MATH 141 Calculus 2 • McGill University, Montreal

Revision sheet: arc length and area of a surface of revolution (MATH 141)

This sheet is not a summary of sections 8.1 and 8.2 of Stewart: you already have the course notes. It answers one question only, what makes students lose marks on arc length and area of a surface of revolution in MATH 141 at McGill University, and which precise gesture avoids each loss.

Every number below is exact and done by hand, as on the exam, where no calculator is allowed. Every length quoted has passed the five second check that settles most errors of this chapter: an arc is longer than its chord, which is longer than its run.

The thread of the chapter

Everything in this chapter is measured ALONG the curve, so every integral carries $ds = \sqrt{1 + (f')^2}\,dx$ , the hypotenuse of $dx$ and $dy$ , never a bare $dx$ . On an exam the root is built to open, and $\sqrt{u^2} = |u|$ : the minus inside becomes a plus outside.

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

Arc length
Length element
Arc length function
Perfect square under a root
Hyperbolic cosine
Catenary
Surface of revolution
Surface area
Frustum of a cone
Gabriel's horn
Improper integral
Comparison test for integrals

The essentials

One triangle, one root

• On a short piece of curve, $dx$ and $dy$ are the legs of a right triangle and $ds$ its hypotenuse: $ds^2 = dx^2 + dy^2$ .
• For $y = f(x)$ , $a \le x \le b$ : $L = \int_a^b \sqrt{1 + (f'(x))^2}\,dx$ . For $x = g(y)$ , $c \le y \le d$ : $L = \int_c^d \sqrt{1 + (g'(y))^2}\,dy$ . The derivative must be continuous on the CLOSED interval.
• Arc length function: $s(x) = \int_a^x \sqrt{1 + (f'(t))^2}\,dt$ , and $s'(x) = \sqrt{1 + (f'(x))^2} \ge 1$ by the Fundamental Theorem.
• Always: arc $\ge$ chord $\ge$ run $= b - a$ . A length below the run is wrong, a negative length is an alarm.
• On an exam the root is designed to open. Expand $1 + (f')^2$ and look at it before choosing any technique.

At the point

P

, the tangent step has legs

dx

and

dy

: the curve's length element is the hypotenuse

ds = \sqrt{1 + (f')^2}\,dx

, always longer than

dx

Most wrong answers in this chapter integrate something without a root, $1 + f'$ , $f'$ or $dx$ alone. The root is never optional; what the exam lets you do is remove it cleanly.

Surfaces: radius times slant

• A band of the surface is a frustum of a cone: area $2\pi r\,ds$ , with $r$ the distance to the AXIS and $ds$ the slant width. So $S = \int 2\pi r\,ds$ .
• About the $x$ -axis: $S = \int_a^b 2\pi f(x)\sqrt{1 + (f'(x))^2}\,dx$ . About the $y$ -axis: $S = \int 2\pi x\,ds$ , in either variable.
• About a line $y = k$ below the curve: $r = y - k$ . About $x = h$ left of the curve: $r = x - h$ .
• Sphere of radius $r$ : $y\sqrt{1 + (y')^2} = r$ , so a zone of thickness $h$ has area $2\pi rh$ and the sphere $4\pi r^2$ .
• The volume uses $dx$ (a disk), the area NEVER does: the band's width is $ds$ .

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: rotating about this axis, the radius is the distance in the second column, and the area is the integral in the third. The examples use arcs from the set of exercises; the red line is the rule students invent.

