Arc length, volumes and surface area

On this page

Integration measures more than area under a curve. Cut a curve, a solid or a surface into tiny pieces whose size you can write down, add the pieces, and let them shrink: the sum becomes an integral. This lesson uses that one idea three times, to find the length of a curve, the volume of a solid of revolution by discs or by cylindrical shells, and the curved surface area of a solid of revolution. In each case the skill is to choose the right small piece and then to keep the algebra exact.

Arc length

A short piece of a curve is close to the hypotenuse of a right-angled triangle with sides δ x and δ y, so its length is

δ s ≈ square root of ((δ x)² + (δ y)²) = square root of (1 + ((δ y)/(δ x))²) δ x.

Adding the pieces and letting δ x → 0 gives the arc length from x = a to x = b:

s = ∫ₐ^b square root of (1 + (dy/(dx))²) dx.

Most curves give an integrand with no elementary antiderivative; then a graphing calculator evaluates the integral numerically. Exam questions that ask for an exact length use curves for which 1 + (dy/(dx))² simplifies to a perfect square or a power of a linear expression.

Find the exact length of the curve y = (2/3)x^(3/2) from x = 0 to x = 3.

Show full solutionHide solution

Simplify the integrand
Method
dy/(dx) = x^(1/2), so 1 + (dy/(dx))² = 1 + x.
Reason
The square root of 1 + x is a power of a linear expression, so it integrates exactly.
Working
Answer
s = ∫₀³ square root of (1 + x) dx; = [(2/3)(1 + x)^(3/2)]₀³; = (2/3)(8-1) = 14/3.

Quick check

The chord from (0,0) to (3,2 square root of 3) has length square root of 21 ≈ 4.58, a little less than 14/3 ≈ 4.67, as it must be for a curve between the same two points.

A student finds the length of y = x² from x = 0 to x = 1 as

s = ∫₀¹(1 + dy/(dx))dx = ∫₀¹(1 + 2x) dx = 2.

Explain why this cannot be right, and find the correct length to 4 significant figures.

Write your answer, then compare it with the solution

Why it fails and the correct length

Show full solutionHide solution

Restore the square and the root
Method
Use square root of (1 + (dy/(dx))²) = square root of (1 + 4x²).
Reason
The length element comes from Pythagoras, so the gradient is squared inside a square root; 1 + dy/(dx) is the length of the two legs of the triangle added together.
Working
Answer
s = ∫₀¹ square root of (1 + 4x²) dx ≈ 1.479
using a graphing calculator. This lies between the chord length square root of 2 ≈ 1.414 and the path along the axes, 1 + 1 = 2.

Common mistake

The student’s answer, 2, is the length of the route from (0,0) along the x-axis to (1,0) and then up to (1,1). A smooth curve between the two points must be shorter than that route and longer than the chord, so a quick bound catches the error.

Discs or shells

When a region is rotated about an axis, you can slice it in two ways:

Discs (or washers) slice perpendicular to the axis. Each slice is a thin disc of radius y about the x-axis, or radius x about the y-axis.
Shells slice parallel to the axis. About the y-axis, a strip at distance x with height y and width δ x sweeps a thin cylindrical shell of circumference 2π x, so its volume is about 2π xy δ x.

Discs about the x-axis: V = π∫ₐ^b y² dx,; Discs about the y-axis: V = π∫c^d x² dy,; Shells about the y-axis: V = 2π∫ₐ^b x y dx.

Shells about the x-axis work the same way, with V = 2π∫ y x dy. Both methods give the same volume; choose the one that uses the curve in the form it is given and avoids splitting the region.

The region bounded by y = x², the line y = 4 and the y-axis, with x ≥ 0, is rotated through 2π about the y-axis. Find the volume by discs and by shells.

Show full solutionHide solution

Discs: slice horizontally
Method
At height y, the disc has radius x = square root of y, for 0 ≤ y ≤ 4.
Reason
Discs about the y-axis need the radius as a function of y.
Working
Answer
V = π∫₀⁴(square root of y)²dy = π∫₀⁴y dy = 8π.
Shells: slice vertically
Method
At distance x, the strip runs from the curve up to y = 4, so its height is 4-x², for 0 ≤ x ≤ 2.
Reason
The shell height is the length of the strip inside the region, not the value of y on the curve.
Working
Answer
V = 2π∫₀²x(4-x²)dx; = 2π[2x²-(1/4)x⁴]₀² = 2π(8-4) = 8π.

