Arc length from the Pythagorean theorem on an infinitesimal triangle, the formula in either variable, why so few arc length integrals are elementary, the surface area of a solid of revolution, and the frustum's slant height.
Subject: Calculus I · 65 slides · symbolic lesson
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Title
Calculus I · Chapter 6 — Applications of Integration
Arc Length of a Curve and Surface Area
Objectives
Five outcomes. The derivation is one triangle, and most of the difficulty is in the integrals it produces.
OpenStax Calculus Volume 1, §6.4 Arc Length of a Curve and Surface Area §6.4, pp. 582-593 — the section these objectives are drawn from
Warm-up
Sections 6.1 to 6.3 all used the same discipline: describe one representative piece, then integrate. The pieces were strips, slabs and shells.
Discussion prompt
For the length of a curve, what is the representative piece, and how long is it?
Hint: A very short piece of a smooth curve looks straight.
Answer:
A short piece of curve is almost a straight segment, and it is the hypotenuse of a tiny right triangle whose legs are the horizontal and vertical changes. The Pythagorean theorem gives its length.
\[ ds = \sqrt{(dx)^{2}+(dy)^{2}} = \sqrt{1+\left(\frac{dy}{dx}\right)^{2}}\,dx \]
Factoring dx out of the root is the whole derivation, and integrating adds the pieces. The formula is short; what is not short is the integrals it produces, and a surprising number of them have no elementary answer at all.
Concept
Over a tiny interval a smooth curve is indistinguishable from a straight segment, whose length the Pythagorean theorem gives. Integrating that length over the curve's extent gives the arc length.
arc length — The length of a curve, computed as the integral of the square root of one plus the derivative squared, over the interval the curve spans.
\[ L = \int_{a}^{b}\sqrt{1+\left[f'(x)\right]^{2}}\,dx \]
The square root is what makes these integrals hard. Squaring a derivative and adding one rarely produces something with an elementary antiderivative, so most arc lengths must be computed numerically.
Figure (svg): A short piece of curve as the hypotenuse of a tiny right triangle
OpenStax Calculus Volume 1, §6.4 Arc Length of a Curve and Surface Area §6.4, pp. 668-678
Section
Section 1
Concept
The representative piece is a short segment whose horizontal leg is the differential and whose vertical leg is the derivative times it. Its length follows immediately.
the differential of arc length — The length of a short piece of curve, equal to the square root of the sum of the squares of the horizontal and vertical changes.
\[ ds = \sqrt{1+\left[f'(x)\right]^{2}}\,dx \]
The derivation also explains the formula's shape: the one comes from the horizontal leg and the derivative squared from the vertical one, so a flat curve gives a root of one and a length equal to the interval.
Figure (svg): A short piece of curve as the hypotenuse of a tiny right triangle
OpenStax Calculus Volume 1, §6.4 Arc Length of a Curve and Surface Area §6.4, pp. 668-678 — arc length of the curve y = f(x)
Picture it
The derivation in a picture.
Figure (svg): A short piece of curve as the hypotenuse of a tiny right triangle
Factoring the differential out of the square root converts the two-differential expression into something integrable in one variable, and that is the only algebraic step involved.
Worked example
Example 6.19. A three-halves power, chosen to work.
\[ \text{Find the length of } y=x^{3/2} \text{ on } [0,1]. \]
Differentiate
Why: The power rule.
\[ y' = (\frac{3}{2}) x ^{\frac{1}{2}} \]
Square it
Why: The half power squares away.
\[ 9 x / 4 \]
Form the integrand
Why: Add one and take the root.
\[ \sqrt{1 + 9 x / 4} \]
Substitute
Why: The radicand is linear.
\[ u = 1 + 9 x / 4 \]
Integrate and evaluate
Why: The power rule with a three-halves exponent.
\[ (\frac{8}{27}) [(\frac{13}{4}) ^{\frac{3}{2}} - 1] \]
Figure (svg): A short piece of curve as the hypotenuse of a tiny right triangle
\[ L = \frac{8}{27}\left[\left(\frac{13}{4}\right)^{3/2}-1\right] \approx 1.44 \]
Verify: sanity-check against the straight-line distance
Why: The curve runs from the origin to the point one comma one, a straight-line distance of about 1.414 — and the arc, being curved, must be slightly longer, which 1.44 is. That check is available for every arc length: the answer must exceed the distance between the endpoints. Note why this integral worked: squaring a half power gave a linear radicand, and the three-halves exponent was chosen precisely for that.
