the axis and one perpendicular to it. Whichever rectangle has endpoints that are described by a single pair of curves across the whole region is the rectangle to use. If both work, set up both integrals symbolically and evaluate the shorter one. Exercise 9 below asks you to do exactly this.

Worked Examples

The region between the two curves $y = 4 - x^{2}$ and $y = x^{2} - 4$ for $x \geq 0$ is revolved about the $y$ -axis. Find the volume of the resulting solid by the shell method.

The region for x at least zero caught between the downward parabola y = 4 minus x squared and the upward parabola y = x squared minus 4, with the typical vertical rectangle MN drawn inside it. — The region between $y = 4 - x^{2}$ and $y = x^{2} - 4$ for $x \geq 0$ , with the typical rectangle $M N$ .

Solution

Let $y_{1} = 4 - x^{2}$ and $y_{2} = x^{2} - 4$ . The two curves meet where $4 - x^{2} = x^{2} - 4$ , that is at $x = \pm 2$ , so for $x \geq 0$ the region runs from $x = 0$ to $x = 2$ . The height of the typical shell $M N$ is

\begin{aligned} MN &= y_{1}-y_{2}\\ &= \left(4-x^{2}\right)-\left(x^{2}-4\right)\\ &= 8-2x^{2}. \end{aligned}

So the volume of the shell shown in the figure is

d V = 2 π ρ d A,

where

d A = M N d x = (8 - 2 x^{2}) d x

and

ρ = the distance of the thin rectangle from the axis of rotation = x .

Therefore

d V = 2 π x (8 - 2 x^{2}) d x

and the total volume is

\begin{aligned} V &= \int_{0}^{2}dV=\int_{0}^{2}2\pi x\left(8-2x^{2}\right)dx\\ &= 4\pi\int_{0}^{2}\left(4x-x^{3}\right)dx\\ &= 4\pi\left[2x^{2}-\frac{1}{4}x^{4}\right]_{0}^{2}\\ &= 4\pi\left(8-\frac{16}{4}\right)=16\pi. \end{aligned}

The region inside the circle $x^{2} + y^{2} = a^{2}$ is revolved about the $y$ -axis. Find the volume of the resulting solid, which is a sphere, by the shell method.

The disk bounded by the circle x squared plus y squared equals a squared, with a thin vertical rectangle at distance x from the y-axis reaching from the lower semicircle to the upper one. — The region inside the circle $x^{2} + y^{2} = a^{2}$ and the thin rectangle that generates a shell.

Solution

We write the equation of the circle as

y_{1} = \sqrt{a^{2} - x^{2}} and y_{2} = - \sqrt{a^{2} - x^{2}} .

Then

\begin{aligned} dA &= \text{area of the thin rectangle}\\ &= (y_{1}-y_{2})\,dx\\ &= 2\sqrt{a^{2}-x^{2}}\,dx \end{aligned}

and the volume of the shell is $d V = 2 π ρ d A$ , where

ρ = distance of the thin rectangle from the axis of rotation = x .

Therefore

d V = 2 π x (2 \sqrt{a^{2} - x^{2}}) d x .

Because $x$ , the distance of the thin rectangle from the $y$ -axis, varies between $0$ and $a$ , the total volume is

\begin{aligned} V &= \int_{0}^{a}dV\\ &= \int_{0}^{a}2\pi x\left(2\sqrt{a^{2}-x^{2}}\right)dx\\ &= 2\pi\int_{0}^{a}2x\sqrt{a^{2}-x^{2}}\,dx. \end{aligned}

Let $u = a^{2} - x^{2}$ . Then $d u = - 2 x d x$ , so $2 x d x = - d u$ . When $x = 0$ we have $u = a^{2}$ , and when $x = a$ we have $u = 0$ . Therefore

\begin{aligned} V &= -2\pi\int_{a^{2}}^{0}\sqrt{u}\,du\\ &= 2\pi\int_{0}^{a^{2}}\sqrt{u}\,du\\ &= 2\pi\left[\frac{2}{3}u^{\frac{3}{2}}\right]_{0}^{a^{2}}\\ &= \frac{4\pi}{3}\left(a^{2}\right)^{\frac{3}{2}}=\frac{4}{3}\pi a^{3}, \end{aligned}

which is the familiar formula for the volume of a sphere of radius $a$ . Note how the minus sign from $d u$ was absorbed: reversing the order of the limits of integration flips the sign back.