Quick check

The two methods agree. The solid is a cylinder of radius 2 and height 4, volume 16π, with the region under the curve removed, so 8π is exactly half of it.

An arch-shaped curve rises from (0, 0) to a maximum height of 0.25 at x = 0.5 and returns to (1, 0). The area under it is lightly shaded. A narrow vertical band between x = 0.7 and x = 0.75 is marked as the strip. A dashed horizontal line at y = 0.15 crosses the arch at two points, about x = 0.18 and x = 0.82. — The region under y = x − x² for 0 ≤ x ≤ 1. Rotating the narrow shaded strip at distance x from the y-axis sweeps a thin cylindrical shell of radius x, height x − x² and thickness δx. A horizontal line at height y meets the curve twice, so discs about the y-axis would need both branches.

Figure data: A thin strip becomes a cylindrical shell
Series	x	y
y = x − x²	0.000	0.000000
y = x − x²	0.025	0.024375
y = x − x²	0.050	0.047500
y = x − x²	0.075	0.069375
y = x − x²	0.100	0.090000
y = x − x²	0.125	0.109375
y = x − x²	0.150	0.127500
y = x − x²	0.175	0.144375
y = x − x²	0.200	0.160000
y = x − x²	0.225	0.174375
y = x − x²	0.250	0.187500
y = x − x²	0.275	0.199375
y = x − x²	0.300	0.210000
y = x − x²	0.325	0.219375
y = x − x²	0.350	0.227500
y = x − x²	0.375	0.234375
y = x − x²	0.400	0.240000
y = x − x²	0.425	0.244375
y = x − x²	0.450	0.247500
y = x − x²	0.475	0.249375
y = x − x²	0.500	0.250000
y = x − x²	0.525	0.249375
y = x − x²	0.550	0.247500
y = x − x²	0.575	0.244375
y = x − x²	0.600	0.240000
y = x − x²	0.625	0.234375
y = x − x²	0.650	0.227500
y = x − x²	0.675	0.219375
y = x − x²	0.700	0.210000
y = x − x²	0.725	0.199375
y = x − x²	0.750	0.187500
y = x − x²	0.775	0.174375
y = x − x²	0.800	0.160000
y = x − x²	0.825	0.144375
y = x − x²	0.850	0.127500
y = x − x²	0.875	0.109375
y = x − x²	0.900	0.090000
y = x − x²	0.925	0.069375
y = x − x²	0.950	0.047500
y = x − x²	0.975	0.024375
y = x − x²	1.000	0.000000
Height y = 0.15	0.000	0.150000
Height y = 0.15	1.000	0.150000

Figure data: A thin strip becomes a cylindrical shell
Series	x	y
y = x − x²	0.000	0.000000
y = x − x²	0.025	0.024375
y = x − x²	0.050	0.047500
y = x − x²	0.075	0.069375
y = x − x²	0.100	0.090000
y = x − x²	0.125	0.109375
y = x − x²	0.150	0.127500
y = x − x²	0.175	0.144375
y = x − x²	0.200	0.160000
y = x − x²	0.225	0.174375
y = x − x²	0.250	0.187500
y = x − x²	0.275	0.199375
y = x − x²	0.300	0.210000
y = x − x²	0.325	0.219375
y = x − x²	0.350	0.227500
y = x − x²	0.375	0.234375
y = x − x²	0.400	0.240000
y = x − x²	0.425	0.244375
y = x − x²	0.450	0.247500
y = x − x²	0.475	0.249375
y = x − x²	0.500	0.250000
y = x − x²	0.525	0.249375
y = x − x²	0.550	0.247500
y = x − x²	0.575	0.244375
y = x − x²	0.600	0.240000
y = x − x²	0.625	0.234375
y = x − x²	0.650	0.227500
y = x − x²	0.675	0.219375
y = x − x²	0.700	0.210000
y = x − x²	0.725	0.199375
y = x − x²	0.750	0.187500
y = x − x²	0.775	0.174375
y = x − x²	0.800	0.160000
y = x − x²	0.825	0.144375
y = x − x²	0.850	0.127500
y = x − x²	0.875	0.109375
y = x − x²	0.900	0.090000
y = x − x²	0.925	0.069375
y = x − x²	0.950	0.047500
y = x − x²	0.975	0.024375
y = x − x²	1.000	0.000000
Height y = 0.15	0.000	0.150000
Height y = 0.15	1.000	0.150000

The region between y = x-x² and the x-axis is rotated through 2π about the y-axis. Find the volume, and explain why shells are easier than discs here.