OpenStax Calculus Volume 1, §6.4 Arc Length of a Curve and Surface Area §6.4, pp. 672-674
Fill the middle
Arc length from the Pythagorean theorem.
Fill in the blanks
ds = \sqrt1+\left[f'(x)\right]^___}\,dx
Why: The one comes from the horizontal leg after the differential is factored out. Without it a horizontal line would have zero length, which is the quickest way to check the formula.
Worked example
Checkpoint 6.19. Where the formula comes from.
\[ \text{Approximate the length of } y=x^{2} \text{ on } [0,2] \text{ with chords.} \]
Divide the interval
Why: Into n equal pieces.
Join consecutive points
Why: Straight chords.
Compute each chord's length
Why: Pythagoras.
\[ \sqrt{\,dx ^{2} + \,dy ^{2}} \]
Sum and refine
Why: More chords.
\[ 4.472, 4.616, 4.639 \]
Take the limit
Why: The integral.
\[ \text{about } 4.647 \]
Figure (svg): Arc length as a limit of polygonal approximations
\[ L = \int_{0}^{2}\sqrt{1+4x^{2}}\,dx \approx 4.647 \]
Verify: explain why the approximations are always low
Why: A straight chord is the shortest path between its endpoints, so the polygon is always shorter than the curve — every finite approximation underestimates, and the sequence increases toward the true length. That one-sidedness is unlike Section 5.1's rectangles, which could err either way, and it means a chord sum is always a guaranteed lower bound on the length.
OpenStax Calculus Volume 1, §6.4 Arc Length of a Curve and Surface Area §6.4, pp. 674-676
Trap
\[ L = \int_{0}^{1}\sqrt{\left[f'(x)\right]^{2}}\,dx = \int_{0}^{1}|f'(x)|\,dx \]
Take the root of the derivative squared alone
Why: The student drops the constant term.
That integrates the vertical change only, ignoring the horizontal. For a horizontal line it would give zero length.
\[ L = \int_{0}^{1}\sqrt{1+\left[f'(x)\right]^{2}}\,dx \]
Both legs of the triangle appear
Why: The one is the horizontal leg, squared.
The check is a horizontal line: its length over an interval must equal the interval's width, and only the correct formula gives that. Testing a formula on a case whose answer is known is worth the few seconds.
Ranking
From a chord to the integral.
Put in order
Why: Step c is the only algebraic move and it is what turns an expression in two differentials into one that can be integrated. Everything else is geometry or Chapter 5 routine.
Two truths and a lie
All three are about the formula.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. That expression accounts only for the vertical change, and it would say a horizontal line has no length at all. Both legs of the triangle contribute, which is what the one plus the derivative squared records.
Prediction
Commit before reasoning.
Predict first
Applying the formula to a line of slope m over an interval of width w gives what?
Correct: The Pythagorean distance.
\[ L = w\sqrt{1+m^{2}} = \sqrt{w^{2}+(mw)^{2}} \]
Why: The derivative is the constant m, so the integrand is a constant root and the integral is that root times the width — which is exactly the straight-line distance between the endpoints, horizontal change and vertical change combined. Recovering elementary geometry is the strongest check that a formula in this chapter has been set up correctly.
Section
Section 2
Concept
A curve can be described as a function of either variable, and the two arc length integrands are generally different. Where one has a vertical tangent, the other is smooth.
arc length in y — The same formula with the roles of the variables exchanged: the integrand is the square root of one plus the derivative of x with respect to y, squared.
\[ L = \int_{c}^{d}\sqrt{1+\left[g'(y)\right]^{2}}\,dy \]
A vertical tangent makes the derivative in x unbounded and the integral improper, while the same point is perfectly ordinary in the other description. Switching costs nothing and can remove the difficulty entirely.
Figure (svg): Arc length in the other variable: sometimes far easier
OpenStax Calculus Volume 1, §6.4 Arc Length of a Curve and Surface Area §6.4, pp. 674-682 — arc length of the curve x = g(y)
Picture it
One awkward, one routine.
Figure (svg): Arc length in the other variable: sometimes far easier
The left description has an unbounded derivative at the origin and the right does not. The curve is unchanged; only the choice of which variable is the input differs.
