Edexcel International A Level Maths · P4 · Pure Mathematics 4
Volumes of revolution
28 free, high-quality past-paper-style questions on Volumes of revolution for Edexcel IAL Maths, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Maths notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.28 new this month · 230 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
1IALM-P4-6-B-01P4 · A2Easy5 marks
The region R is bounded by the curve with equation y = x2 + 2, the coordinate axes and the line x = 1. The region R is rotated through 360° about the x-axis to form a solid.
Figure: the region R
(a)
Use algebraic integration to find the exact volume of the solid.
(4)
(b)
Give the volume to 3 significant figures.
(1)
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Answer
Mark
Notes
Part (a)
uses V = π∫y2 dx with y2 = (x2 + 2)2
M1
y2 = x4 + 4x2 + 4
A1
π[x5/5 + 4x3/3 + 4x]01
A1
π(1/5 + 4/3 + 4) = 83π/15
A1
Part (b)
17.4
B1
Common error: Squaring x² + 2 as x⁴ + 4. The middle term 4x² must be there: (a + b)² = a² + 2ab + b².
The curve C has equation y = ex + e−x. The region R is bounded by C, the coordinate axes and the line x = ln 2. The region R is rotated through 360° about the x-axis.
Figure: the region R
(a)
Show that y2 = e2x + 2 + e−2x.
(1)
(b)
Find the exact volume of the solid generated, giving your answer in the form π(a + b ln 2), where a and b are rational.
(4)
(c)
Give your answer to part (b) to 3 significant figures.
In this question r and h are positive constants. Use integration to prove each of the standard volume formulae below.
Figure: part (a): the region bounded by L, the x-axis and x = h
(a)
The line L passes through the origin O and the point (h, r). The region bounded by L, the x-axis and the line x = h is rotated through 360° about the x-axis. Show that the volume of the cone formed is ⅓πr2h.
(3)
(b)
The upper half of the circle x2 + y2 = r2 and the x-axis enclose a region that is rotated through 360° about the x-axis. Show that the volume of the sphere formed is 4πr3/3.
(3)
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Answer
Mark
Notes
Part (a)
L: y = rx/h
B1
V = π∫0h (r2/h2)x2 dx = π(r2/h2)[x3/3]0h
M1
= ⅓πr2h cso
A1
Part (b)
y2 = r2 − x2 with limits −r and r
B1
π[r2x − x3/3]−rr
M1
π(⅔r3 + ⅔r3) = 4πr3/3 cso
A1
Exam tip: In a proof, keep r and h as letters all the way through and state the limits explicitly; the result must come out exactly, not from a numerical example.
Figure 1 shows part of the curve C with equation y = x(3 − x). The curve meets the x-axis at the origin O and at the point A. The shaded region R is bounded by C and the x-axis. The region R is rotated through 360° about the x-axis to form a solid.
(a)
Write down the coordinates of A.
(1)
(b)
Use algebraic integration to find the exact volume of the solid.
(4)
(c)
Give the volume to 3 significant figures.
(1)
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Answer
Mark
Notes
Part (a)
A(3, 0)
B1
Part (b)
V = π∫03x2(3 − x)2 dx
M1
y2 = 9x2 − 6x3 + x4
A1
π[3x3 − 3x4/2 + x5/5]03
M1
π(81 − 243/2 + 243/5) = 81π/10
A1
Part (c)
25.4
B1
Common error: Squaring only part of the product. The whole of y must be squared: (x(3 − x))² = x²(3 − x)² = 9x² − 6x³ + x⁴.
Figure 1 shows part of the curve C with equation y = 1/√(2x + 1), x ≥ 0. The shaded region R is bounded by C, the coordinate axes and the line x = 4. The region R is rotated through 360° about the x-axis to form the solid S.
(a)
Show that the volume of S is π ln 3.
(4)
(b)
The plane x = 1, shown dashed in Figure 1, cuts S into two pieces. Show that the two pieces have equal volumes.