Axis of rotation	Radius r	Area S
$x$ -axis	$y$	$\int 2\pi y\,ds$
Example: $y = \sqrt x$ , $0 \le x \le 2$ : $S = \pi\int_0^2 \sqrt{4x + 1}\,dx = \frac{13\pi}{3}$ .
$y$ -axis	$x$	$\int 2\pi x\,ds$
Example: $y = x^2$ , $0 \le x \le 1$ : $S = \int_0^1 2\pi x\sqrt{1 + 4x^2}\,dx = \frac{\pi}{6}(5\sqrt 5 - 1)$ .
line $y = k$ below the arc	$y - k$	$\int 2\pi (y - k)\,ds$
Example: About $y = -1$ : $S = \int 2\pi y\,ds + 2\pi L$ , for instance $\frac{47\pi}{16} + 2\pi \cdot \frac{17}{12} = \frac{277\pi}{48}$ .
line $x = h$ left of the arc	$x - h$	$\int 2\pi (x - h)\,ds$
Example: Segment $x = 3$ , $0 \le y \le 2$ , about $x = 1$ : $r = 2$ , $S = \int_0^2 4\pi\,dy = 8\pi$ , a cylinder.
$y$ -axis	$y$	$\int 2\pi y\,ds$ wrong radius
Example: Cylinder $x = 1$ , $0 \le y \le 3$ , about the $y$ -axis: $\int_0^3 2\pi y\,dy = 9\pi$ , while the true area is $2\pi \cdot 1 \cdot 3 = 6\pi$ . What to do: Draw the radius as a segment perpendicular to the axis, from the axis to the curve, and write its length.

The length element $ds$ is the same in every line: it belongs to the curve. Only the radius changes with the axis.

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 one plus f prime instead of the root

the whole question

What not to write

“For $y = x$ on $[0, 1]$ , $L = \int_0^1 (1 + f'(x))\,dx = 2$ .”

What to write

“ $L = \int_0^1 \sqrt{1 + 1^2}\,dx = \sqrt 2$ , the length of the diagonal of the unit square.”

Why: $\int_a^b (1 + f')\,dx = (b - a) + f(b) - f(a)$ is run plus rise, the staircase path, and it needs no calculus: a sure sign it is not a length. The square root of a sum is not the sum of the roots.

2. Keeping the minus sign when the square opens

3 to 4 marks: method and answer

What not to write

“ $y = \frac{x^3}{3} + \frac{1}{4x}$ : $\sqrt{1 + (x^2 - \frac{1}{4x^2})^2} = x^2 - \frac{1}{4x^2}$ , so $L = [\frac{x^3}{3} + \frac{1}{4x}]_1^3 = \frac{17}{2}$ .”

What to write

“ $1 + (x^2 - \frac{1}{4x^2})^2 = (x^2 + \frac{1}{4x^2})^2$ , so $L = [\frac{x^3}{3} - \frac{1}{4x}]_1^3 = \frac{53}{6}$ .”

Why: $1 + (A - B)^2 = (A + B)^2$ when $AB = \frac14$ : the minus becomes a plus. The wrong line integrates $f'$ and returns the RISE $f(3) - f(1) = \frac{17}{2}$ , shorter than the chord $\frac{\sqrt{305}}{2}$ , which is impossible for an arc.

3. Using 2 pi y dx for an area, the volume reflex

the whole question, since every number after it is wrong

What not to write

“The cone from $y = \frac34 x$ , $0 \le x \le 4$ , has lateral area $\int_0^4 2\pi \cdot \frac34 x\,dx = 12\pi$ .”

What to write

“ $S = \int_0^4 2\pi \cdot \frac34 x \cdot \frac54\,dx = 15\pi = \pi r\ell$ , with $r = 3$ and $\ell = 5$ .”

The band of the cone runs along the slant, of length

5

, not along the axis, of length

4

2\pi y\,dx

gives

12\pi

, the true area is

15\pi

Why: A band of the surface has slant width $ds$ , and $\frac{ds}{dx} = \sqrt{1 + (f')^2}$ does not tend to $1$ as the bands get thinner. For a volume, the slant changes a slice only at second order, which is why $dx$ works there and only there.

4. Taking y as the radius for a rotation about the y-axis

the whole question, and a far harder integral on the way

What not to write

“ $y = x^2$ , $0 \le x \le 1$ , about the $y$ -axis: $S = \int_0^1 2\pi x^2\sqrt{1 + 4x^2}\,dx$ .”

What to write

“The radius to the $y$ -axis is $x$ : $S = \int_0^1 2\pi x\sqrt{1 + 4x^2}\,dx = \frac{\pi}{6}(5\sqrt 5 - 1)$ .”