Worked example
Example 6.21. A vertical tangent avoided.
\[ \text{Find the length of } y=x^{2/3} \text{ on } [0,1]. \]
Differentiate in x
Why: The power rule.
\[ y' = (\frac{2}{3}) x ^{-\frac{1}{3}} \]
Note the difficulty
Why: It is unbounded at zero.
Rewrite the curve in y
Why: Solve for x.
\[ x = y ^{\frac{3}{2}} \]
Differentiate in y
Why: The power rule.
\[ x' = (\frac{3}{2}) y ^{\frac{1}{2}} \]
Form the integrand and integrate
Why: A linear radicand.
\[ \text{about } 1.44 \]
Figure (svg): Arc length in the other variable: sometimes far easier
\[ L = \int_{0}^{1}\sqrt{1+\tfrac94 y}\,dy \approx 1.44 \]
Verify: notice that the two curves are the same one
Why: The curve given as y equal to x to the two thirds on the interval from 0 to 1 is exactly the curve x equal to y to the three halves on the same interval — the same set of points, and this section's first worked example computed its length as 1.44 in the other orientation. So the answers agree, as they must. What changed was only which variable was treated as the input, and that turned an improper integral into a routine one.
OpenStax Calculus Volume 1, §6.4 Arc Length of a Curve and Surface Area §6.4, pp. 677-679
Fill the middle
A curve rewritten as a function of y.
Fill in the blanks
x=y^1/2 \;\Longrightarrow\; \frac______ = \frac32 y^___}
Why: The derivative must be taken with respect to the variable of integration. Squaring this half power gives a linear radicand, which is exactly why this orientation is easier.
Worked example
Checkpoint 6.21. Neither description is smooth throughout.
\[ \text{How would you find the length of a circle's upper half?} \]
Write it as a function of x
Why: The upper semicircle.
\[ y = \sqrt{r ^{2} - x ^{2}} \]
Differentiate
Why: The chain rule.
\[ y' = -x / \sqrt{r ^{2} - x ^{2}} \]
Note the endpoints
Why: Vertical tangents at both.
Consider the other variable
Why: The same problem at the top.
Note the resolution
Why: Split, or use parametric form.
Figure (svg): The solution to Worked example a curve with a vertical tangent inside shown as a ladder of expressions, one row per legal move
\[ L = \int_{-r}^{r}\frac{r\,dx}{\sqrt{r^{2}-x^{2}}} = \pi r \]
Verify: check the answer against known geometry
Why: Half a circle of radius r has length pi times r, and the integral gives exactly that — it is the arcsine form from Section 5.7, evaluated between the endpoints where the arcsine reaches plus and minus a right angle. So the improper integral converges to the right thing. Note that this is the first place a non-function curve appears: no single description in either variable covers a whole circle, which is what parametric equations exist for.
OpenStax Calculus Volume 1, §6.4 Arc Length of a Curve and Surface Area §6.4, pp. 679-681
Error analysis
A student computes arc length after switching variables.
Annotate
On: \( x=y^{3/2} \;\Longrightarrow\; L = \int\sqrt{1+\left[\tfrac23 x^{-1/3}\right]^{2}}\,dy \)
This is Section 5.5's variable-consistency rule once more: integrand, derivative, differential and limits must all refer to one variable. Switching some but not all leaves an expression with no meaning.
Sorting
Look for vertical tangents.
Sort into buckets
Sort each curve by the better description.
The rule is to describe the curve so that its derivative stays bounded. A vertical tangent in one orientation is a horizontal tangent in the other, and horizontal tangents cause no trouble at all.
Two truths and a lie
All three are about choosing.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. The integrands are generally quite different — one may be improper where the other is a routine substitution — even though both integrals evaluate to the same number. That the answers must agree while the work need not is what makes choosing worthwhile.
Prediction
Commit before reasoning.
Predict first
A curve has a vertical tangent at a point. Why does describing it in the other variable remove the difficulty?
Correct: Because it becomes a horizontal tangent.
\[ \frac{dy}{dx}\to\infty \;\Longleftrightarrow\; \frac{dx}{dy}\to 0 \]
Why: A vertical tangent means the derivative with respect to x is unbounded, making the integrand blow up. Exchanging the roles of the variables turns that same point into a place where the derivative is zero, and a zero derivative gives an integrand of exactly one — the smallest and most harmless value it can take. The curve is unchanged; only its description is.