(2)
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Answer
Mark
Notes
Part (a)
V = π∫04 1/(2x + 1) dx
M1
π[½ ln(2x + 1)]
A1
π(½ ln 9 − ½ ln 1)
M1
½ ln 9 = ln 3 and ln 1 = 0, so V = π ln 3 cso
A1
Part (b)
π∫01 1/(2x + 1) dx = ½π ln 3
M1
this is half of π ln 3, so the other piece also has volume ½π ln 3
A1
Common error: Integrating 1/(2x + 1) as ln(2x + 1). Divide by the coefficient of x: the integral is ½ ln(2x + 1).
Figure 1 shows part of the curve with equation y = 2 cos x. The curve meets the y-axis at the point A and the positive x-axis at the point B. The shaded region R is bounded by the curve and the coordinate axes.
(a)
Write down the coordinates of A and of B.
(2)
(b)
The region R is rotated through 360° about the x-axis. Use the identity cos 2x ≡ 2 cos2x − 1 to show that the volume of the solid formed is π2.
(4)
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Answer
Mark
Notes
Part (a)
A(0, 2)
B1
B(π/2, 0)
B1
Part (b)
V = π∫0π/2 4 cos2x dx
M1
4 cos2x = 2 + 2 cos 2x
M1
π[2x + sin 2x]0π/2
A1
π((π + sin π) − 0) = π2 cso
A1
Exam tip: Rewrite cos²x with the double-angle identity before integrating. Here the sin 2x term is 0 at both limits, x = 0 and x = π/2.
A flower pot is modelled as the solid formed when the shaded region R in Figure 1 is rotated through 360° about the x-axis. The region R is bounded by the straight line L, the x-axis, the y-axis and the line x = 12. The two ends of L lie on grid points, shown as dots in Figure 1. Lengths are in centimetres. The pot has a closed base at x = 0 and an open top at x = 12, and the thickness of its wall is ignored.
(a)
Use Figure 1 to find an equation of L.
(2)
(b)
Use algebraic integration to find the capacity of the pot, giving your answer as an exact multiple of π.
(3)
(c)
A gardener says that the pot holds more than 1.5 litres of soil. Use your answer to part (b) to decide whether the gardener is correct.
(1)
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Answer
Mark
Notes
Part (a)
L passes through (0, 4) and (12, 8): gradient (8 − 4)/12 = 1/3
M1
y = 4 + x/3
A1
Part (b)
V = π∫012 (4 + x/3)2 dx
M1
π[16x + 4x2/3 + x3/27]012
A1
π(192 + 192 + 64) = 448π cm3
A1
Part (c)
448π ≈ 1407 cm3 = 1.41 litres, which is less than 1.5 litres, so the gardener is not correct
B1
Exam tip: On a grid, read two exact points on the line and state them, here (0, 4) and (12, 8), before finding the gradient and intercept.
The region R is bounded by the curve with equation y = tan x, 0 ≤ x < π/2, the x-axis and the line x = π/3. The solid S is formed when R is rotated through 360° about the x-axis.
Figure: the region R; x = π/2 is an asymptote
(a)
Starting from sin2x + cos2x ≡ 1, show that tan2x ≡ sec2x − 1.
(1)
(b)
Find the exact volume of S, giving your answer in the form π(p√3 + qπ), where p and q are rational.
(4)
(c)
Give the volume to 3 significant figures.
(1)
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Answer
Mark
Notes
Part (a)
divides every term by cos2x: tan2x + 1 ≡ sec2x, and rearranges
B1
Part (b)
V = π∫0π/3 (sec2x − 1) dx
M1
π[tan x − x]
A1
substitutes the limits: π((√3 − π/3) − 0)
M1
π(√3 − π/3), i.e. p = 1, q = −⅓
A1
Part (c)
2.15
B1
Common error: Trying to integrate tan²x directly. Replace it by sec²x − 1, whose integral, tan x − x, follows straight from the derivative of tan x.
The curve C has equation y = xe−x, x ≥ 0. The region R is bounded by C, the x-axis and the line x = 1.