The region inside the curve

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

is revolved about the $y$ -axis. Find the volume of the resulting ellipsoid.

Solution

The ellipsoid obtained by revolving the region inside the ellipse about the y-axis, with a cylindrical shell of radius x standing inside it. — The ellipsoid generated by revolving the region inside the ellipse about the $y$ -axis.

If we solve the equation of the ellipse for $y$ , we get

y_{1} = b \sqrt{1 - \frac{x^{2}}{a^{2}}}, y_{2} = - b \sqrt{1 - \frac{x^{2}}{a^{2}}} .

Then

\begin{aligned} dA &= \text{area of the thin rectangle}\\ &= (y_{1}-y_{2})\,dx\\ &= 2b\sqrt{1-\frac{x^{2}}{a^{2}}}\,dx \end{aligned}

and the volume of the shell is

\begin{aligned} dV &= 2\pi x\,dA\\ &= 2\pi x\left(2b\sqrt{1-\frac{x^{2}}{a^{2}}}\right)dx. \end{aligned}

Because $x$ , the distance between the thin rectangle and the $y$ -axis, varies between $0$ and $a$ , the total volume is

V = \int_{0}^{a} d V = \int_{0}^{a} 4 π b x \sqrt{1 - \frac{x^{2}}{a^{2}}} d x .

Let $u = 1 - \frac{x^{2}}{a^{2}}$ . Then

d u = - \frac{2}{a^{2}} x d x, so x d x = - \frac{a^{2}}{2} d u .

We have $u = 1$ when $x = 0$ , and $u = 0$ when $x = a$ . Therefore

\begin{aligned} V &= 4\pi b\int_{1}^{0}\sqrt{u}\underbrace{\left(-\frac{a^{2}}{2}\,du\right)}_{x\,dx}\\ &= -2\pi a^{2}b\left[\frac{2}{3}u^{\frac{3}{2}}\right]_{1}^{0}\\ &= -2\pi a^{2}b\left(0-\frac{2}{3}\right)\\ &= \frac{4}{3}\pi a^{2}b. \end{aligned}

Two sign changes occur here and they cancel: the substitution contributes the factor $- \frac{a^{2}}{2}$ , and the bracket ${[\frac{2}{3} u^{3 / 2}]}_{1}^{0} = - \frac{2}{3}$ is itself negative because the limits run downward. If you prefer to avoid negative-looking intermediate steps, flip the limits first:

V = 4 π b \cdot \frac{a^{2}}{2} \int_{0}^{1} \sqrt{u} d u = 2 π a^{2} b \cdot \frac{2}{3} = \frac{4}{3} π a^{2} b .

As a check, setting $a = b$ gives $\frac{4}{3} π a^{3}$ , the volume of a sphere, which agrees with the previous example.

Given that a disk with radius $1$ and centre $(5, 0)$ is revolved about the $y$ -axis, compute the volume of the resulting doughnut-shaped solid.

The disk of radius 1 centred at the point (5,0) together with the doughnut-shaped torus it sweeps out when revolved about the y-axis. — The disk of radius $1$ centred at $(5, 0)$ and the torus it generates.

Solution

The equation of a circle with radius $1$ and centre $(5, 0)$ is

(x - 5)^{2} + y^{2} = 1.

This equation gives us two functions

y_{1} = \sqrt{1 - (x - 5)^{2}} and y_{2} = - \sqrt{1 - (x - 5)^{2}} .