Section
Section 3
Concept
The integrand is a square root of one plus a squared derivative, and such expressions almost never have elementary antiderivatives. Curves with computable arc length are the exception.
non-elementary arc length — Most arc length integrals have no antiderivative in the elementary functions. The sine's arc length defines the elliptic integral, which is studied as a function in its own right.
\[ \int\sqrt{1+\cos^{2}x}\,dx \quad \text{is elliptic, not elementary} \]
Textbook exercises are chosen from the small family that works — three-halves powers, and a few contrived combinations — which can leave a misleading impression of how typical they are.
Figure (svg): Why so few arc length integrals are elementary
OpenStax Calculus Volume 1, §6.4 Arc Length of a Curve and Surface Area §6.4, pp. 676-684 — the difficulty of arc length integrals
Picture it
From trivial to non-elementary.
Figure (svg): Why so few arc length integrals are elementary
The fourth row is the ordinary case. Even a sine's length required mathematicians to define a new class of functions, which is a fair indication of how special the workable examples are.
Worked example
Example 6.20. The exponent chosen to make the radicand linear.
\[ \text{Why is the arc length of } y=x^{3/2} \text{ elementary?} \]
Differentiate
Why: The power rule.
\[ y' = (\frac{3}{2}) x ^{\frac{1}{2}} \]
Square it
Why: The half exponent doubles.
\[ (\frac{9}{4}) x \]
Note the result
Why: Linear in x.
Recognise the integrand
Why: A root of a linear function.
Generalise
Why: Any exponent whose double is 1.
Figure (svg): Why so few arc length integrals are elementary
\[ \left[\tfrac32x^{1/2}\right]^{2} = \tfrac94x \]
Verify: test the pattern on a nearby exponent
Why: For y equal to x squared the derivative is 2x, whose square is 4x squared — giving a root of one plus four x squared, which needs a trigonometric substitution and produces an inverse hyperbolic function. For y equal to x cubed the radicand is one plus nine x to the fourth, which is not elementary at all. So moving the exponent by a small amount destroys the property, and the three-halves power is genuinely special rather than representative.
OpenStax Calculus Volume 1, §6.4 Arc Length of a Curve and Surface Area §6.4, pp. 675-677
Sorting
Square the derivative and look at the radicand.
Sort into buckets
Sort each curve's arc length integral.
The third is elementary but only just: it needs a trigonometric substitution beyond this course and produces an inverse hyperbolic function. The boundary between workable and not is narrow, and it does not correspond to how simple a curve looks.
Worked example
Checkpoint 6.20. The ordinary case.
\[ \text{Find the length of } y=\sin x \text{ on } [0,\pi]. \]
Differentiate
Why: Standard.
\[ y' = \cos x \]
Form the integrand
Why: Square and add one.
\[ \sqrt{1 + \cos ^{2} x} \]
Attempt an antiderivative
Why: No substitution applies.
Identify the integral
Why: A named class.
Compute numerically
Why: Section 5.1's methods.
\[ \text{about } 3.820 \]
Figure (svg): The solution to Worked example a length that must be numerical shown as a ladder of expressions, one row per legal move
\[ L = \int_{0}^{\pi}\sqrt{1+\cos^{2}x}\,dx \approx 3.820 \]
Verify: sanity-check against bounds
Why: The straight-line distance from the origin to the point pi comma zero is about 3.142, so the arc must be longer — and the curve is bounded above by the two-segment path up to the peak and back down, which has length about 3.724 by Pythagoras. The true value of 3.820 exceeds both, which reveals the second bound was itself an underestimate as a chord path must be. Numerical methods give the answer to any precision even though no formula exists.
OpenStax Calculus Volume 1, §6.4 Arc Length of a Curve and Surface Area §6.4, pp. 677-679
Trap
\[ \int\sqrt{1+\cos^{2}x}\,dx: \; \text{try } u=\cos x, \text{ then } u=\sin x, \ldots \]
Keep searching for a substitution
Why: The student assumes a technique must exist.
No elementary antiderivative exists. This is an elliptic integral, and its impossibility was proved rather than merely observed.
\[ \text{compute numerically: } L \approx 3.820 \]
Recognise the shape and switch to a numerical method
Why: Most arc lengths must be computed this way.
The signal is a root of one plus a squared trigonometric or non-linear expression. Recognising it early saves the search, and Section 5.1's rectangles give as many digits as wanted.
Fill the middle
The three-halves power, chosen so the radicand is linear.