Figure: the region R
(a)
Use integration by parts to find ∫x2e−2x dx.
(5)
(b)
Hence find the exact volume of the solid formed when R is rotated through 360° about the x-axis. Give your answer in the form π(e2 + a)/(be2), where a and b are integers.
(3)
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Answer
Mark
Notes
Part (a)
first application with u = x2: −½x2e−2x + ∫xe−2x dx
M1
correct expression
A1
second application: ∫xe−2x dx = −½xe−2x − ¼e−2x
M1
correct
A1
−¼e−2x(2x2 + 2x + 1) + c
A1
Part (b)
V = π∫01x2e−2x dx, using y2 = x2e−2x
M1
uses the answer to (a) with both limits: π(−(5/4)e−2 + ¼)
M1
π(e2 − 5)/(4e2)
A1
Common error: Forgetting the lower limit. At x = 0 the bracket is −¼, not 0, because e⁰ = 1, so it adds ¼ to the answer.
The curve C has equation y = 3/√(x(x + 3)), x > 0. The region R is bounded by C, the x-axis and the lines x = 1 and x = 3.
Figure: the region R
(a)
Express 9/(x(x + 3)) in partial fractions.
(2)
(b)
The region R is rotated through 360° about the x-axis. Use your answer to part (a) to find the exact volume of the solid generated, giving your answer in the form kπ ln 2, where k is an integer.
(5)
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Answer
Mark
Notes
Part (a)
9 ≡ A(x + 3) + Bx, with a value substituted or coefficients compared
The curve C has equation y = 2 sin x √(cos x), 0 ≤ x ≤ π/2. The region R is bounded by C and the x-axis. The region R is rotated through 360° about the x-axis.
Figure: the region R between C and the x-axis
(a)
Use the substitution u = sin x to show that the volume of the solid is 4π∫01u2 du.
(4)
(b)
Hence show that the solid has the same volume as a sphere of radius 1.
(2)
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Answer
Mark
Notes
Part (a)
y2 = 4 sin2x cos x, so V = π∫0π/2 4 sin2x cos x dx
B1
du = cos x dx
M1
limits: x = 0 gives u = 0 and x = π/2 gives u = 1
B1
4π∫01u2 du cso
A1
Part (b)
4π[u3/3]01 = 4π/3
M1
sphere: (4/3)π × 13 = 4π/3, the same
A1
Exam tip: When a substitution is given, change all three parts of the integral: the integrand, the dx and the limits.
The curve C has equation y = x2 + 1 and the line L has equation y = 3 − x. The finite region R is bounded by C and L.
Figure: the finite region R between C and L
(a)
Find the coordinates of the points where C and L meet.
(3)
(b)
The region R is rotated through 360° about the x-axis. Explain why the volume of the solid is not π∫−21 (2 − x − x2)2 dx.
(1)
(c)
Find the exact volume of the solid.
(5)
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Answer
Mark
Notes
Part (a)
x2 + 1 = 3 − x ⇒ x2 + x − 2 = 0
M1
(x + 2)(x − 1) = 0, x = −2 or 1
A1
(−2, 5) and (1, 2)
A1
Part (b)
each cross-section is a washer of area π(3 − x)2 − π(x2 + 1)2, which is not π(difference of radii)2
B1
Part (c)
V = π∫−21 [(3 − x)2 − (x2 + 1)2] dx, line as the outer radius
M1
integrand 8 − 6x − x2 − x4
A1
π[8x − 3x2 − x3/3 − x5/5]
M1
π[(8 − 3 − ⅓ − ⅕) − (−16 − 12 + 8/3 + 32/5)]
M1
117π/5
A1
Exam tip: Check which graph is further from the axis with a test value: at x = 0 the line gives 3 and the curve gives 1, so the line is the outer radius.
Figure 1 shows the curve C with equation y = 2√x, x ≥ 0, and the line L with equation y = x. C and L meet at the origin O and at the point P. The shaded region R is bounded by C and L.
