OCR A Level Further Mathematics A
Y545 Additional Pure Mathematics
Arc lengths and surface areas of revolution
Be able to find arc lengths and areas of surface of revolution for
curves defined in cartesian or parametric form.
Definitions
• Cartesian and parametric form. A plane curve may be written in cartesian form 𝑦 = f (𝑥),
or in parametric form 𝑥 = 𝑥(𝑡), 𝑦 = 𝑦(𝑡), where the point (𝑥, 𝑦) is traced out as the parameter
𝑡 varies.
• Arc length. The arc length of a curve is the distance measured along the curve between two
of its points.
• Surface of revolution. When an arc of a curve is rotated about a fixed line (the axis of
revolution, here the 𝑥-axis), it sweeps out a surface of revolution.
• Surface area of revolution. This is the area of the curved surface so generated — the lateral
area, excluding any flat ends. The radius of rotation is the distance from the curve to the axis:
𝑦 for the 𝑥-axis, 𝑥 for the 𝑦-axis.
• Mean value of a function. The mean value of f on [𝑎, 𝑏] is the constant 𝑦¯ whose integral over
[𝑎, 𝑏] equals that of f; replacing the curve by the line 𝑦 = 𝑦¯ models it by its average height.
• Frustum and cylinder. Rotating a straight line segment about the 𝑥-axis produces a frustum
of a cone, or a cylinder when the segment is parallel to the axis. The curved surface area of a
frustum with end radii 𝑟 1 , 𝑟 2 and slant height ℓ is 𝜋(𝑟 1 + 𝑟 2 )ℓ.
• Hyperbolic functions. cosh 𝑥 =
of 𝑦 = cosh 𝑥 is a catenary.
𝑒𝑥 + 𝑒−𝑥
𝑒𝑥 − 𝑒−𝑥
1
, sinh 𝑥 =
, sech 𝑥 =
. The graph
2
2
cosh 𝑥
1
Formulae
Element of arc length
√︄
d𝑦
1+
d𝑥
d𝑠 =
√︄
2
d𝑥
d𝑡
d𝑥 =
2
d𝑦
+
d𝑡
2
d𝑡
Arc length
∫
√︄
d𝑦
1+
d𝑥
𝑠=
2
∫
𝑠=
d𝑥
(cartesian)
√︄
d𝑥
d𝑡
2
d𝑦
+
d𝑡
2
d𝑡
(parametric)
∫
𝑆 = 2𝜋
Surface area of revolution
𝑟 d𝑠, where 𝑟 is the distance to the axis: 𝑟 = 𝑦 for the
𝑥-axis, 𝑟 = 𝑥 for the 𝑦-axis.
√︄
∫
(cartesian)
𝑆 𝑥 = 2𝜋
𝑦
∫
√︂
𝑆 𝑥 = 2𝜋
(parametric)
𝑦
d𝑦
1+
d𝑥
2
2
2
d𝑥
d𝑡
Mean value of a function f over [𝑎, 𝑏]
+
d𝑦
d𝑡
𝑆 𝑦 = 2𝜋
d𝑥,
𝑥
𝑆 𝑦 = 2𝜋
d𝑡,
𝑥
d𝑥
1+
d𝑦
√︂
∫
1
𝑦¯ =
𝑏−𝑎
Hyperbolic results (for the 2020 question)
Specimen — Question 1
√︄
∫
d𝑥
d𝑡
2
2
+
d𝑦
d𝑦
d𝑡
2
d𝑡
∫ 𝑏
f (𝑥) d𝑥
𝑎
d
cosh 𝑥 = sinh 𝑥,
d𝑥
cosh2 𝑥 − sinh2 𝑥 = 1
(parametric)
A curve is given by 𝑥 = 𝑡 2 − 2 ln 𝑡, 𝑦 = 4𝑡 for 𝑡 > 0. When the arc of the curve between the points
where 𝑡 = 1 and 𝑡 = 4 is rotated through 2𝜋 radians about the 𝑥-axis, a surface of revolution is formed with
surface area 𝐴.