Fill in the blanks
\left[\tfrac32x^1\right]^___ = \tfrac94 x^___}
Why: Squaring a half power gives a first power, so the radicand is linear and a simple substitution finishes it. That is the entire reason this exponent appears in textbook exercises.
Two truths and a lie
All three are about the difficulty.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. The sine is about as simple as a curve gets and its length is not elementary; the exponential is likewise. Simplicity of the curve and simplicity of its arc length integral are close to unrelated, which is worth knowing before spending long on a search.
Prediction
Commit before reasoning.
Predict first
Why do arc length integrals so often fail to be elementary when volume integrals rarely do?
Correct: Because of the square root.
\[ \int\left[f\right]^{2}dx \text{ vs } \int\sqrt{1+\left[f'\right]^{2}}dx \]
Why: A volume integrand squares the boundary, which turns a polynomial into a polynomial and a square root into something rational — it simplifies. An arc length integrand squares the DERIVATIVE, adds one, and takes a root, which almost always produces something outside the elementary functions. The same curve can have a one-line volume of revolution and a length with no closed form at all.
Section
Section 4
Concept
Revolving a short arc about an axis sweeps a narrow band whose area is its circumference times its arc length. Integrating gives the surface area.
surface of revolution — The surface swept when a curve is revolved about an axis. Its area is the integral of the circumference times the differential of arc length.
\[ S = 2\pi\int_{a}^{b}f(x)\sqrt{1+\left[f'(x)\right]^{2}}\,dx \]
The integrand is the arc length integrand times a circumference, so surface integrals inherit all of arc length's difficulty and add a factor. They are the hardest integrals in this chapter.
Figure (svg): Surface area against volume: the same solid, different integrands
OpenStax Calculus Volume 1, §6.4 Arc Length of a Curve and Surface Area §6.4, pp. 680-690 — area of a surface of revolution
Picture it
The same solid, two integrands.
Figure (svg): Surface area against volume: the same solid, different integrands
The volume integrand is a polynomial when the boundary is; the surface integrand carries a square root regardless. That asymmetry is why a solid can have an easy volume and an impossible surface area.
Worked example
Example 6.23. The band's area.
\[ \text{Find the surface area when } y=\sqrt{x} \text{ on } [1,4] \text{ is revolved about the } x\text{-axis.} \]
Differentiate
Why: The power rule.
\[ y' = \frac{1}{2 \sqrt{x}} \]
Form the arc length factor
Why: Square and add one.
\[ \sqrt{1 + \frac{1}{4 x}} \]
Multiply by the circumference
Why: Twice pi times the radius.
Simplify the product
Why: The roots combine.
\[ 2 \pi \sqrt{x + \frac{1}{4}} \]
Integrate from 1 to 4
Why: A power rule substitution.
\[ \text{about } 30.85 \]
Figure (svg): Surface area against volume: the same solid, different integrands
\[ S = 2\pi\int_{1}^{4}\sqrt{x+\tfrac14}\,dx \approx 30.85 \]
Verify: notice the simplification and check the size
Why: Multiplying the square root of x by the square root of one plus one over four x combined into a single root of x plus a quarter — which is what made the integral elementary. Without that cancellation the integrand would have been much worse. As a size check, the surface lies between the cylinder of radius 1 and the one of radius 2 over the same length, whose lateral areas are about 18.8 and 37.7, and 30.85 sits between them.
OpenStax Calculus Volume 1, §6.4 Arc Length of a Curve and Surface Area §6.4, pp. 685-687
Fill the middle
A band's area is circumference times arc length.
Fill in the blanks
S = 2\pi\int f(x)\sqrtf'(x)}\right]^___}\,dx
Why: The band's width along the surface is the differential of arc length, not the horizontal differential. Omitting the factor always underestimates, since the root is at least one.
Worked example
Checkpoint 6.23. Different difficulty.
\[ \text{Compare the volume and surface integrals for } y=x^{2} \text{ on } [0,1] \text{ about the } x\text{-axis.} \]
Write the volume integrand
Why: Disks.
\[ \pi x ^{4} \]
Evaluate
Why: The power rule.
\[ \frac{\pi}{5} \]
Write the surface integrand
Why: Circumference times arc length.
\[ 2 \pi x ^{2} \sqrt{1 + 4 x ^{2}} \]
Attempt it
Why: A trigonometric substitution.
Compare
Why: One line against a page.