(a)
Find the coordinates of P.
(2)
(b)
The region R is rotated through 360° about the x-axis. Use algebraic integration to find the exact volume of the solid generated.
(4)
(c)
The region bounded by C, the x-axis and the line x = 4 is rotated through 360° about the x-axis to form the solid T. Show that the volume of the solid in part (b) is one third of the volume of T.
(2)
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Answer
Mark
Notes
Part (a)
2√x = x ⇒ 4x = x2
M1
x = 4: P(4, 4)
A1
Part (b)
V = π∫04 [(2√x)2 − x2] dx, with C as the outer radius
M1
π∫04 (4x − x2) dx
A1
π[2x2 − x3/3]04
M1
π(32 − 64/3) = 32π/3
A1
Part (c)
volume of T = π∫04 4x dx = 32π
M1
(32π/3) ÷ 32π = 1/3
A1
Common error: Using π∫(2√x − x)² dx. Each slice is a washer, so subtract the squares of the radii, π(R² − r²), not the square of their difference.
Figure 1 shows part of the curve with equation y = ex and the line with equation y = e. The curve crosses the y-axis at (0, 1) and meets the line at the point Q. The shaded region R is bounded by the curve, the line and the y-axis.
(a)
Find the coordinates of Q.
(1)
(b)
The region R is rotated through 360° about the x-axis. Find the exact volume of the solid generated, giving your answer in the form π(e2 + a)/b, where a and b are integers.
(4)
(c)
The region bounded by the curve, the coordinate axes and the line x = 1 is rotated through 360° about the x-axis to form the solid T. Find the volume of T, and show that the solid in part (b) and T together have the volume of a cylinder of radius e and length 1.
(2)
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Answer
Mark
Notes
Part (a)
ex = e ⇒ Q(1, e)
B1
Part (b)
V = π∫01 (e2 − e2x) dx
M1
π[e2x − ½e2x]01
A1
π[(e2 − ½e2) − (0 − ½)]
M1
π(e2 + 1)/2
A1
Part (c)
volume of T = π∫01 e2x dx = π(e2 − 1)/2
B1
π(e2 + 1)/2 + π(e2 − 1)/2 = πe2 = π × e2 × 1
B1
Exam tip: The line y = e rotated about the x-axis sweeps out a cylinder, so its part of the integrand is just the constant e².
Figure 1 shows the curve C with parametric equations x = sec t, y = tan t, 0 ≤ t ≤ π/3. C meets the x-axis at the point where t = 0. The shaded region R is bounded by C, the x-axis and the line x = 2, which meets C at the point where t = π/3. The region R is rotated through 360° about the x-axis to form a solid.
(a)
Show that the volume of the solid is π∫0π/3 tan3t sec t dt.
(3)
(b)
Use the substitution u = sec t to find the exact volume of the solid.
(3)
(c)
Show that a Cartesian equation of C is x2 − y2 = 1, and hence verify your answer to part (b) by integrating with respect to x.
(2)
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Answer
Mark
Notes
Part (a)
dx/dt = sec t tan t
B1
V = π∫y2(dx/dt) dt with t-limits 0 and π/3
M1
π∫0π/3 tan2t × sec t tan t dt = π∫0π/3 tan3t sec t dt cso
A1
Part (b)
tan2t = u2 − 1, du = sec t tan t dt, limits u = 1 and u = 2
M1
π∫12 (u2 − 1) du = π[u3/3 − u]12
A1
π((8/3 − 2) − (1/3 − 1)) = 4π/3
A1
Part (c)
sec2t − tan2t = 1 gives x2 − y2 = 1
B1
π∫12 (x2 − 1) dx = π[x3/3 − x]12 = 4π/3
B1
Exam tip: Match the limits to the figure: x runs from 1 to 2 while t runs from 0 to π/3, and dx/dt > 0 on that interval, so the integral comes out positive.
Figure 1 shows the curve with equation y = x√(sin x), 0 ≤ x ≤ π. The shaded region R is bounded by the curve and the x-axis. The region R is rotated through 360° about the x-axis to form a solid.