Given that 𝐴 = 𝑘 𝜋, where 𝑘 is an integer,
• write down an integral which gives 𝐴 and
• find the value of 𝑘.
Solution.
[4]
d𝑥
2
d𝑦
= 2𝑡 − and
= 4, so
d𝑡
𝑡
d𝑡
2 2 2
2
d𝑥
d𝑦
4
+
= 2𝑡 − 2𝑡 + 16 = 4𝑡 2 + 8 + 2 = 2𝑡 + 2𝑡 .
d𝑡
d𝑡
𝑡
Hence, with 𝑦 = 4𝑡,
∫ 4
𝐴 = 2𝜋
1
𝐴 = 2𝜋
∫ 4 ∫ 4
√︃
4𝑡 4𝑡 2 + 8 + 𝑡42 d𝑡 = 2𝜋
4𝑡 2𝑡 + 2𝑡 d𝑡 = 2𝜋
8𝑡 2 + 8 d𝑡.
1
4
8𝑡 3
608 32
+ 8𝑡 = 2𝜋
−
= 2𝜋 · 192 = 384𝜋,
3
3
3
1
1
2
so 𝑘 = 384.
2019 — Question 7
(cartesian)
1
13
31
The points 𝑃 12 , 24
and 𝑄 32 , 24
lie on the curve 𝑦 = 31 𝑥 3 + .
4𝑥
The area of the surface generated when arc 𝑃𝑄 is rotated completely about the 𝑥-axis is denoted by 𝐴.
[4]
(a) Find the exact value of 𝐴. Give your answer as a rational multiple of 𝜋.
Student X finds an approximation to 𝐴 by modelling the arc 𝑃𝑄 as the straight line segment 𝑃𝑄,
then rotating this line segment completely about the 𝑥-axis to form a surface.
(b) Find the approximation to 𝐴 obtained by student X. Give your answer as a rational multiple of 𝜋. [4]
Student Y finds a second approximation to 𝐴 by modelling the original curve as the line 𝑦 = 𝑀,
1
where 𝑀 is the mean value of the function f (𝑥) = 31 𝑥 3 + , then rotating this line completely about
4𝑥
the 𝑥-axis to form a surface.
(c) Find the approximation to 𝐴 obtained by student Y. Give your answer correct to four decimal places.
[4]
2
2
1
1
1
d𝑦
d𝑦
1
4
2
2
=𝑥 + +
= 𝑥 − 2 , so 1 +
Solution. (a)
= 𝑥 + 2 . Then
d𝑥
d𝑥
2 16𝑥 4
4𝑥
4𝑥
∫ 3/2 3
∫ 3/2 5
1
1
𝑥
1
𝑥
𝑥
2
𝐴 = 2𝜋
+
𝑥 + 2 d𝑥 = 2𝜋
+ +
d𝑥
3
4𝑥
3
3 16𝑥 3
4𝑥
1/2
1/2
6
3/2
𝑥
155 155𝜋
𝑥2
1
1145
95
= 2𝜋
= 2𝜋 ·
+
−
= 2𝜋
+
=
.
18
6
1152 1152
144
72
32𝑥 2 1/2
31/24 − 13/24
3
= and length |𝑃𝑄| =
(b) The chord 𝑃𝑄 has gradient
1
4
segment gives a frustum of slant height |𝑃𝑄| with end radii 𝑦 𝑃 , 𝑦 𝑄 , so
√︂
12 +
2
3
4
=
55𝜋
31
5
11 5
𝐴 ≈ 2𝜋 · 12 13
+
24
24 · 4 = 2𝜋 · 12 · 4 = 24 .
(c) The mean value of f on
1 3
2 , 2 is
3/2
4
∫ 3/2 3
𝑥
1
𝑥
1
5
1
𝑀=
+
d𝑥 =
+ ln 𝑥
=
+ ln 3.
3
4𝑥
12 4
12 4
1/2
1/2
Modelling the curve as 𝑦 = 𝑀 and rotating gives a cylinder of radius 𝑀 and length 1, so
5
1
𝐴 ≈ 2𝜋𝑀 · 1 = 2𝜋
+ ln 3 = 4.3437 (4 d.p.).