Figure (svg): The solution to Worked example surface against volume for one solid shown as a ladder of expressions, one row per legal move
\[ V = \frac{\pi}{5}, \qquad S = 2\pi\int_{0}^{1}x^{2}\sqrt{1+4x^{2}}\,dx \]
Verify: state the general lesson
Why: The volume was a one-line polynomial integral and the surface required a technique beyond this course. That gap is typical rather than special: squaring a boundary simplifies it while squaring a derivative and taking a root does not. Anyone estimating how long a problem will take should look at whether a square root survives into the integrand, since that single feature usually decides it.
OpenStax Calculus Volume 1, §6.4 Arc Length of a Curve and Surface Area §6.4, pp. 687-689
Error analysis
A student computes a surface area.
Annotate
On: \( S = 2\pi\int_{1}^{4}\sqrt{x}\,dx \)
The omission always underestimates, since the missing factor is at least one. It is the surface-area counterpart of using a vertical height where a slant height belongs, and it comes from the same confusion.
Matching
Each part of the surface integrand.
Match the pairs
Why: The middle two together give the arc length of the band's width, and the first gives how far it travels around. Their product is the band's area, exactly as an unrolled strip's would be.
Sorting
Look for a surviving square root.
Sort into buckets
Sort each computation for a curve given as a polynomial.
The pattern is clean: anything measuring length along the curve carries the root, and anything measuring area or volume across it does not. Knowing which kind of quantity is wanted predicts the difficulty before any setup.
Prediction
Commit before reasoning.
Predict first
Leaving the arc length factor out of a surface integral has what effect?
Correct: It always underestimates.
\[ \sqrt{1+\left[f'\right]^{2}} \ge 1, \text{ with equality only where } f'=0 \]
Why: The factor is the square root of one plus a squared quantity, so it is at least one and exceeds one wherever the curve is not horizontal. Dropping it therefore always reduces the integrand and the answer. The systematic direction of the error is a useful diagnostic: a surface area that comes out suspiciously close to a cylinder's suggests the factor was left out.
Section
Section 5
Concept
The band swept by a short arc is a frustum, whose area involves its slant height. That slant height is the arc length, which is where the square root comes from.
frustum — A slice of a cone between two parallel circles. Its lateral surface area is twice pi times the average radius times the slant height — never the vertical height.
\[ S_{\text{frustum}} = 2\pi \bar{r}\,s \]
Using the vertical height instead is the section's characteristic error, and it is the same mistake as omitting the arc length factor — the two are different descriptions of one confusion.
Figure (svg): A frustum: the surface swept by a straight segment revolved about an axis
OpenStax Calculus Volume 1, §6.4 Arc Length of a Curve and Surface Area §6.4, pp. 682-690 — the frustum and slant height
Picture it
Radii, slant height, and area.
Figure (svg): A frustum: the surface swept by a straight segment revolved about an axis
The slant height runs along the sloping side, and it is longer than the vertical height whenever the cone is not a cylinder. Only the slant height measures what the surface actually spans.
Worked example
Example 6.24. Deriving a known formula.
\[ \text{Find the lateral surface of a cone of radius } R \text{ and height } h \text{ by revolving a line.} \]
Write the line
Why: From the apex.
\[ y = R x / h\text{ on } [0, h] \]
Differentiate
Why: A constant slope.
\[ y' = \frac{R}{h} \]
Form the arc length factor
Why: Constant.
\[ \sqrt{1 + R ^{2} / h ^{2}} \]
Recognise it
Why: The slant height over the height.
\[ \frac{s}{h}\text{ where } s ^{2} = R ^{2} + h ^{2} \]
Integrate
Why: The circumference times a constant.
Figure (svg): A frustum: the surface swept by a straight segment revolved about an axis
\[ S = \pi R\sqrt{R^{2}+h^{2}} = \pi Rs \]
Verify: compare with the classical formula
Why: Elementary geometry gives a cone's lateral surface as pi times the radius times the slant height, which is exactly what came out — and the slant height appeared naturally as the arc length factor times the vertical extent. Using the vertical height instead would give pi R h, which is smaller and wrong. That the formula reproduces school geometry is the strongest possible check on the method.
OpenStax Calculus Volume 1, §6.4 Arc Length of a Curve and Surface Area §6.4, pp. 686-688
Fill the middle
A frustum's lateral surface.
Fill in the blanks
S = 2\pi\bars\,___
Why: The slant height is what the surface spans. Using the vertical height instead always underestimates, and the error grows as the cone flattens.