Therefore, the height of the typical thin rectangle is

\begin{aligned} MN &= y_{1}-y_{2}\\ &= \sqrt{1-(x-5)^{2}}-\left(-\sqrt{1-(x-5)^{2}}\right)\\ &= 2\sqrt{1-(x-5)^{2}}, \end{aligned}

the area of a thin rectangle is

\begin{aligned} dA &= (MN)\,dx\\ &= 2\sqrt{1-(x-5)^{2}}\,dx, \end{aligned}

and the volume of the shell is

\begin{aligned} dV &= 2\pi\rho\,dA\\ &= 2\pi x\left(2\sqrt{1-(x-5)^{2}}\right)dx\\ &= 4\pi x\sqrt{1-(x-5)^{2}}\,dx. \end{aligned}

Because $ρ = x$ , the distance between the thin rectangle and the axis, varies between $4$ and $6$ , the total volume is

V = \int_{4}^{6} d V = \int_{4}^{6} 4 π x \sqrt{1 - (x - 5)^{2}} d x .

Let $u = x - 5$ , so that $x = u + 5$ and $d u = d x$ . The limits $x = 4$ and $x = 6$ become $u = - 1$ and $u = 1$ :

\begin{aligned} V &= \int_{-1}^{1}4\pi(u+5)\sqrt{1-u^{2}}\,du\\ &= 4\pi\int_{-1}^{1}u\sqrt{1-u^{2}}\,du+20\pi\int_{-1}^{1}\sqrt{1-u^{2}}\,du. \end{aligned}

Because the function $f (u) = u \sqrt{1 - u^{2}}$ is odd, that is

f (- u) = (- u) \sqrt{1 - (- u)^{2}} = - u \sqrt{1 - u^{2}} = - f (u),

and the interval $[- 1, 1]$ is symmetric about the origin, we know

\int_{- 1}^{1} u \sqrt{1 - u^{2}} d u = 0.

On the other hand, $y = \sqrt{1 - u^{2}}$ is the upper half of the unit circle $u^{2} + y^{2} = 1$ , so $\int_{- 1}^{1} \sqrt{1 - u^{2}} d u$ is exactly the area of a semicircle of radius $1$ , as the figure below shows.

The upper half of the unit circle drawn over the interval from u equals minus 1 to u equals 1, with the semicircular region beneath it shaded. — The integral $\int_{- 1}^{1} \sqrt{1 - u^{2}} d u$ is equal to the area of a semicircle of radius 1.

Therefore

\int_{- 1}^{1} \sqrt{1 - u^{2}} d u = \frac{1}{2} π (1)^{2} = \frac{π}{2} .

The volume of the doughnut-shaped solid is

\begin{aligned} V &= 4\pi\underbrace{\int_{-1}^{1}u\sqrt{1-u^{2}}\,du}_{=0}+20\pi\underbrace{\int_{-1}^{1}\sqrt{1-u^{2}}\,du}_{=\pi/2}\\ &= 0+20\pi\cdot\frac{\pi}{2}\\ &= 10\pi^{2}. \end{aligned}

The Torus and Pappus's Theorem

If a circle is revolved about an axis that does not cut it, the doughnut-shaped solid is called a torus. If the radius of the circle is $r$ and the distance between the centre of the circle and the axis is $a$ , with $a > r$ , then repeating the computation of the last example with $u = x - a$ gives

\bbox[8px, #E6F0FA, border: 3px solid #0066CC]{V=\left(\pi r^{2}\right)(2\pi a)=2\pi^{2}r^{2}a.}

A torus formed by revolving a circle of radius r about an axis lying at distance a from its centre, annotated with the volume formula. — A torus and its volume.

Read that formula as a product: it is the area $π r^{2}$ of the disk being revolved, multiplied by $2 π a$ , the distance travelled by the centre of the disk as it goes once around the axis. Because the disk is symmetric, its centre is its centroid, and this is an instance of a general result.