12 4
3
5
. Rotating the
4
2020 — Question 5
(cartesian)
A designer intends to manufacture a product using a 3-D printer. The product will take the form of a
surface 𝑆 which must meet a number of design specifications. The designer chooses to model 𝑆 with the
equation 𝑧 = 𝑦 cosh 𝑥 for − ln 20 ≤ 𝑥 ≤ ln 20, −2 ≤ 𝑦 ≤ 2.
(a)
(i) In the Printed Answer Booklet, on the axes provided, sketch the section of 𝑆 given by 𝑦 = 1.
[1]
(ii) One of the design specifications of the product is that this section should have a length no
greater than 20 units.
Determine whether the product meets this requirement according to the model.
[4]
(b)
(i) In the Printed Answer Booklet, on the axes provided, sketch the contour of 𝑆 given by 𝑧 = 1.
[1]
(ii) When this contour is rotated through 2𝜋 radians about the 𝑥-axis, the surface 𝑇 is generated.
The surface area of 𝑇 is denoted by 𝐴.
Show that 𝐴 can be written in the form
∫ ln 20
1 √︁
𝑘𝜋
cosh4 𝑥 + cosh2 𝑥 − 1 d𝑥
3
cosh 𝑥
0
[5]
for some integer 𝑘 to be determined.
(iii) A second design specification is that the surface area of 𝑇 must not be greater than 20 square
units.
Use your calculator to decide whether the product meets this requirement according to the
model.
[2]
Solution. (a)(i) The section is 𝑧 = cosh 𝑥: a catenary with minimum (0, 1), symmetric about the 𝑧-axis,
rising to cosh(ln 20) = 10.025 at each end.
𝑧
10.025
1
𝑥
− ln 20
ln 20
d𝑧
= sinh 𝑥 and 1 + sinh2 𝑥 = cosh2 𝑥, so
d𝑥
∫ ln 20
ln 20
1
length =
cosh 𝑥 d𝑥 = sinh 𝑥 − ln 20 = 2 sinh(ln 20) = 20 −
= 19.95.
20
− ln 20
(a)(ii) With 𝑧 = cosh 𝑥,
Since 19.95 < 20, the requirement is met.
1
(b)(i) The contour 𝑧 = 1 is 𝑦 =
= sech 𝑥: maximum (0, 1), symmetric, decreasing to sech(ln 20) ≈
cosh 𝑥
0.0998 at each end.
4
𝑦
1
𝑥
− ln 20
(b)(ii) With 𝑦 = sech 𝑥,
ln 20
d𝑦
sinh 𝑥
, so
=−
d𝑥
cosh2 𝑥
d𝑦
1+
d𝑥
2
=
cosh4 𝑥 + sinh2 𝑥
cosh4 𝑥
√︂
2
2
using sinh 𝑥 = cosh 𝑥−1. Hence 𝑦 1 +
d𝑦
d𝑥
2
=
is even,
∫ ln 20
𝐴 = 2𝜋
√︁
cosh4 𝑥 + cosh2 𝑥 − 1
cosh3 𝑥
− ln 20
=
1
cosh4 𝑥 + cosh2 𝑥 − 1
cosh4 𝑥
√︁
3
cosh 𝑥
d𝑥 = 4𝜋
,
cosh4 𝑥 + cosh2 𝑥 − 1, and since the integrand
∫ ln 20 √︁
cosh4 𝑥 + cosh2 𝑥 − 1
cosh3 𝑥
0
d𝑥,
so 𝑘 = 4.
(b)(iii) Evaluating, 𝐴 = 4𝜋 × 1.5647 . . . ≈ 19.7 < 20, so the requirement is met.
2022 — Question 7
(parametric)
3
(a) Differentiate 16 + 𝑡 2 2 with respect to 𝑡.
∫ 3
Let 𝐼𝑛 =
[1]
√︁
𝑡 𝑛 16 + 𝑡 2 d𝑡 for integers 𝑛 ≥ 1.
0
(b) Show that, for 𝑛 ≥ 3, (𝑛 + 2)𝐼𝑛 = 125 × 3 𝑛−1 − 16(𝑛 − 1)𝐼𝑛−2 .