Why: The wrong integral is the area about the $x$ -axis, another surface. Draw the radius perpendicular to the axis before writing: for the $y$ -axis it is horizontal, of length $x$ . About $y = -1$ it is $y + 1$ , not $y$ and not $y - 1$ .

5. Removing the square root of a square without an absolute value

2 marks, and a negative length handed in

What not to write

“ $y = \ln|\cos x|$ on $[\frac{2\pi}{3}, \pi]$ : $\sqrt{\sec^2 x} = \sec x$ , so $L = \int_{2\pi/3}^{\pi}\sec x\,dx = -\ln(2 + \sqrt 3)$ .”

What to write

“On $[\frac{2\pi}{3}, \pi]$ , $\cos x < 0$ , so $\sqrt{\sec^2 x} = -\sec x$ and $L = \ln(2 + \sqrt 3)$ .”

Why: $\sqrt{u^2} = |u|$ . Decide the sign of $u$ on the interval BEFORE removing the root. With $\cosh$ there is nothing to decide, since $\cosh u > 0$ always, which is why chains are the easy case.

6. Integrating in y with bounds in x

2 to 3 marks

What not to write

“ $y = x^{2/3}$ from $(0, 0)$ to $(8, 4)$ is $x = y^{3/2}$ , so $L = \int_0^8 \sqrt{1 + \frac{9y}{4}}\,dy$ .”

What to write

“In $y$ the bounds are $0$ and $4$ : $L = \int_0^4 \sqrt{1 + \frac{9y}{4}}\,dy = \frac{8}{27}(10\sqrt{10} - 1)$ .”

Why: Derivative, differential and bounds must all be in the same variable. Switching to $y$ is the right move here, because $\frac{dy}{dx}$ is infinite at $x = 0$ while $\frac{dx}{dy}$ is finite, but only if the bounds switch too.

7. Multiplying two antiderivatives in a surface integral

the whole question

What not to write

“ $S = 2\pi\int y\,ds = 2\pi\left(\int y\,dx\right)\left(\int \sqrt{1 + (y')^2}\,dx\right)$ , the area under the curve times its length.”

What to write

“Expand the product $y\sqrt{1 + (y')^2}$ first, then integrate term by term.”

Why: The integral of a product is not the product of the integrals. For $y = \frac{x^3}{6} + \frac{1}{2x}$ on $[1, 2]$ the true area is $\frac{47\pi}{16}$ ; the shortcut does not even give the right units.

8. Deducing a finite area from a finite volume

the whole question, and the comparison marks

What not to write

“Rotating $y = \frac1x$ , $x \ge 1$ , gives a finite volume $\pi$ , so the surface area is finite too.”

What to write

“For $x \ge 1$ , $\frac{2\pi}{x}\sqrt{1 + \frac{1}{x^4}} \ge \frac{2\pi}{x} \ge 0$ and $\int_1^\infty \frac{dx}{x}$ diverges, so by comparison $S = \infty$ .”

Gabriel's horn: the disks of radius

\frac1x

add up to a volume of

\pi

, while the bands of radius

\frac1x

add up to an infinite area.

Why: Far out, the curve is nearly flat: the area behaves like $2\pi\int y\,dx$ , the volume like $\pi\int y^2\,dx$ , and squaring a small $y$ helps convergence. For $y = x^{-p}$ the volume is finite for $p > \frac12$ , the area only for $p > 1$ .

Which method to choose

Which root opens, by the FORM of the derivative

Compute the derivative, expand $1 + (\text{derivative})^2$ , and read its form before choosing any technique

If $f' = A - B$ with $AB = \frac14$ → $1 + (f')^2 = (A + B)^2$ : integrate $A + B$

Example: $f' = \frac{x^2}{2} - \frac{1}{2x^2}$ gives $\frac{x^2}{2} + \frac{1}{2x^2}$
If $f' = \pm\tan x$ (curves $\ln\cos x$ , $\ln|\cos x|$ ) → $\sqrt{\sec^2 x} = |\sec x|$ : check the sign of $\cos x$ , then $\int\sec x$