Pappus's theorem for volumes. If a plane region of area $A$ is revolved about a line in its plane that does not intersect the region, and $\bar{d}$ is the distance from the centroid of the region to that line, then the volume of the resulting solid of revolution is

V = 2 π \bar{d} A .

The shell method is essentially a proof of this theorem in disguise. Writing $V = \int_{a}^{b} 2 π ρ d A$ and factoring out the constant $2 π$ gives $V = 2 π \int ρ d A$ , and the integral $\int ρ d A$ divided by the total area $A$ is precisely the definition of the centroid's distance $\bar{d}$ from the axis. So $V = 2 π \bar{d} A$ each time. In the torus example, $A = π (1)^{2} = π$ and $\bar{d} = 5$ , so Pappus predicts $V = 2 π (5) (π) = 10 π^{2}$ , matching the integral we computed.

Pappus's theorem is a shortcut only when the centroid is already known, typically by symmetry. For a region whose centroid you would have to compute first, the shell integral is usually the faster route.

Exercises

The region bounded by $y = x^{2}$ , $y = 0$ and $x = 2$ is revolved about the $y$ -axis. Use the shell method to find the volume of the resulting solid.

Solution

The shell at distance $x$ from the axis has radius $ρ = x$ , height $f (x) = x^{2}$ and thickness $d x$ , so $d V = 2 π x \cdot x^{2} d x$ . Since $x$ runs from $0$ to $2$ ,

\begin{aligned} V &= \int_{0}^{2}2\pi x\cdot x^{2}\,dx\\ &= 2\pi\int_{0}^{2}x^{3}\,dx\\ &= 2\pi\left[\frac{x^{4}}{4}\right]_{0}^{2}\\ &= 2\pi\cdot\frac{16}{4}=8\pi. \end{aligned}

The region bounded by $y = \sqrt{x}$ , $y = 0$ and $x = 4$ is revolved about the $y$ -axis. Find the volume by the shell method.

Solution

Here $ρ = x$ and the shell height is $\sqrt{x}$ , so $d V = 2 π x \sqrt{x} d x = 2 π x^{3 / 2} d x$ with $0 \leq x \leq 4$ :

\begin{aligned} V &= 2\pi\int_{0}^{4}x^{3/2}\,dx\\ &= 2\pi\left[\frac{2}{5}x^{5/2}\right]_{0}^{4}\\ &= 2\pi\cdot\frac{2}{5}\cdot32\\ &= \frac{128\pi}{5}. \end{aligned}

Here $4^{5 / 2} = {(\sqrt{4})}^{5} = 2^{5} = 32$ .

The region bounded by $y = \frac{1}{x}$ , $y = 0$ , $x = 1$ and $x = 3$ is revolved about the $y$ -axis. Find the volume.

Solution

The factor $x$ supplied by the circumference cancels the $1 / x$ in the height, which is what makes this one so short:

\begin{aligned} V &= \int_{1}^{3}2\pi x\cdot\frac{1}{x}\,dx\\ &= 2\pi\int_{1}^{3}dx\\ &= 2\pi\left[x\right]_{1}^{3}=4\pi. \end{aligned}

Notice how badly washers would have fared here: the region's horizontal cross sections change formula at $y = \frac{1}{3}$ and again at $y = 1$ , so the washer setup would need two integrals.

The region bounded by $y = x$ and $y = x^{2}$ is revolved about the $y$ -axis. Find the volume of the resulting solid.

Solution

The curves meet where $x = x^{2}$ , that is at $x = 0$ and $x = 1$ , and on $[0, 1]$ we have $x \geq x^{2}$ . So the shell height is $x - x^{2}$ and

\begin{aligned} V &= \int_{0}^{1}2\pi x\left(x-x^{2}\right)dx\\ &= 2\pi\int_{0}^{1}\left(x^{2}-x^{3}\right)dx\\ &= 2\pi\left[\frac{x^{3}}{3}-\frac{x^{4}}{4}\right]_{0}^{1}\\ &= 2\pi\left(\frac{1}{3}-\frac{1}{4}\right)=2\pi\cdot\frac{1}{12}=\frac{\pi}{6}. \end{aligned}

The region bounded by $y = 2 x - x^{2}$ and the $x$ -axis is revolved about the $y$ -axis. Find the volume of the resulting solid.