Worked example
Checkpoint 6.24. Slant against vertical.
\[ \text{For a cone with } R=3 \text{ and } h=4, \text{ compare the correct and incorrect surfaces.} \]
Find the slant height
Why: Pythagoras.
\[ s = 5 \]
Compute the correct surface
Why: pi times R times s.
\[ 15 \pi,\text{ about } 47.1 \]
Compute the wrong one
Why: Using the height instead.
\[ 12 \pi,\text{ about } 37.7 \]
Find the shortfall
Why: The ratio.
\[ 20 \% l o w \]
Note when it worsens
Why: A shallower cone.
Figure (svg): The solution to Worked example how much the error costs shown as a ladder of expressions, one row per legal move
\[ 15\pi \text{ against } 12\pi \]
Verify: see how the error scales
Why: For a cone of radius 3 and height 1 the slant height is about 3.16 while the height is 1, so the wrong formula would be about 68 percent low — the error grows without bound as the cone flattens. Conversely for a very tall thin cone the slant height and height are nearly equal and the error is small. So the mistake is worst precisely where surfaces are widest and matter most, which is a bad combination in any practical setting.
OpenStax Calculus Volume 1, §6.4 Arc Length of a Curve and Surface Area §6.4, pp. 688-690
Trap
\[ S = 2\pi\bar{r}h \]
Use the frustum's vertical height
Why: The student takes the height of the band across the axis.
The surface spans along the slope, not across the axis. The vertical height is shorter whenever the side is not vertical.
\[ S = 2\pi\bar{r}s, \quad s = \sqrt{(\Delta r)^{2}+h^{2}} \]
Use the slant height, which is the arc length
Why: That is what the surface actually spans.
Testing on a cone settles it: the classical formula uses the slant height, and only that choice reproduces it. Checking a new formula against a case whose answer is already known is worth the moment it takes.
Sorting
Which length does each quantity use?
Sort into buckets
Sort each formula.
A cylinder is the case where the two coincide, since its side is vertical — which is exactly why cylinders make a poor test case for telling the two apart. A cone shows the difference immediately.
Two truths and a lie
All three are about the slant height.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. They are equal only for a cylinder, whose side is vertical. For a cone of radius 3 and height 4 the slant height is 5, and the resulting surface differs by 20 percent — larger for flatter cones.
Prediction
Commit before reasoning.
Predict first
Why does the surface area formula contain the same square root as arc length?
Correct: Because the band's extent is the arc length.
\[ dS = 2\pi f(x)\,ds, \quad ds = \sqrt{1+\left[f'\right]^{2}}\,dx \]
Why: The band swept by a short arc is a frustum, and a frustum's area is its average circumference times its slant height — which is exactly the arc length of the piece that swept it. So the arc length factor is not an extra complication bolted on: it is the frustum formula, written in the notation of this chapter. That identification is what makes the surface formula memorable rather than arbitrary.
Comparison
Fill the blanks. Only one of these carries a square root.
Comparison matrix
| Quantity | Representative piece | Integrand |
|---|---|---|
| Area under a curve | a strip | f(x) |
| Volume of revolution | a disk | pi f(x)^2 |
| Arc length | a short segment | sqrt(1 + [f'(x)]^2) |
| Surface of revolution | a frustum band | 2 pi f(x) sqrt(1 + [f'(x)]^2) |
The bottom two rows carry the root and the top two do not, which predicts their difficulty before any setup. Anything measuring length along the curve inherits it.
Pattern
Given a curve whose length or surface area is wanted.
Step five is not a failure. Most arc length integrands genuinely have no elementary antiderivative, and recognising that early saves a long and fruitless search.
Stewart, Calculus: Early Transcendentals 8e, §8.1 Arc Length §8.1, pp. 544-550
Check
The formula.
Check your understanding
What is the arc length integrand for y = f(x)?
Answer: A
Why: The Pythagorean theorem on a short segment, with the differential factored out.
Check
Difficulty.
Check your understanding
Why do so few arc length integrals have elementary antiderivatives?
Answer: A
Why: Even the sine's arc length is an elliptic integral rather than an elementary one.
Check
Surface area.
Check your understanding
A frustum's lateral surface uses which length?
Answer: A
Why: The surface spans along the slope, and the slant height is the arc length of the segment that swept it.