Example: $y = \ln\cos x$ on $[0, \frac{\pi}{3}]$ : $L = \ln(2 + \sqrt 3)$
If $f' = \sinh$ (curves $a\cosh\frac{x}{a}$ ) → $\sqrt{\cosh^2} = \cosh$ , no sign to check; bounds are logarithms

Example: $y = \cosh x$ on $[0, \ln 3]$ : $L = \sinh(\ln 3) = \frac43$
If $f'$ infinite at an endpoint → switch to $x = g(y)$ , bounds in $y$

Example: $y = x^{2/3}$ becomes $x = y^{3/2}$ , $0 \le y \le 4$
If $1 + (f')^2 = 1 + cu$ with $u$ linear, or the radius supplies $u'$ → one substitution, $u = 1 + (f')^2$

Example: $y = x^{3/2}$ : $\sqrt{1 + \frac{9x}{4}}$ ; $y = x^3$ about the $x$ -axis: $u = 1 + 9x^4$
If $\sqrt{1 + u^2}$ with nothing to cancel → trigonometric substitution $u = \tan\theta$ , then $\int\sec^3\theta$

Example: $\int_0^1 \sqrt{1 + y^2}\,dy = \frac12(\sqrt 2 + \ln(1 + \sqrt 2))$

If no branch applies and the question says set up, write the integral with its bounds and stop: $\int_0^\pi \sqrt{1 + \cos^2 x}\,dx$ , the length of one arch of $\sin x$ , has no elementary antiderivative.

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.

Computing the area of a surface of revolution

When to use it: Any question that says find the area of the surface obtained by rotating a curve about an axis

1 Name the axis and write the radius as a length: $r = y$ , $r = x$ , $r = y + 1$ . Draw it perpendicular to the axis if in doubt.
2 Choose the variable in which the derivative is finite on the closed interval, and write the bounds in that variable.
3 Compute $1 + (\text{derivative})^2$ , expand it, and remove the root cleanly, with its absolute value if needed.
4 Write $S = \int 2\pi r\,ds$ in full, simplify the product $r \cdot \sqrt{\cdots}$ , then integrate.
5 Check the order of magnitude: $S$ lies between $2\pi r_{\min}L$ and $2\pi r_{\max}L$ .

Concluding sentence

“Rotating about the $y$ -axis, the radius is $x$ and $ds = \sqrt{1 + 4x^2}\,dx$ , so $S = \int_0^1 2\pi x\sqrt{1 + 4x^2}\,dx$ ; with $u = 1 + 4x^2$ , $S = \frac{\pi}{6}(5\sqrt 5 - 1)$ .”

The trap: Writing $2\pi y\,dx$ out of habit from the volume chapter, or $r = y$ for a rotation about the $y$ -axis. Step 1 and the word slant in step 4 prevent both.

Marking: Typically 1 mark for the radius, 1 for ds with the right bounds, 1 for the simplification of the root, 1 for the integration, and the order of magnitude check protects the other four.

Check before you hand in

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

Arc, chord, run

Compute the chord between the endpoints: the length must exceed it, and the chord must exceed $b - a$ . It catches the minus sign kept under the root, which returns the rise.

$y = \frac{x^3}{3} + \frac{1}{4x}$ on $[1, 3]$ : run $2$ $<$ chord $\frac{\sqrt{305}}{2}$ $<$ $L = \frac{53}{6}$ ; the wrong $\frac{17}{2}$ is below the chord.
Test the formula on a segment or a circle

When unsure of a formula, apply it to a curve whose answer is known without calculus.

$y = \sqrt{4 - x^2}$ on $[0, 1]$ : $2\arcsin\frac12 = \frac{\pi}{3} = 2 \cdot \frac{\pi}{6}$ , radius times angle.
The surface between two bounds

The area of a surface about the $x$ -axis lies between $2\pi y_{\min}L$ and $2\pi y_{\max}L$ .