Solution

The parabola meets the $x$ -axis where $2 x - x^{2} = x (2 - x) = 0$ , that is at $x = 0$ and $x = 2$ , and $2 x - x^{2} \geq 0$ on $[0, 2]$ . Therefore

\begin{aligned} V &= \int_{0}^{2}2\pi x\left(2x-x^{2}\right)dx\\ &= 2\pi\int_{0}^{2}\left(2x^{2}-x^{3}\right)dx\\ &= 2\pi\left[\frac{2x^{3}}{3}-\frac{x^{4}}{4}\right]_{0}^{2}\\ &= 2\pi\left(\frac{16}{3}-4\right)=2\pi\cdot\frac{4}{3}=\frac{8\pi}{3}. \end{aligned}

The region bounded by $y = x^{2}$ , $y = 0$ and $x = 1$ is revolved about the vertical line $x = 2$ . Find the volume of the resulting solid.

Solution

The region lies to the left of the axis $x = 2$ , so the shell radius is $ρ = 2 - x$ , which is positive for every $x$ in $[0, 1]$ . The shell height is $x^{2}$ , so

\begin{aligned} V &= \int_{0}^{1}2\pi(2-x)x^{2}\,dx\\ &= 2\pi\int_{0}^{1}\left(2x^{2}-x^{3}\right)dx\\ &= 2\pi\left[\frac{2x^{3}}{3}-\frac{x^{4}}{4}\right]_{0}^{1}\\ &= 2\pi\left(\frac{2}{3}-\frac{1}{4}\right)=2\pi\cdot\frac{5}{12}=\frac{5\pi}{6}. \end{aligned}

A useful check on the sign of the radius: at $x = 0$ the shell has radius $2$ , the full distance from the axis, and at $x = 1$ it has radius $1$ . Writing $x - 2$ instead would give negative radii and a negative volume.

The region bounded by $y = \sqrt{x}$ , $y = 0$ and $x = 4$ is revolved about the vertical line $x = - 1$ . Find the volume of the resulting solid.

Solution

The region lies to the right of the axis $x = - 1$ , so the shell radius is $ρ = x - (- 1) = x + 1$ . The shell height is $\sqrt{x}$ and $x$ runs from $0$ to $4$ :

\begin{aligned} V &= \int_{0}^{4}2\pi(x+1)\sqrt{x}\,dx\\ &= 2\pi\int_{0}^{4}\left(x^{3/2}+x^{1/2}\right)dx\\ &= 2\pi\left[\frac{2}{5}x^{5/2}+\frac{2}{3}x^{3/2}\right]_{0}^{4}\\ &= 2\pi\left(\frac{2}{5}\cdot32+\frac{2}{3}\cdot8\right)\\ &= 2\pi\left(\frac{64}{5}+\frac{16}{3}\right)=2\pi\cdot\frac{192+80}{15}=\frac{544\pi}{15}. \end{aligned}

Sanity check: the answer should exceed the volume $\frac{128 π}{5} = \frac{384 π}{15}$ found in Exercise 2 for the same region about the $y$ -axis, since every shell now has a radius one unit larger. It does.

The region bounded by $y = e^{- x^{2}}$ , $y = 0$ , $x = 0$ and $x = 1$ is revolved about the $y$ -axis. Find the volume, and explain why the washer method is not a practical option here.

Solution

By the shell method, $ρ = x$ and the height is $e^{- x^{2}}$ :

V = \int_{0}^{1} 2 π x e^{- x^{2}} d x .