[5]
(c) The curve 𝐶 is defined parametrically by 𝑥 = 𝑡 4 cos 𝑡, 𝑦 = 𝑡 4 sin 𝑡, for 0 ≤ 𝑡 ≤ 3. The length of 𝐶 is
denoted by 𝐿.
Show that 𝐿 = 𝐼3 . (You are not required to evaluate this integral.)
[4]
√
3/2 3
1/2
d
Solution. (a)
16 + 𝑡 2
= 2 16 + 𝑡 2
· 2𝑡 = 3𝑡 16 + 𝑡 2 .
d𝑡
√
√
3/2
(b) Write 𝑡 𝑛 16 + 𝑡 2 = 𝑡 𝑛−1 · 𝑡 16 + 𝑡 2 and integrate by parts, taking 𝑣 = 13 16 + 𝑡 2
(from part (a)):
h
3𝐼𝑛 = 𝑡
2 3/2
𝑛−1
16 + 𝑡
2 1/2
2
3/2 i 3
0
∫ 3
− (𝑛 − 1)
3/2
𝑡 𝑛−2 16 + 𝑡 2
d𝑡.
0
2 1/2
Since 16 + 𝑡
= 16 16 + 𝑡
+ 𝑡 2 16 + 𝑡
, the remaining integral equals 16𝐼𝑛−2 + 𝐼𝑛 , and the
𝑛−1
3/2
boundary term is 3
· 25 = 125 × 3 𝑛−1 . Thus
3𝐼𝑛 = 125 × 3 𝑛−1 − (𝑛 − 1) (16𝐼𝑛−2 + 𝐼𝑛 ) =⇒ (𝑛 + 2)𝐼𝑛 = 125 × 3 𝑛−1 − 16(𝑛 − 1)𝐼𝑛−2 .
5
d𝑥
d𝑦
= 4𝑡 3 cos 𝑡 − 𝑡 4 sin 𝑡 and
= 4𝑡 3 sin 𝑡 + 𝑡 4 cos 𝑡, so
d𝑡
d𝑡
2 2
d𝑥
d𝑦
+
= 16𝑡 6 + 𝑡 8 = 𝑡 6 16 + 𝑡 2 ,
d𝑡
d𝑡
√︂ 2
√
2
d𝑦
d𝑥
3 16 + 𝑡 2 for 0 ≤ 𝑡 ≤ 3, and
hence
+
=
𝑡
d𝑡
d𝑡
(c)
𝐿=
∫ 3 √︁
𝑡 3 16 + 𝑡 2 d𝑡 = 𝐼3 .
0
2023 — Question 2
(cartesian)
√
A curve has equation 𝑦 = 1 + 𝑥 2 , for 0 ≤ 𝑥 ≤ 1, where both the 𝑥- and 𝑦-units are in cm. The area of the
surface generated when this curve is rotated fully about the 𝑥-axis is 𝐴 cm2 .
∫ 1 √︁
1 + 𝑘𝑥 2 d𝑥 for some integer 𝑘 to be determined.
(a) Show that 𝐴 = 2𝜋
[4]
0
A small component for a car is produced in the shape of this surface. The curved surface area of the
component must be 8 cm2 , accurate to within one percent. The engineering process produces such
components with a curved surface area accurate to within one half of one percent.
(b) Determine whether all components produced will be suitable for use in the car.
[2]
2
𝑥2
𝑥
d𝑦
1 + 2𝑥 2
d𝑦
=1+
=√
, so 1 +
=
, and
Solution. (a)
d𝑥
d𝑥
1 + 𝑥2
1 + 𝑥2
1 + 𝑥2
√︄
d𝑦
𝑦 1+
d𝑥
2
=
√︁
√︂
1 + 𝑥2 ·
1 + 2𝑥 2 √︁
= 1 + 2𝑥 2 .
1 + 𝑥2
∫ 1 √︁
Therefore 𝐴 = 2𝜋
1 + 2𝑥 2 d𝑥, giving 𝑘 = 2.