$y = \frac{x^3}{6} + \frac{1}{2x}$ on $[1, 2]$ : $L = \frac{17}{12}$ , $y$ between $\frac23$ and $\frac{19}{12}$ , and $\frac{47\pi}{16}$ lies between $2\pi \cdot \frac23 \cdot \frac{17}{12}$ and $2\pi \cdot \frac{19}{12} \cdot \frac{17}{12}$ .
Signs and the root of a square

A length or an area is positive. A negative answer means an absolute value was dropped, never that the arc runs backwards.

$\int_{2\pi/3}^{\pi}\sec x\,dx = -\ln(2 + \sqrt 3) < 0$ : the root was $-\sec x$ on that interval.

The typical problem, taken apart

Length and surface of the same arc, with a root built to open

Let $y = \frac{x^3}{3} + \frac{1}{4x}$ for $1 \le x \le 3$ . Find the length of this arc, then the area of the surface obtained by rotating it about the $x$ -axis.

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

Step 1

$f'(x) = x^2 - \frac{1}{4x^2}$ . With $A = x^2$ and $B = \frac{1}{4x^2}$ , $AB = \frac14$ .

Why

Naming $A$ and $B$ and checking their product takes ten seconds and tells you the root will open before you expand anything.

Step 2

$1 + (f')^2 = x^4 - \frac12 + \frac{1}{16x^4} + 1 = x^4 + \frac12 + \frac{1}{16x^4} = \left(x^2 + \frac{1}{4x^2}\right)^2$ , and the bracket is positive, so $ds = \left(x^2 + \frac{1}{4x^2}\right)dx$ .

Why

The minus sign in $f'$ becomes a plus sign in $ds$ . The positivity of the bracket is what allows removing the root without an absolute value, and it is worth saying.

Step 3

$L = \int_1^3 \left(x^2 + \frac{1}{4x^2}\right)dx = \left[\frac{x^3}{3} - \frac{1}{4x}\right]_1^3 = \left(9 - \frac{1}{12}\right) - \left(\frac13 - \frac14\right) = \frac{53}{6}$ .

Why

The antiderivative is not $f$ : its second term has the opposite sign. Chord check: $\frac{\sqrt{305}}{2} < \frac{53}{6}$ , and the run is only $2$ .

Step 4

$S = 2\pi\int_1^3 \left(\frac{x^3}{3} + \frac{1}{4x}\right)\left(x^2 + \frac{1}{4x^2}\right)dx = 2\pi\int_1^3 \left(\frac{x^5}{3} + \frac{x}{3} + \frac{1}{16x^3}\right)dx$ .

Why

The radius is $y$ because the axis is the $x$ -axis, and the width is $ds$ found above. Expanding the product first is the only safe way: $\frac{x}{12} + \frac{x}{4} = \frac{x}{3}$ .

Step 5

$S = 2\pi\left[\frac{x^6}{18} + \frac{x^2}{6} - \frac{1}{32x^2}\right]_1^3 = 2\pi\left(\frac{364}{9} + \frac43 + \frac{1}{36}\right) = 2\pi \cdot \frac{1505}{36} = \frac{1505\pi}{18}$ .

Why

Order of magnitude: $y$ runs from $\frac{7}{12}$ to about $9$ along a length of $\frac{53}{6}$ , so $S$ must lie between $2\pi \cdot \frac{7}{12} \cdot \frac{53}{6}$ and $2\pi \cdot 9.1 \cdot \frac{53}{6}$ , and it does.

The conclusion, written out

“ $L = \frac{53}{6}$ and $S = \frac{1505\pi}{18}$ .”

The classic mistake on this problem: Writing $\sqrt{1 + (f')^2} = x^2 - \frac{1}{4x^2}$ , which gives the rise $\frac{17}{2}$ instead of the length, then $2\pi\int y\,dx$ for the area, which ignores the slant of the bands.