(a)
Use integration by parts to find ∫x2 sin x dx.
(4)
(b)
Hence find the exact volume of the solid.
(2)
(c)
Give the volume to 3 significant figures.
(1)
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Answer
Mark
Notes
Part (a)
with u = x2: −x2 cos x + ∫2x cos x dx
M1
correct expression
A1
∫2x cos x dx = 2x sin x − ∫2 sin x dx = 2x sin x + 2 cos x
M1
−x2 cos x + 2x sin x + 2 cos x + c
A1
Part (b)
V = π∫0πx2 sin x dx, using y2 = x2 sin x
M1
π[(π2 + 0 − 2) − (0 + 0 + 2)] = π(π2 − 4)
A1
Part (c)
18.4
B1
Common error: Dropping the lower limit. At x = 0 the bracket is 2 cos 0 = 2, not 0, so 2 must be subtracted.
A wooden peg is modelled as the solid formed when a region R is rotated through 360° about the x-axis. Figure 1 shows the profile on a centimetre grid. The region R is bounded by the y-axis, the x-axis, the line y = 3 for 0 ≤ x ≤ 6, and the curve with equation y = √(k(9 − x)) for 6 ≤ x ≤ 9, where k is a positive constant. The line meets the curve at a grid point, and the curve meets the x-axis at a grid point; both points are shown as dots in Figure 1.
(a)
Use Figure 1 to find the value of k.
(2)
(b)
Show that the volume of the peg is 135π/2 cm3.
(5)
(c)
The wood has density 0.65 g cm−3. Find the mass of the peg, giving your answer to 3 significant figures.
(2)
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Answer
Mark
Notes
Part (a)
the curve passes through (6, 3): 3 = √(3k)
M1
k = 3
A1
Part (b)
cylinder part: π × 32 × 6 = 54π
B1
curved end: π∫69 3(9 − x) dx
M1
3π[9x − x2/2]69
A1
3π(81/2 − 36) = 27π/2
A1
54π + 27π/2 = 135π/2 cso
A1
Part (c)
0.65 × 135π/2
M1
138 g
A1
Exam tip: Split a composite solid where the profile changes, here at x = 6. A part with constant radius is a cylinder, so πr²h is quicker than integrating.
A fishing float is a solid of revolution. Figure 1 shows its profile on a centimetre grid, with the x-axis along its axis of symmetry. The radius y cm of the float was measured at x = 0, 4, 8, 12, 16 and 20. The measurements lie on grid points, shown as dots in Figure 1.
(a)
Use the trapezium rule with all six measurements to estimate the volume of the float. Give your answer to 3 significant figures.
(4)
(b)
A designer models the profile by the curve with equation y2 = cx(20 − x), 0 ≤ x ≤ 20, shown dashed in Figure 1. The constant c is chosen so that the greatest radius of the model is 6 cm. Find c.
(2)
(c)
Find the volume of the model, giving your answer as an exact multiple of π.
(3)
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Answer
Mark
Notes
Part (a)
values of y2: 0, 25, 36, 36, 25, 0
B1
V ≈ π∫020y2 dx with strip width 4
M1
π × ½ × 4 × [0 + 0 + 2(25 + 36 + 36 + 25)] = 488π
M1
1530 cm3
A1
Part (b)
y is greatest where x(20 − x) is greatest, at x = 10
M1
100c = 36, c = 0.36
A1
Part (c)
V = π∫020 0.36(20x − x2) dx
M1
0.36π[10x2 − x3/3]020
A1
0.36π(4000 − 8000/3) = 480π cm3
A1
Exam tip: For a volume, apply the trapezium rule to y², not to y: square every measurement first, then multiply the estimate by π.
A designer models a ball used in a field game. In the first model, the ball is the solid formed when the region between the curve y = 9 − x2/25, −15 ≤ x ≤ 15, and the x-axis is rotated through 360° about the x-axis. In the second model, the ball is formed by rotating the region between the x-axis and the upper half of the ellipse with parametric equations x = 15 cos t, y = 9 sin t, 0 ≤ t ≤ π, through 360° about the x-axis. Lengths are in centimetres.