0
∫ 1 √︁
(b) Evaluating, 𝐴 = 2𝜋
1 + 2𝑥 2 d𝑥 = 7.9876 . . . ≈ 7.99 cm2 . The specification accepts areas within
0
1% of 8, i.e. [ 7.92, 8.08 ]. The process produces areas within 0.5% of 𝐴, i.e. [ 0.995𝐴, 1.005𝐴 ] =
[ 7.95, 8.03 ]. Since [ 7.95, 8.03 ] lies entirely within [ 7.92, 8.08 ], every component produced meets the
specification, so yes, all components will be suitable.
Three harder problems
(same techniques, no new theory — but trickier)
6
Challenge 1
(cartesian — rotation about the 𝑦-axis)
𝑥2 1
− ln 𝑥, for 1 ≤ 𝑥 ≤ 2. The arc is rotated through 2𝜋 radians about the
The curve 𝐶 has equation 𝑦 =
4
2
𝑦-axis, generating a surface of area 𝑆.
(a) Explain why it is not practical to find 𝑆 by first writing 𝑥 as a function of 𝑦.
(b) By integrating with respect to 𝑥, show that 𝑆 =
10𝜋
.
3
[1]
[5]
Solution. (a) The equation 𝑦 = 41 𝑥 2 − 12 ln 𝑥 mixes a power of 𝑥 with ln 𝑥, and cannot be rearranged to
∫ √︁
give 𝑥 explicitly as a function of 𝑦. So the closed form 𝑥 = 𝑔(𝑦) needed for 𝑆 = 2𝜋 𝑥 1 + (d𝑥/d𝑦) 2 d𝑦
is unavailable — we must instead integrate with respect to 𝑥.
(b) The key point: for rotation about the√︂𝑦-axis the radius is 𝑥, but the element d𝑠 may still be taken in
2
∫
∫2
1
d𝑦 𝑥
d𝑦
terms of 𝑥. Thus 𝑆 = 2𝜋 𝑥 d𝑠 = 2𝜋 1 𝑥 1 + d𝑥
= − , so
d𝑥. Now
d𝑥 2 2𝑥
2
2
d𝑦
𝑥2 1
1
1
𝑥
1+
=
+ + 2 =
+
,
d𝑥
4
2 4𝑥
2 2𝑥
and the radius times the slant collapses to a polynomial:
√︄
2
d𝑦
𝑥
1
𝑥2 + 1
𝑥 1+
=𝑥
+
=
.
d𝑥
2 2𝑥
2
∫ 2
Hence 𝑆 = 2𝜋
1
Challenge 2
3
2
10𝜋
𝑥
14 4
𝑥2 + 1
=
d𝑥 = 𝜋
+𝑥 = 𝜋
−
.
2
3
3
3
3
1
(cartesian — rotation about the 𝑦-axis)
The arc of the curve 𝑦 = 𝑥 2 for 0 ≤ 𝑥 ≤ 1 is rotated through 2𝜋 radians about the 𝑦-axis, generating a
surface of area 𝑆 𝑦 .
(a) Show that 𝑆 𝑦 =
𝜋 √
5 5−1 .
6
[4]
(b) The same arc is instead rotated about the 𝑥-axis, generating a surface of area 𝑆 𝑥 . Without evaluating
any integral, explain carefully why 𝑆 𝑦 > 𝑆 𝑥 .
[3]
Solution. (a) For rotation about the 𝑦-axis the radius is 𝑥, and with
d𝑦
= 2𝑥,
d𝑥
∫ 1 √︁
𝑆 𝑦 = 2𝜋
𝑥 1 + 4𝑥 2 d𝑥.
0
The substitution 𝑢 = 1 + 4𝑥 2 (d𝑢 = 8𝑥 d𝑥, limits 𝑢 = 1 to 𝑢 = 5) gives
∫
i5 𝜋 √
1 5√
𝜋 2h
𝑆 𝑦 = 2𝜋 ·
𝑢 d𝑢 = · 𝑢 3/2 = 5 5 − 1 .