Learn by heart

• $ds^2 = dx^2 + dy^2$ ; $L = \int \sqrt{1 + (f')^2}\,dx$ or $\int \sqrt{1 + (g')^2}\,dy$ , bounds in the same variable.
• $f' = A - B$ with $AB = \frac14$ : $1 + (f')^2 = (A + B)^2$ . The minus inside, the plus outside.
• $\sqrt{u^2} = |u|$ . $1 + \tan^2 = \sec^2$ , $1 + \sinh^2 = \cosh^2$ , and $\cosh > 0$ .
• $\int\sec x\,dx = \ln|\sec x + \tan x| + C$ ; $\sinh(\ln a) = \frac{a - 1/a}{2}$ , $\cosh(\ln a) = \frac{a + 1/a}{2}$ .
• $s(x) = \int_a^x \sqrt{1 + (f'(t))^2}\,dt$ and $s'(x) = \sqrt{1 + (f'(x))^2}$ .
• $S = \int 2\pi r\,ds$ : $r = y$ about the $x$ -axis, $r = x$ about the $y$ -axis, the distance to the line otherwise.
• Sphere zone: $2\pi rh$ . Cone: $\pi r\ell$ . Arc $\ge$ chord $\ge$ run.
• Gabriel's horn: volume $\pi$ , area infinite. Finite volume never implies finite area.

Frequently asked questions

How do I find the arc length of a curve by hand in MATH 141?

Differentiate, then expand one plus the derivative squared before doing anything else. On an exam without a calculator it is built to be a perfect square, or one plus tangent squared, or one plus hyperbolic sine squared, so the square root opens. Remove it with its absolute value, integrate, and check that the length is longer than the straight chord between the endpoints.

Why is the surface area formula not just 2 pi y dx like the volume?

Because a thin band of the surface is slanted: its width is the length element ds, not dx, and ds is dx times the square root of one plus the derivative squared. That factor does not shrink to one as the bands get thinner. For a volume the slant only changes a slice by a negligible amount, which is why dx works there. A cone shows it: dx gives pi r h, the true area is pi r times the slant.

When should I compute arc length in y instead of x?

When the derivative dy over dx is infinite somewhere on the interval, typically a vertical tangent at an endpoint, as for y equals x to the power two thirds at the origin. Write x as a function of y, check that dx over dy is finite, and change the bounds to values of y. The derivative, the differential and the bounds must all be in the same variable.

How can Gabriel's horn have a finite volume but an infinite surface area?

The volume adds up disks whose area is pi over x squared, and that integral converges. The surface adds up bands whose area is at least two pi over x, and the integral of one over x diverges. Squaring the small radius makes the volume converge while the area does not. There is no contradiction: the paint that fills the horn gets thinner than any coat as the tube narrows.

Do I need to know hyperbolic functions for MATH 141?

You need what the exam gives you: cosh and sinh defined with exponentials, their derivatives, and the identity cosh squared minus sinh squared equals one, which you can prove from the definitions in two lines. That is enough for every hanging cable question. The bounds are then logarithms, and e to the ln of a is a, so the answer comes out exact.

Practise it

Corrected exercises: Arc length and surface area, 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

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Revision sheet: arc length and area of a surface of revolution (MATH 141)

The essentials

One triangle, one root

Surfaces: radius times slant

The rules in table form

The radius, axis by axis

The mistakes that cost marks

1. Writing one plus f prime instead of the root

2. Keeping the minus sign when the square opens

3. Using 2 pi y dx for an area, the volume reflex

4. Taking y as the radius for a rotation about the y-axis

5. Removing the square root of a square without an absolute value

6. Integrating in y with bounds in x

7. Multiplying two antiderivatives in a surface integral

8. Deducing a finite area from a finite volume

Which method to choose

Which root opens, by the FORM of the derivative

How the answer is expected to be written

Computing the area of a surface of revolution

Check before you hand in

The typical problem, taken apart

Length and surface of the same arc, with a root built to open

Learn by heart

Frequently asked questions

How do I find the arc length of a curve by hand in MATH 141?

Why is the surface area formula not just 2 pi y dx like the volume?

When should I compute arc length in y instead of x?

How can Gabriel's horn have a finite volume but an infinite surface area?

Do I need to know hyperbolic functions for MATH 141?

Corrected exercises: Arc length and surface area, MATH 141 at McGill

See also

Looking for a MATH 141 tutor in Montreal?