Real world
A roofing contractor is quoting for a curved barrel roof whose cross-section is an arc. The client has given the building's width and the roof's rise at the centre, and the quote depends on the area of material needed.
Discussion prompt
Explain why the width is not enough, what quantity is actually needed, and how the estimate is produced.
Hint: The material lies along the curve, not across the building.
Answer:
The roofing material spans along the curve, so the quantity needed is the arc length of the cross-section, multiplied by the building's length. The building's width is the horizontal extent and is strictly less than the arc — quoting on width alone always under-orders material.
\[ \text{material} = L\int_{-w/2}^{w/2}\sqrt{1+\left[f'(x)\right]^{2}}\,dx \]
The integral is almost certainly not elementary. A circular arc gives the semicircle integrand of the second idea, which does have a closed form; a parabolic profile gives the root of one plus a linear-squared term, needing a technique beyond this course; anything else is numerical. In practice the contractor's software integrates numerically, which gives whatever precision is wanted.
The shortfall is the same error as using a vertical height for a slant height. For a shallow roof — say a rise of one metre over a ten-metre span — the arc is only about 2.6 percent longer than the width, and the error might pass unnoticed. For a rise of three metres it is nearer 20 percent, which on a large roof is a serious under-order.
Note that the error is always in the same direction, since the arc length integrand is at least one. A contractor who quotes on width will always be short, never over — which is exactly the kind of systematic bias worth knowing about before it appears on an invoice.
Commit first
Answer, then rate your confidence honestly.
Predict first
Why does the arc length integrand contain a one under the square root?
Correct: Because it is the horizontal leg.
\[ \sqrt{(dx)^{2}+(dy)^{2}} = \sqrt{1+\left(\tfrac{dy}{dx}\right)^{2}}\,dx \]
Why: The short segment is the hypotenuse of a triangle whose legs are the horizontal and vertical changes. Factoring the horizontal change out of the root leaves one from that leg and the derivative squared from the other. Dropping the one would give a horizontal line zero length, which is the quickest test of any proposed arc length formula.
Explain it
They wrote a surface area integral without the square root factor.
Discussion prompt
In four sentences or fewer, show them what is missing.
Hint: Ask what the band's width actually is.
Answer:
Ask them how wide the band is: it runs along the surface, following the slope, so its width is a piece of ARC and not a horizontal step. That arc is longer than the horizontal step whenever the curve is not flat, which is what the square root measures.
Have them test it on a cone: the classical formula uses the slant height, and dropping the root amounts to using the vertical height instead. For a cone of radius 3 and height 4 that is 20 percent low, and the gap grows as the cone flattens.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For the derivation, draw the triangle and factor the differential out. For the variable, pick the one whose derivative stays bounded. For recognising, look for a root of one plus a squared transcendental. For surfaces, remember the band's width is arc length and multiply by the circumference. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
At the top, draw a short piece of curve with its horizontal and vertical legs marked, write the Pythagorean expression, and factor the differential out in two lines to reach the formula. Beneath, write both variable versions side by side. Below, draw a curve with a vertical tangent and write its two integrands, marking which is improper. In the middle of the page, list four curves with their arc length integrands and mark which are elementary, writing one line on what makes the three-halves power work. In the lower half, draw a frustum with its two radii and slant height labelled, write its surface formula, and beneath it write the surface of revolution integral showing which factor is the circumference and which the arc length. At the bottom, work the cone's lateral surface from the formula and compare with the classical result.
If your frustum drawing marks the vertical height rather than the slant height, redraw it — the slant is the sloping side and it is the longer of the two, and seeing that once is what makes the error hard to repeat.
Recap
Five things, and the third is knowing when to stop looking for a formula.
| If you see | Then |
|---|---|
| A length along a curve | The integrand carries a square root |
| A vertical tangent | Describe the curve in the other variable |
| A root of one plus a squared transcendental | Expect a non-elementary integral |
| A three-halves power | The radicand is linear: it works |
| A surface of revolution | Circumference times arc length |
| A frustum | The slant height, never the vertical one |
| A surface close to a cylinder's | The arc length factor may have been dropped |
Section 6.5 turns from geometry to physics. Work, force on a dam and pumping a tank all use the same discipline — describe one representative piece and integrate — with a physical quantity in place of a geometric one.
OpenStax Calculus Volume 1, §6.4 Arc Length of a Curve and Surface Area §6.4, pp. 582-593 — everything on these slides traces back here
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