1
8 1
4 3
6
7
(b) Both surfaces are generated from the same arc, so they share the very same element of arc length d𝑠
at each point. The only difference is the radius of rotation: it equals 𝑥 for the 𝑦-axis, but 𝑦 = 𝑥 2 for the
𝑥-axis. On 0 ∫< 𝑥 < 1 we have 𝑥 2 < 𝑥, so the radius is strictly smaller everywhere for the 𝑥-axis rotation.
Since 𝑆 = 2𝜋 (radius) d𝑠 with the identical positive d𝑠, it follows that 𝑆 𝑥 < 𝑆 𝑦 .
Challenge 3
(parametric — one arch of a cycloid)
The cycloid 𝐶 is given by 𝑥 = 𝑡 − sin 𝑡, 𝑦 = 1 − cos 𝑡, for 0 ≤ 𝑡 ≤ 2𝜋 (one arch).
[4]
(a) Show that the length of the arch is 8.
(b) The arch is rotated through 2𝜋 radians about the 𝑥-axis. Show that the area of the surface generated
64𝜋
is
.
[5]
3
(c) A student approximates this surface area by replacing the arch with the horizontal line 𝑦 = 𝑦¯ , where
𝑦¯ is the mean value of 𝑦 with respect to 𝑥 over the arch, and rotating this line about the 𝑥-axis. Show
that the student’s approximation is 6𝜋 2 , and state, with a reason, whether it over- or under-estimates
the true area.
[5]
Solution. (a)
d𝑥
d𝑦
= 1 − cos 𝑡 and
= sin 𝑡, so
d𝑡
d𝑡
2 2
d𝑥
d𝑦
+
= (1 − cos 𝑡) 2 + sin2 𝑡 = 2 − 2 cos 𝑡 = 4 sin2 2𝑡 ,
d𝑡
d𝑡
giving d𝑠 = 2 sin 2𝑡 d𝑡 (valid as sin 2𝑡 ≥ 0 on [0, 2𝜋]). Hence
∫ 2𝜋
h
i2𝜋
𝐿=
= (−4) (−1) − (−4) (1) = 8.
2 sin 2𝑡 d𝑡 = − 4 cos 2𝑡
0
0
(b) Here 𝑦 = 1 − cos 𝑡 = 2 sin2 2𝑡 , so
∫ 2𝜋
∫ 2𝜋
∫ 2𝜋
2 𝑡
𝑡
𝑆 = 2𝜋
𝑦 d𝑠 = 2𝜋
2 sin 2 · 2 sin 2 d𝑡 = 8𝜋
sin3 2𝑡 d𝑡.
0
0
0
With 𝑢 = 2𝑡 (d𝑡 = 2 d𝑢, limits 0 to 𝜋), this is
∫ 𝜋
4 64𝜋
𝑆 = 16𝜋
.
sin3 𝑢 d𝑢 = 16𝜋 · =
3
3
0
(c) The base runs from 𝑥 = 0 to 𝑥 = 2𝜋. Since d𝑥 = (1 − cos 𝑡) d𝑡, the mean value of 𝑦 with respect to 𝑥 is
∫ 2𝜋
∫ 2𝜋
1
1
1
3
𝑦¯ =
𝑦 d𝑥 =
(1 − cos 𝑡) 2 d𝑡 =
· 3𝜋 = ,
2𝜋 0
2𝜋 0
2𝜋
2
∫ 2𝜋
∫ 2𝜋
using 0 (1 − cos 𝑡) 2 d𝑡 = 0 1 − 2 cos 𝑡 + cos2 𝑡 d𝑡 = 2𝜋 + 𝜋 = 3𝜋. Rotating 𝑦 = 𝑦¯ gives a cylinder of
radius 32 and length 2𝜋, of curved surface area
2𝜋 · 23 · 2𝜋 = 6𝜋 2 ≈ 59.2.
∫
64𝜋
This under-estimates the true area
≈ 67.0. The reason: the approximation replaces 2𝜋 𝑦 d𝑠 by
3
√︁
∫
2𝜋 𝑦 d𝑥, and since d𝑠 = 1 + (d𝑦/d𝑥) 2 d𝑥 ≥ d𝑥 with 𝑦 ≥ 0, ignoring the slant of the arch’s steep sides
can only lose surface area.
8