MATH 141 Calculus 2 • McGill University, Montreal

Revision sheet: volumes by cylindrical shells (MATH 141)

This sheet is not a summary of section 6.3 of Stewart: you have the course notes. It answers one question, what makes students lose marks on volumes by cylindrical shells in MATH 141 at McGill University, and which precise gesture avoids each loss.

Every volume below is exact and done by hand, as on the exam, and every one of them has been computed a second way, by washers or by linearity, before being written here.

The thread of the chapter

A shell is swept by a strip PARALLEL to the axis: its radius is the DISTANCE from the strip to the axis, its height is the LENGTH of the strip, and the variable is the one perpendicular to the axis. Draw the strip before writing anything; a zero or a negative volume always means a radius that is not a distance.

This chapter is part of MATH 141, Calculus 2 (McGill)

What is on this sheet

1. The essentials
2. The rules in table form
3. The mistakes that cost marks
4. Which method to choose
5. How the answer is expected to be written
6. Check before you hand in
7. The typical problem, taken apart
8. Learn by heart
9. Frequently asked questions

Topics covered

Method of cylindrical shells
Solid of revolution
Axis of rotation
Radius of a shell
Height of a shell
Washer method
Inverse function
Odd and even functions
Linearity of the definite integral
Torus

The essentials

One strip, three readings

• A strip PARALLEL to the axis sweeps a shell; a strip PERPENDICULAR to the axis sweeps a disk or a washer. Same strip, different axis, different method.
• Shell volume: $2\pi \times$ radius $\times$ height $\times$ thickness. Unrolled, the shell is a slab of length $2\pi r$ .
• RADIUS: the distance from the strip to the axis, always positive. Larger coordinate minus smaller one, read on the figure.
• HEIGHT: the length of the strip, top minus bottom for a vertical strip, right minus left for a horizontal one.
• VARIABLE: the one perpendicular to the axis. Vertical axis, vertical strips, $dx$ ; horizontal axis, horizontal strips, $dy$ , every boundary written as $x$ in terms of $y$ .

The same vertical strip under

y = 2 - \frac{x^2}{2}

: about the vertical axis it sweeps a shell of radius

x

and height

f(x)

; about the horizontal axis it sweeps a disk of radius

f(x)

Count the lengths in the integrand: radius, height, thickness. Three lengths make a volume. Two lengths, as in $2\pi\int f(x)\,dx$ , mean the radius has been dropped.

Why shells exist, and when to prefer washers

• Shells never ask for the inverse of the curve. About the $y$ -axis, $y = f(x)$ is used as it is.
• Washers about the $y$ -axis need $x$ as a function of $y$ , and TWO branches when a horizontal line meets the curve twice.
• When both methods are possible, choose the variable in which the boundaries are easy and the integral is short; the axis then names the method.
• Moving the axis a distance $d$ further from the region adds $2\pi d A$ , where $A$ is the area: linearity of the integral, one line of checking.

The rules in table form

Each row reads from left to right: the assumptions, then the result. A red cell is not an answer, it is the finding that the form settles nothing and the instruction to rewrite it. Every case is followed by a worked example.

The radius, axis by axis

Read a line as: for this axis and this position of the region, the strip is of this kind and its radius is the last column. Each example has been computed both ways. The red line is the case students try to force into a formula.

Axis	Region lies	Strip and variable	Radius
$y$ -axis	at $x \ge 0$	vertical, $dx$	$x$
Example: Under $y = 4 - x^2$ , $0 \le x \le 2$ : $2\pi\int_0^2 x(4 - x^2)\,dx = 8\pi$ .
$x = k$	left of the axis	vertical, $dx$	$k - x$
Example: Same region about $x = 3$ : $2\pi\int_0^2 (3 - x)(4 - x^2)\,dx = 24\pi$ .
$x = k$	right of the axis	vertical, $dx$	$x - k$
Example: Same region about $x = -1$ : radius $x + 1$ , $V = \frac{56\pi}{3}$ .
$x$ -axis	at $y \ge 0$	horizontal, $dy$	$y$
Example: Between $x = 1 + y^2$ and $x = 5$ : $2\pi\int_0^2 y(4 - y^2)\,dy = 8\pi$ .
$y = k$	below the axis	horizontal, $dy$	$k - y$
Example: Between $x = y^2$ and $x = 4$ about $y = 3$ : $2\pi\int_{-2}^{2} (3 - y)(4 - y^2)\,dy = 64\pi$ .
$y = k$	above the axis	horizontal, $dy$	$y - k$
Example: Same region about $y = -2$ : radius $y + 2$ , $V = \frac{128\pi}{3}$ .
any	on both sides	either	no single radius no such formula
Example: Under $y = 1 - x^2$ , $-1 \le x \le 1$ , about the $y$ -axis: $2\pi\int_{-1}^{1} x(1 - x^2)\,dx = 0$ , yet the solid has volume $\frac{\pi}{2}$ . What to do: The two sides sweep overlapping solids. Mirror one side onto the other and turn only what reaches furthest from the axis; for a symmetric region, one half.