Substituting $u = x^{2}$ , so that $d u = 2 x d x$ , and noting that $u = 0$ when $x = 0$ and $u = 1$ when $x = 1$ ,

\begin{aligned} V &= \pi\int_{0}^{1}e^{-u}\,du\\ &= \pi\left[-e^{-u}\right]_{0}^{1}\\ &= \pi\left(1-e^{-1}\right)\approx1.986. \end{aligned}

The washer method would require the region's boundary as a function of $y$ , that is $x = \sqrt{- \ln y}$ , and the resulting integral $π \int (- \ln y) d y$ over the appropriate range, split at $y = e^{- 1}$ where the boundary switches from the curve to the line $x = 1$ . The shell integral needed one substitution; the washer setup needs two integrals and a logarithm. The extra factor of $x$ that the shell method supplies is precisely what makes the substitution work.

The region bounded by $y = x^{3}$ , $y = 0$ and $x = 2$ is revolved about the $y$ -axis. Set up the volume integral by both the shell method and the washer method, evaluate both, and say which is easier.

Solution

Shells. A vertical strip at $x$ has height $x^{3}$ and radius $x$ , with $0 \leq x \leq 2$ :

\begin{aligned} V &= \int_{0}^{2}2\pi x\cdot x^{3}\,dx\\ &= 2\pi\int_{0}^{2}x^{4}\,dx\\ &= 2\pi\left[\frac{x^{5}}{5}\right]_{0}^{2}=2\pi\cdot\frac{32}{5}=\frac{64\pi}{5}. \end{aligned}

Washers. A horizontal strip at height $y$ runs from the curve $x = y^{1 / 3}$ out to the line $x = 2$ , so the outer radius is $R = 2$ and the inner radius is $r = y^{1 / 3}$ , with $0 \leq y \leq 8$ :

\begin{aligned} V &= \int_{0}^{8}\pi\left(2^{2}-\left(y^{1/3}\right)^{2}\right)dy\\ &= \pi\int_{0}^{8}\left(4-y^{2/3}\right)dy\\ &= \pi\left[4y-\frac{3}{5}y^{5/3}\right]_{0}^{8}\\ &= \pi\left(32-\frac{3}{5}\cdot32\right)=\pi\cdot\frac{64}{5}=\frac{64\pi}{5}. \end{aligned}

Here $8^{5 / 3} = {(8^{1 / 3})}^{5} = 2^{5} = 32$ . Both give $\frac{64 π}{5}$ , as they must.

Which is easier? The shell integral is easier. It required no inversion of $y = x^{3}$ , no fractional exponents, and its integrand $2 π x^{4}$ is a monomial. The washer setup forced us to solve for $x = y^{1 / 3}$ , to recognise that the outer radius is the constant $2$ rather than a curve, and to handle the exponent $5 / 3$ . If the cubic were replaced by something non-invertible, such as $y = x^{3} + x$ , the washer route would close entirely while the shell route would be unchanged.

The region bounded by $y = \sin x$ and the $x$ -axis for $0 \leq x \leq π$ is revolved about the $y$ -axis. Find the volume.

Solution

Since $\sin x \geq 0$ on $[0, π]$ , the shell height is $\sin x$ and the radius is $x$ :

V = \int_{0}^{π} 2 π x \sin x d x .

Integrate by parts with $u = x$ and $d v = \sin x d x$ , so $d u = d x$ and $v = - \cos x$ :

\begin{aligned} \int_{0}^{\pi}x\sin x\,dx &= \left[-x\cos x\right]_{0}^{\pi}+\int_{0}^{\pi}\cos x\,dx\\ &= \left[-x\cos x+\sin x\right]_{0}^{\pi}\\ &= \left(-\pi(-1)+0\right)-\left(0+0\right)=\pi. \end{aligned}

Therefore

V = 2 π \cdot π = 2 π^{2} \approx 19.74 .

The washer method here would require $x$ as a function of $y$ , which is $x = \arcsin y$ on the left half of the arch and $x = π - \arcsin y$ on the right half, so it would need two branches. Shells avoid the split entirely.