Your choice of method, integral and volume

Show answerHide answer

Discs about the y-axis need x in terms of y, and a horizontal line meets the curve twice, so there are two branches x = (1/2)(1± square root of (1-4y)). Shells use the curve as given:

V = 2π∫₀¹x(x-x²)dx; = 2π(1/3-1/4) = π/6.

Washers between those two branches, from y = 0 to y = 1/4, give the same π/6.

Surface area of revolution

Rotating a short piece of curve of length δ s at height y ≥ 0 about the x-axis sweeps a thin band of radius y and slant width δ s, with area about 2π y δ s. Adding the bands:

S = 2π∫ₐ^b y square root of (1 + (dy/(dx))²) dx.

About the y-axis, the radius is x instead, so S = 2π∫ₐ^b x square root of (1 + (dy/(dx))²) dx for x ≥ 0. The band’s width is the slant length δ s, not δ x: a sloping band is wider than its shadow on the axis.

Common mistakes

Writing 1 + dy/(dx) or 1 + (dy/(dx)) without the square, or forgetting the square root.
Using 2π∫ y dx for surface area, which leaves out the slant factor.
Taking the shell height as the y-value on the curve when the region lies between the curve and another line.
Using x-limits in an integral with respect to y, or the reverse.
Splitting a region into awkward branches when the other slicing would use the curve directly.

Before you move on

The arc y = square root of x, 0 ≤ x ≤ 2, is rotated through 2π about the x-axis. Find the exact curved surface area. Then use the same formula to show that a sphere of radius r has surface area 4π r².

Write your answer, then compare it with the solution

Your integral, its value and the sphere check

Hint 1

Put y inside the root

y square root of (1 + (y')²) = square root of (y² + y²(y')²). For y = square root of x, this becomes square root of (x + 1/4).

Show full solutionHide solution

Simplify and integrate
Method
y' = 1/(2 square root of x), so y square root of (1 + (y')²) = square root of x square root of (1 + 1/(4x)) = square root of (x + 1/4).
Reason
The root of a linear expression integrates directly.
Working
Answer
S = 2π∫₀² square root of (x + 1/4) dx; = 2π · (2/3)[(9/4)^(3/2)-(1/4)^(3/2)]; = 2π · 2/3 · 26/8 = 13π/3.
Check on a sphere
Method
Rotate y = square root of (r²-x²), -r ≤ x ≤ r.
Reason
A known answer tests the formula and the algebra.
Working
Answer
y' = -x/y, so y square root of (1 + x²/y²) = square root of (y² + x²) = r. Then S = 2π∫₋ᵣ^rr dx = 4π r².

Quick check

Using 2π∫ y dx, without the arc-length factor, would undercount every sloping band; for the sphere it gives π²r² ≈ 9.87r² instead of 4π r² ≈ 12.57r². The sphere is a useful check on any surface-area formula you set up.

Syllabus and review details

GCE A-Level H2 Further Mathematics 2027 · 2027

Last reviewed: 2026-11-06

Read offline with Premium

Guided steps

Reading settings

Arc length

An exact arc length

Simplify the integrand

An arc shorter than a straight line?

Restore the square and the root

Discs or shells

One bowl, two methods

Discs: slice horizontally

Shells: slice vertically

Rotating about the y-axis

Surface area of revolution

Common mistakes

Before you move on

A curved surface, checked on a sphere

Simplify and integrate

Check on a sphere

Lesson done

Arc length

An exact arc length

An arc shorter than a straight line?

Discs or shells

One bowl, two methods

Rotating about the y-axis

Surface area of revolution

Common mistakes

Before you move on

A curved surface, checked on a sphere

Questions