The radius column never contains a bare $x - k$ for an axis on the right, and never a negative number: if the integrand changes sign on the interval, the radius is wrong.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Writing the radius as a coordinate difference in the wrong order

2 to 3 marks for the set-up, even when the final absolute value is right

What not to write

“About $x = 3$ : $V = 2\pi\int_0^2 (x - 3)(4 - x^2)\,dx = -24\pi$ , so the volume is $24\pi$ .”

What to write

“The region lies left of $x = 3$ , so the radius is $3 - x$ : $V = 2\pi\int_0^2 (3 - x)(4 - x^2)\,dx = 24\pi$ .”

Why: A radius is a distance, larger coordinate minus smaller one. The negative sign is not a detail to erase at the end: it is the proof, visible on the paper, that the set-up is wrong, and markers grade the set-up.

2. Reading the line x = -1 as subtract 1

the whole question

What not to write

“About $x = -1$ the radius is $x - 1$ : $V = 2\pi\int_0^2 (x - 1)(4 - x^2)\,dx = -\frac{8\pi}{3}$ .”

What to write

“The distance from $x$ to $-1$ is $x - (-1) = x + 1$ : $V = 2\pi\int_0^2 (x + 1)(4 - x^2)\,dx = \frac{56\pi}{3}$ .”

Why: Write the subtraction with the axis value in parentheses, $x - (-1)$ , before simplifying. The wrong radius is not only of the wrong sign, it is of the wrong size, so no absolute value can rescue it.

3. Dropping the radius, or using pi instead of 2 pi

3 marks: the formula is the question

What not to write

“ $V = 2\pi\int_0^3 x(x - 3)^2\,dx = \frac{27\pi}{2}$ ” or “ $V = \pi\int_0^3 x \cdot x(x - 3)^2\,dx = \frac{81\pi}{10}$ .”

What to write

“Radius $x$ , height $x(x - 3)^2$ , circumference $2\pi x$ : $V = 2\pi\int_0^3 x \cdot x(x - 3)^2\,dx = \frac{81\pi}{5}$ .”

Why: The first answer is in square units: only two lengths are multiplied. The second has the right units but half the value, because the unrolled shell is $2\pi r$ long; the $\pi r^2$ of a disk area has leaked in.

4. Using vertical strips for shells about the x-axis

the whole question: it is the volume of another solid

What not to write

“ $R$ : $y = \sqrt{x - 1}$ , $y = 0$ , $x = 5$ , about the $x$ -axis: $V = 2\pi\int_1^5 x\sqrt{x - 1}\,dx = \frac{544\pi}{15}$ .”

What to write

“About the $x$ -axis the shell strips are horizontal: $x$ from $1 + y^2$ to $5$ , radius $y$ , so $V = 2\pi\int_0^2 y(4 - y^2)\,dy = 8\pi$ .”

Why: A shell strip is parallel to its axis. Vertical strips with radius $x$ are shells about the $y$ -axis, and $\frac{544\pi}{15}$ is exactly that solid's volume. Draw the strip before choosing the variable.

5. Taking the curve for the height of a horizontal strip

2 marks, and a solid that is not the one asked

What not to write

“The strip at height $y$ ends on $x = 1 + y^2$ , so its height is $1 + y^2$ : $V = 2\pi\int_0^2 y(1 + y^2)\,dy = 12\pi$ .”

What to write

“The strip runs from $x = 1 + y^2$ to $x = 5$ : its length is $5 - (1 + y^2) = 4 - y^2$ , and $V = 8\pi$ .”

Why: $1 + y^2$ is the distance from the $y$ -axis to the curve: it measures the region on the OTHER side of the curve. Indeed $8\pi + 12\pi = 20\pi$ , the full cylinder of radius $2$ and length $5$ .

6. Integrating across an axis that cuts the region

3 to 4 marks

What not to write

“ $R$ under $y = \frac{(x + 1)(2 - x)}{2}$ , $-1 \le x \le 2$ , about the $y$ -axis: $V = 2\pi\int_{-1}^{2} x \cdot \frac{(x + 1)(2 - x)}{2}\,dx = \frac{9\pi}{4}$ .”