Figure: the two profile curves (lengths in cm)
(a)
Show that the volume of the first model is 1296π cm3.
(5)
(b)
Show that the volume of the second model is 1620π cm3.
(4)
(c)
A ball of this type has a volume of 4.9 litres. State, with a reason, which model gives the better estimate of its volume.
(2)
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Answer
Mark
Notes
Part (a)
V = π∫−1515 (9 − x2/25)2 dx
M1
integrand 81 − 18x2/25 + x4/625
A1
π[81x − 6x3/25 + x5/3125]
M1
at 15: 1215 − 810 + 243 = 648, and at −15: −648
A1
π(648 + 648) = 1296π cso
A1
Part (b)
dx/dt = −15 sin t; x = −15 at t = π and x = 15 at t = 0
B1
V = π∫π0 81 sin2t (−15 sin t) dt = 1215π∫0π sin3t dt
M1
sin3t = sin t − sin t cos2t, integral [−cos t + ⅓cos3t] from 0 to π = 4/3
M1
1215π × 4/3 = 1620π cso
A1
Part (c)
first model 4.07 litres, second model 5.09 litres
M1
the second (ellipse) model, because 5.09 is closer to 4.9 than 4.07 is
A1
Exam tip: When x decreases as t increases, set the t-limits to match x from −15 to 15 (t from π to 0); the negative dx/dt then gives a positive volume.
The curve C has equation y = 4/x, x > 0. The point P(2, 2) lies on C, and the line T is the tangent to C at P. The region R is bounded by C, T and the line x = 1.
Figure: the region R between C, T and the line x = 1
(a)
Show that T has equation y = 4 − x.
(3)
(b)
Show that 4/x − (4 − x) = (x − 2)2/x, and hence explain why C lies on or above T for all x > 0.
(2)
(c)
The region R is rotated through 360° about the x-axis. Find the exact volume of the solid generated.
(5)
(d)
By considering two cones, verify the value of π∫12 (4 − x)2 dx found in part (c).
(2)
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Answer
Mark
Notes
Part (a)
dy/dx = −4/x2
M1
gradient at P: −4/4 = −1
A1
y − 2 = −(x − 2), so y = 4 − x cso
A1
Part (b)
(4 − 4x + x2)/x = (x − 2)2/x
M1
a square is ≥ 0 and x > 0, so the difference is never negative
A1
Part (c)
V = π∫12 (16/x2 − (4 − x)2) dx, curve as the outer radius
M1
π∫12 16/x2 dx = π[−16/x]12 = 8π
A1
π∫12 (4 − x)2 dx = π[−(4 − x)3/3]12
A1
substitutes the limits: π(−8/3 + 9) = 19π/3
M1
5π/3
A1
Part (d)
T meets the x-axis at x = 4: cone of radius 3, height 3 minus cone of radius 2, height 2
M1
⅓π(27 − 8) = 19π/3, as before
A1
Exam tip: Where a line is a tangent, 'curve minus line' has a repeated factor, here (x − 2)². Spotting it tells you which graph is outer without testing points.
A ceramic bowl is modelled as a solid of revolution. The x-axis is the axis of symmetry of the bowl, and the lowest point of the inside surface is at the origin O. The inside surface is formed by rotating the curve y = √(18x), 0 ≤ x ≤ 8, through 360° about the x-axis, and the outside surface by rotating the curve y = √(18(x + 0.5)), −0.5 ≤ x ≤ 8, through 360° about the x-axis. The rim of the bowl is the flat ring in the plane x = 8 between the two surfaces. Lengths are in centimetres.
Figure: cross-section of the bowl through its axis; the ceramic is shaded
(a)
Show that the capacity of the bowl (the volume inside the inside surface up to the plane x = 8) is 576π cm3.
(3)
(b)
Find the exact volume of ceramic used to make the bowl.