The region inside the circle $x^{2} + y^{2} = 4$ with $x \geq 1$ is revolved about the $y$ -axis. Find the volume of the resulting solid, which is the ring left after drilling a cylindrical hole of radius $1$ straight through the centre of a sphere of radius $2$ .

Solution

A vertical strip at $x$ runs from $y = - \sqrt{4 - x^{2}}$ up to $y = \sqrt{4 - x^{2}}$ , so its height is $2 \sqrt{4 - x^{2}}$ , and $x$ runs from $1$ to $2$ . Hence

V = \int_{1}^{2} 2 π x \cdot 2 \sqrt{4 - x^{2}} d x = 2 π \int_{1}^{2} 2 x \sqrt{4 - x^{2}} d x .

Let $u = 4 - x^{2}$ , so $d u = - 2 x d x$ and $2 x d x = - d u$ . When $x = 1$ we get $u = 3$ , and when $x = 2$ we get $u = 0$ . Therefore

\begin{aligned} V &= -2\pi\int_{3}^{0}\sqrt{u}\,du\\ &= 2\pi\int_{0}^{3}\sqrt{u}\,du\\ &= 2\pi\left[\frac{2}{3}u^{3/2}\right]_{0}^{3}\\ &= \frac{4\pi}{3}\cdot3\sqrt{3}=4\sqrt{3}\,\pi\approx21.77. \end{aligned}

Check: the whole sphere of radius $2$ has volume $\frac{4}{3} π (2)^{3} = \frac{32 π}{3} \approx 33.51$ , so the drilled ring is indeed a substantial but proper part of it.

A disk of radius $r$ with centre $(a, 0)$ , where $a > r > 0$ , is revolved about the $y$ -axis. Show by the shell method that the volume of the resulting torus is $V = 2 π^{2} r^{2} a$ , and confirm that this agrees with Pappus's theorem.

Solution

The boundary circle is $(x - a)^{2} + y^{2} = r^{2}$ , so the vertical strip at $x$ has height

2 \sqrt{r^{2} - (x - a)^{2}},

and $x$ runs from $a - r$ to $a + r$ . The shell radius is $ρ = x$ , which is positive throughout because $a > r$ . Therefore

V = \int_{a - r}^{a + r} 2 π x \cdot 2 \sqrt{r^{2} - (x - a)^{2}} d x = 4 π \int_{a - r}^{a + r} x \sqrt{r^{2} - (x - a)^{2}} d x .

Let $u = x - a$ , so $x = u + a$ and $d u = d x$ ; the limits become $u = - r$ and $u = r$ :

\begin{aligned} V &= 4\pi\int_{-r}^{r}(u+a)\sqrt{r^{2}-u^{2}}\,du\\ &= 4\pi\int_{-r}^{r}u\sqrt{r^{2}-u^{2}}\,du+4\pi a\int_{-r}^{r}\sqrt{r^{2}-u^{2}}\,du. \end{aligned}

The first integrand is odd, since replacing $u$ by $- u$ changes the sign of the factor $u$ and leaves $\sqrt{r^{2} - u^{2}}$ unchanged, and the interval $[- r, r]$ is symmetric about the origin, so that integral is $0$ . The second integral is the area of a semicircle of radius $r$ , namely $\frac{1}{2} π r^{2}$ . Hence

V = 0 + 4 π a \cdot \frac{π r^{2}}{2} = 2 π^{2} r^{2} a .

Agreement with Pappus. The revolved region is a disk of area $A = π r^{2}$ , and by symmetry its centroid is its centre, at distance $\bar{d} = a$ from the axis. Pappus's theorem gives

V = 2 π \bar{d} A = 2 π a \cdot π r^{2} = 2 π^{2} r^{2} a,

the same answer. Setting $r = 1$ and $a = 5$ recovers $V = 2 π^{2} (1) (5) = 10 π^{2}$ , the value computed in the last worked example of this section.

Building a Cylindrical Shell

Regions Between Two Curves

Shell or Washer? How to Choose

Worked Examples

The Torus and Pappus's Theorem

Exercises

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