What to write

“The left piece, mirrored, lies inside the right piece, so the solid is swept by the right piece alone: $V = 2\pi\int_0^2 x \cdot \frac{(x + 1)(2 - x)}{2}\,dx = \frac{8\pi}{3}$ .”

Mirrored across the

y

-axis, the left piece (dashed) lies entirely under the right piece: turning the whole region gives the same solid as turning the right piece alone.

Why: On the left of the axis the factor $x$ is negative, so the left piece is SUBTRACTED; adding it instead, $\frac{8\pi}{3} + \frac{5\pi}{12} = \frac{37\pi}{12}$ , counts the overlap twice. Mirror the pieces and keep, at each height, what reaches furthest from the axis.

7. Checking with washers but forgetting the inner branch

a correct answer replaced by a wrong one, and 10 minutes

What not to write

“Check by washers for $y = 2x^2 - x^4$ about the $y$ -axis: $\pi\int_0^1 \left(1 + \sqrt{1 - y}\right)dy = \frac{5\pi}{3}$ , so my shell answer $\frac{4\pi}{3}$ is wrong.”

What to write

“The line $y = c$ meets the hump twice, $x^2 = 1 \pm \sqrt{1 - c}$ , so the washer area is $\pi(x_{out}^2 - x_{in}^2) = 2\pi\sqrt{1 - y}$ and $V = \frac{4\pi}{3}$ .”

Why: When a horizontal line crosses the boundary twice, the slice has a hole. The outer branch alone fills the funnel-shaped hollow of the solid, of volume $\frac{\pi}{3}$ .

8. Losing the parentheses around the bottom curve

2 marks

What not to write

“Between $y = x$ and $y = x^2 - 2x$ the height is $x - x^2 - 2x$ , and $V = 2\pi\int_0^3 x(-x - x^2)\,dx = -\frac{117\pi}{2}$ .”

What to write

“Height $= x - (x^2 - 2x) = 3x - x^2$ , positive on $(0, 3)$ , and $V = 2\pi\int_0^3 x(3x - x^2)\,dx = \frac{27\pi}{2}$ .”

Why: Write the bottom curve in parentheses every time, then check that the height is positive at one point inside the interval. That the region dips below the $x$ -axis changes nothing for a vertical axis.

Which method to choose

Shells or washers: decide on the FORM of the region

Look at the boundary curves and ask in which variable they are easy to write, before looking at the axis

Left: the line

y = c

meets the hump twice, so washers need two branches and shells win. Right: it meets

y = \ln x

once, at

x = e^c

, so washers work.

If curves $y = f(x)$ that cannot be solved for $x$ , vertical axis → shells in x: vertical strips, no inverse needed

Example: $y = x^3 + x$ about the $y$ -axis: $2\pi\int_0^1 x(x^3 + x)\,dx = \frac{16\pi}{15}$
If curves $y = f(x)$ that cannot be solved for $x$ , horizontal axis → disks or washers in x

Example: $y = x^3 + x$ about the $x$ -axis: $\pi\int_0^1 (x^3 + x)^2\,dx = \frac{92\pi}{105}$
If a horizontal line meets the curve twice, vertical axis → shells: one height per strip instead of two branches

Example: $y = 2x^2 - x^4$ about the $y$ -axis: $\frac{4\pi}{3}$ in one line
If boundaries given as $x = g(y)$ , horizontal axis → shells in y: horizontal strips, right minus left

Example: $x = y^2$ and $x = 4$ about $y = 3$ : $64\pi$
If both variables are easy → take the shorter integral: fewer pieces, fewer integrations by parts

Example: $y = \ln x$ about the $y$ -axis: washers $\pi\int_0^1 (e^2 - e^{2y})\,dy$ beat shells $2\pi\int_1^e x\ln x\,dx$

The axis never chooses the method by itself: it only tells you that, for a given variable, one method uses strips parallel to it and the other slices perpendicular to it.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Setting up a volume by shells

When to use it: Any question that says use cylindrical shells, or where you have chosen them

1 Sketch the region and the axis, and draw ONE strip parallel to the axis. Name the variable: $x$ for vertical strips, $y$ for horizontal ones.
2 Write the radius as a distance, with the axis value in parentheses: $3 - x$ , $x - (-1)$ , $4 - y$ .
3 Write the height as a length: top minus bottom, or right minus left, each curve in the variable of integration.
4 Give the bounds of the variable from the intersections, then write $V = \int 2\pi\,(\text{radius})(\text{height})\,d(\text{variable})$ .
5 Integrate, and check that the integrand was positive on the interval and that the answer has the units of a volume.