(4)
(c)
The ceramic has density 2.4 g cm−3. Find the mass of the bowl, to the nearest 10 g.
(2)
(d)
The bowl is held with its axis vertical and its opening uppermost, and soup is poured in to a depth of d cm, measured along the axis from O. Find the exact value of d for which the bowl is filled to half its capacity.
(3)
(e)
Give one limitation of this model of the bowl.
(1)
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Answer
Mark
Notes
Part (a)
π∫08 18x dx
M1
π[9x2]08
A1
9 × 64π = 576π cso
A1
Part (b)
outer solid: π∫−0.58 18(x + 0.5) dx
M1
π[9(x + 0.5)2]−0.58 = 9 × 8.52π = 650.25π
A1
subtracts the capacity: 650.25π − 576π
M1
297π/4 cm3 (= 74.25π)
A1
Part (c)
2.4 × 297π/4 = 559.8…
M1
560 g
A1
Part (d)
π∫0d 18x dx = 9πd2
M1
9πd2 = 288π
M1
d = 4√2 (≈ 5.66)
A1
Part (e)
e.g. a real bowl has a flat foot so it can stand, not a curved point; the walls may not have exactly this shape or thickness; the ceramic may not be of uniform density
B1
Exam tip: For a hollow object, find the volume enclosed by the outside surface and subtract the volume enclosed by the inside surface; do not try to integrate the wall directly.
Figure 1 shows the curve C with parametric equations x = t − sin t, y = 1 − cos t, 0 ≤ t ≤ 2π. The curve meets the x-axis at the origin O and at the point B. The shaded region R is bounded by C and the x-axis.
(a)
Find the coordinates of B.
(2)
(b)
The region R is rotated through 360° about the x-axis. Show that the volume of the solid generated is π∫02π (1 − cos t)3 dt.
(3)
(c)
Use the identity cos3t ≡ cos t(1 − sin2t) to show that ∫02π cos3t dt = 0.
(2)
(d)
Hence find the exact volume of the solid.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
y = 0 ⇒ cos t = 1 ⇒ t = 2π (as t ≠ 0 at B)
M1
B(2π, 0)
A1
Part (b)
dx/dt = 1 − cos t
B1
V = π∫y2(dx/dt) dt with limits t = 0 and t = 2π
M1
(1 − cos t)2(1 − cos t) = (1 − cos t)3 cso
A1
Part (c)
∫(cos t − cos t sin2t) dt = sin t − ⅓ sin3t
M1
both terms are 0 at t = 0 and at t = 2π, so the integral is 0
A1
Part (d)
(1 − cos t)3 = 1 − 3 cos t + 3 cos2t − cos3t
M1
3 cos2t = (3/2)(1 + cos 2t), so the integral is 2π − 0 + 3π − 0
M1
5π2
A1
Exam tip: Over a whole number of periods, the integrals of cos t, cos 2t and cos³t are all 0, so only the constant terms contribute.
Figure 1 shows the curves with equations y = sin x and y = cos x, 0 ≤ x ≤ π/2, which meet at the point P. The region S, shaded gold, is bounded by the two curves and the y-axis. The region R, shaded blue, is bounded by the two curves and the x-axis.
(a)
Find the exact coordinates of P.
(2)
(b)
Show that the volume of the solid formed when R is rotated through 360° about the x-axis is π(π − 2)/4.
(5)
(c)
Find the exact volume of the solid formed when S is rotated through 360° about the x-axis.
(3)
(d)
Together, R and S form the region between the curve y = cos x and the x-axis for 0 ≤ x ≤ π/2. Use this fact to check your answers to parts (b) and (c).
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
sin x = cos x ⇒ tan x = 1
M1
P(π/4, √2/2)
A1
Part (b)
V = π∫0π/4 sin2x dx + π∫π/4π/2 cos2x dx
M1
uses sin2x = ½(1 − cos 2x) and cos2x = ½(1 + cos 2x)
Common error: Taking R as the whole region under y = sin x up to π/2. R lies under the lower curve on each side of P, so split the integral at x = π/4.