Concluding sentence

“Vertical strips, radius $3 - x$ , height $4 - x^2$ , $0 \le x \le 2$ , so $V = 2\pi\int_0^2 (3 - x)(4 - x^2)\,dx = 24\pi$ .”

The trap: Starting from a memorized $2\pi\int x f(x)\,dx$ and patching it for the axis. The phrase above names each factor, which is where the marks are when the arithmetic slips.

Marking: Typically 1 mark for the strip and the variable, 1 for the radius, 1 for the height, 1 for the bounds, and the rest for the integration.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

Positive integrand, positive volume

Evaluate radius times height at one point inside the interval. It must be positive; a zero or negative result means a radius that is not a distance or a height in the wrong order.

$(3 - x)(4 - x^2)$ at $x = 1$ : $2 \cdot 3 = 6 > 0$ . With $(x - 3)$ it would be $-6$ .
Three lengths in the integrand

Radius, height and thickness: three lengths make a volume. Two lengths mean a factor is missing.

$2\pi\int_0^3 x(x - 3)^2\,dx$ has only height and thickness: square units, so the radius $x$ is missing.
The bounding cylinder

Enclose the region in a rectangle; turned about the axis, it gives a cylinder or a cylindrical tube whose volume must exceed yours.

$y = x(x - 3)^2$ on $[0, 3]$ fits in $[0, 3] \times [0, 4]$ : $\frac{81\pi}{5} = 16.2\pi < 36\pi$ .
The shift check

Moving the axis a distance $d$ further from the region adds $2\pi d A$ . Two volumes of the same region about two parallel axes check each other.

Under $4 - x^2$ : $8\pi$ about the $y$ -axis, $A = \frac{16}{3}$ , so $8\pi + 2\pi \cdot \frac{16}{3} = \frac{56\pi}{3}$ about $x = -1$ .

The typical problem, taken apart

Between two curves, about a vertical line off the region

The region $R$ is bounded by $y = \sqrt{x}$ and $y = x^2$ . Find the volume of the solid obtained by turning $R$ about the line $x = -1$ .

No calculator. Every choice must be justified as on a MATH 141 final.

The region between

y = x^2

and

y = \sqrt{x}

, and one vertical strip: its distance to the axis

x = -1

x + 1

, not

x - 1

Step 1

$\sqrt{x} = x^2$ gives $x = x^4$ , so $x = 0$ or $x = 1$ . At $x = \frac{1}{4}$ : $\sqrt{x} = \frac{1}{2} > \frac{1}{16}$ , so $y = \sqrt{x}$ is on top.

Why

The intersections are the bounds, and the test point decides the order of the height. Both come before any formula.

Step 2

Axis vertical and both curves easy in $x$ : shells, vertical strips, variable $x$ . Radius $x - (-1) = x + 1$ , height $\sqrt{x} - x^2$ .

Why

The choice is justified in one sentence: no inverse needed, one height per strip. Writing the subtraction with the axis in parentheses is what prevents the x - 1 slip.

Step 3

$V = 2\pi\int_0^1 (x + 1)(\sqrt{x} - x^2)\,dx = 2\pi\int_0^1 \left(x^{3/2} - x^3 + x^{1/2} - x^2\right)dx = 2\pi\left(\frac{2}{5} - \frac{1}{4} + \frac{2}{3} - \frac{1}{3}\right)$ .

Why

Expanding the product first turns the integral into four powers of x, each done by the power rule: no technique beyond the table.

Step 4

$\frac{2}{5} - \frac{1}{4} + \frac{1}{3} = \frac{24 - 15 + 20}{60} = \frac{29}{60}$ , so $V = 2\pi \cdot \frac{29}{60} = \frac{29\pi}{30}$ .

Why

One common denominator, written out, so that a marker can follow the arithmetic and give the method marks even after a slip.

Step 5

Shift check: about the $y$ -axis, $2\pi\int_0^1 x(\sqrt{x} - x^2)\,dx = 2\pi\left(\frac{2}{5} - \frac{1}{4}\right) = \frac{3\pi}{10}$ , and $A = \frac{2}{3} - \frac{1}{3} = \frac{1}{3}$ , so $\frac{3\pi}{10} + 2\pi \cdot \frac{1}{3} = \frac{9\pi + 20\pi}{30} = \frac{29\pi}{30}$ .