A craft workshop makes wooden beads. Each bead starts as a ball, and a cylindrical channel is bored through the centre of the ball. Figure 1 shows a cross-section of one bead through the axis of its channel, on a millimetre grid. The outline of the ball is the circle x2 + y2 = 65, and the walls of the channel are the lines y = 4 and y = −4. The line y = 4 meets the circle at A and B. The bead is the solid formed when the shaded region R, bounded by the line y = 4 and the minor arc AB, is rotated through 360° about the x-axis.
(a)
Find the coordinates of A and of B, and hence write down the length of the bead.
(2)
(b)
Use algebraic integration to show that the volume of the bead is 1372π/3 mm3.
(5)
(c)
The workshop also makes a bead of the same length, 14 mm, from a ball of radius 10 mm. Find the exact radius of its channel, and show by integration that this bead has the same volume as the bead in Figure 1.
(3)
(d)
By considering the integrand, explain why every bead of length 14 mm made in this way has volume 1372π/3 mm3, whatever the radius of the ball.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
x2 + 16 = 65 ⇒ x = ±7
M1
A(−7, 4), B(7, 4); length 14 mm
A1
Part (b)
V = π∫−77 [(65 − x2) − 42] dx
M1
π∫−77 (49 − x2) dx
A1
π[49x − x3/3]
M1
π[(343 − 343/3) − (−343 + 343/3)]
M1
= 1372π/3 mm3 cso
A1
Part (c)
channel radius2 = 102 − 72 = 51, radius √51 mm
B1
V = π∫−77 [(100 − x2) − 51] dx
M1
= π∫−77 (49 − x2) dx, the same integral as in part (b), so the volume is 1372π/3 mm3
A1
Part (d)
for a ball of radius r (> 7), channel radius2 = r2 − 49, so the integrand is (r2 − x2) − (r2 − 49)
M1
= 49 − x2, and the limits are ±7: neither depends on r, so the volume is always 1372π/3 mm3
A1
Exam tip: When a cylindrical channel of radius a runs along the axis, every washer has the same inner radius a, so the integrand is (outer radius)² − a².
Figure 1 shows the curve C with equation y = x(4 − x2)1/4, 0 ≤ x ≤ 2. The curve has a maximum point M and meets the x-axis at O and at the point B(2, 0). The shaded region R is bounded by C and the x-axis.
(a)
Find the exact x-coordinate of M.
(4)
(b)
The region R is rotated through 360° about the x-axis. Use the substitution x = 2 sin θ to show that the volume of the solid formed is 16π∫0π/2 sin2θ cos2θ dθ.
(4)
(c)
Hence show that the volume of the solid is π2.
(4)
Show mark scheme
Answer
Mark
Notes
Part (a)
product and chain rules: dy/dx = (4 − x2)1/4 + x × ¼(4 − x2)−3/4 × (−2x)
M1
correct derivative
A1
dy/dx = 0 ⇒ (4 − x2) − x2/2 = 0
M1
x = √(8/3) = 2√6/3
A1
Part (b)
y2 = x2(4 − x2)1/2
B1
dx = 2 cos θ dθ, with limits θ = 0 and θ = π/2
M1
(4 − 4 sin2θ)1/2 = 2 cos θ for 0 ≤ θ ≤ π/2
M1
π∫ 4 sin2θ × 2 cos θ × 2 cos θ dθ = 16π∫0π/2 sin2θ cos2θ dθ cso
A1
Part (c)
sin2θ cos2θ = ¼ sin22θ
M1
= ⅛(1 − cos 4θ)
M1
2π[θ − ¼ sin 4θ]0π/2
A1
2π × π/2 = π2 cso
A1
Common error: Writing √(4 − 4 sin²θ) as 2 − 2 sin θ. Use 1 − sin²θ = cos²θ first: √(4 cos²θ) = 2 cos θ, which is positive for 0 ≤ θ ≤ π/2.
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