Why

Thirty seconds, and the answer is confirmed by a computation that shares none of the arithmetic of step 4.

The conclusion, written out

“ $V = 2\pi\int_0^1 (x + 1)(\sqrt{x} - x^2)\,dx = \frac{29\pi}{30}$ .” Washers give the same, with outer radius $\sqrt{y} + 1$ and inner radius $y^2 + 1$ .

The classic mistake on this problem: Writing the radius $x - 1$ , which gives $2\pi\left(\frac{2}{5} - \frac{1}{4} - \frac{2}{3} + \frac{1}{3}\right) = -\frac{11\pi}{30}$ ; or taking the height $x^2 - \sqrt{x}$ , which flips the sign of the whole answer.

Learn by heart

• Strip PARALLEL to the axis: shell. Strip PERPENDICULAR: disk or washer.
• $V = \int 2\pi\,(\text{radius})(\text{height})\,d(\text{variable})$ , the variable perpendicular to the axis.
• Radius = distance to the axis: larger coordinate minus smaller. $x - (-1) = x + 1$ .
• Height = length of the strip: top minus bottom, right minus left, bottom curve in parentheses.
• Axis cutting the region: no single formula, turn only the side that reaches furthest.
• Shifting the axis by $d$ away from the region adds $2\pi d A$ .
• Torus: $V = 2\pi^2 R r^2$ , with $R$ measured to the CENTRE of the cross-section.

Frequently asked questions

When should I use cylindrical shells instead of washers?

Use shells when the washers would force you to solve the curve for the other variable, for example a cubic, or when a horizontal line meets the curve twice and the washer would need two branches. When both variables are easy, choose the shorter integral. The axis alone does not decide: it only tells you which strip each method uses.

How do I find the radius of a shell when the axis is x = -1 or x = 3?

The radius is the distance from the strip to the axis, always positive: the larger coordinate minus the smaller one. For an axis at x equals 3 with the region on its left, the radius is 3 minus x. For an axis at x equals minus 1 with the region on its right, it is x minus minus 1, that is x plus 1, not x minus 1.

Why did I get a negative volume with the shell method?

Because the radius or the height is not a length. Either the radius was written as x minus k for an axis to the right of the region, or the height was written bottom minus top, or the parentheses around the bottom curve were lost. Taking the absolute value at the end does not fix the set-up, and the marker grades the set-up.

Do shells and washers always give the same volume?

Yes, when both are set up correctly they describe the same solid and give exactly the same number. If they disagree, one of them is wrong, most often a washer that forgot its inner radius or a shell whose height was read as the curve instead of the length of the strip. Neither method is an approximation.

What if the axis of rotation passes through the region?

Then the two sides of the axis sweep overlapping solids, and integrating across the axis subtracts one side from the other. Mirror one side onto the other and turn only what reaches furthest from the axis at each height. For a region symmetric about the axis, turn one half only, and never double it.

Practise it

Corrected exercises: Volumes by cylindrical shells, MATH 141 at McGill

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

Previous sheet Volumes: disks and washers Next sheet Work

Looking for a MATH 141 tutor in Montreal?

Get in touch for a first session. Volumes of revolution are where a first integral calculus course stops being a list of techniques and starts asking you to read a picture.

Book a session WhatsApp

Site by Studio Squalli

Revision sheet: volumes by cylindrical shells (MATH 141)

The essentials

One strip, three readings

Why shells exist, and when to prefer washers

The rules in table form

The radius, axis by axis

The mistakes that cost marks

1. Writing the radius as a coordinate difference in the wrong order

2. Reading the line x = -1 as subtract 1

3. Dropping the radius, or using pi instead of 2 pi

4. Using vertical strips for shells about the x-axis

5. Taking the curve for the height of a horizontal strip

6. Integrating across an axis that cuts the region

7. Checking with washers but forgetting the inner branch

8. Losing the parentheses around the bottom curve

Which method to choose

Shells or washers: decide on the FORM of the region

How the answer is expected to be written

Setting up a volume by shells

Check before you hand in

The typical problem, taken apart

Between two curves, about a vertical line off the region

Learn by heart

Frequently asked questions

When should I use cylindrical shells instead of washers?

How do I find the radius of a shell when the axis is x = -1 or x = 3?

Why did I get a negative volume with the shell method?

Do shells and washers always give the same volume?

What if the axis of rotation passes through the region?

Corrected exercises: Volumes by cylindrical shells, MATH 141 at McGill

See also

Looking for a MATH 141 tutor in Montreal?