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Topic 5.20 · HL only

Volume of revolutions + HL area problems: notes and practice questions

Summary
  • Calculate the area between curves by integrating the difference between the upper and lower functions with respect to x or y.
  • If limits are not provided, find them by solving for intersection points.
  • For volumes of revolution, integrate π\pi times the square of the function(s) with respect to the axis of rotation.
  • When rotating the area between two curves, use the washer method: π∫(Router2−Rinner2)d(variable)\pi \int (R_{outer}^2 - R_{inner}^2) d(variable).
  • Always sketch the graphs to identify relevant regions and functions.
  • Pay attention to whether an exact value or decimal approximation is required.

How it is examined

Rotating about the yy-axis means writing xx in terms of yy and changing the limits to yy values, and doing neither is the common failure. The π\pi gets dropped often enough that it is worth a marking point of its own. Volume between two curves is π∫(y12−y22) dx\pi\int (y_1^2 - y_2^2)\,\mathrm{d}x, not π∫(y1−y2)2 dx\pi\int (y_1 - y_2)^2\,\mathrm{d}x. 6 to 8 marks.

Given in the booklet

V=π∫aby2 dxV = \pi\displaystyle\int_a^b y^2\,\mathrm{d}x and V=π∫abx2 dyV = \pi\displaystyle\int_a^b x^2\,\mathrm{d}y are given.

Key ideas
  • Area of the region enclosed by a curve and the yy-axis in a given interval.
  • Volumes of revolution about the xx-axis or yy-axis.

Linking questions

  • Other contexts: industrial design.

Practice questions

6 questions · 1 easy · 3 medium · 2 hard
Showing 6 of 6

Question 1

EasyPaper 2 · calculator4 marks

Consider the function f(x)=(8x−3)e−2xf(x) = (8x - 3)e^{- 2x}. If the region enclosed between the function, x-axis and y-axis is rotated through 2π2\pi about the x-axis, find the volume of revolution.

Question 2

MediumPaper 1 · no calculator7 marks

A solid is formed by rotating the curve with equation x2=8yx^2 = 8y for y≥0y \ge 0 by 360∘360^\circ about the yy-axis. This forms a solid paraboloid.

A cylindrical hole of radius 44 is drilled through the center of the paraboloid, along the yy-axis. The resulting solid is a ring of height hh.

This information is shown in the following diagrams.

Diagram showing a cross-section of a paraboloid with a cylindrical hole and a 3D view of the resulting ring

The volume of the ring is 144π144\pi.

Find the value of hh.

Question 3

HardPaper 1 · no calculator20 marks
(a)

Consider the family of integrals defined by In=∫xne−x dxI_n = \int x^n e^{-x} \, dx for n∈N0n \in \mathbb{N}_0, where N0={0,1,2,...}\mathbb{N}_0 = \{0, 1, 2, ...\}.

(a) By using integration by parts, show that In=−xne−x+nIn−1I_n = -x^n e^{-x} + n I_{n-1} for n≥1n \ge 1.

[3]
(b)

(b) Hence, find an explicit expression for ∫x3e−x dx\int x^3 e^{-x} \, dx.

[4]
(c)

(c) The region RR is enclosed by the graph of y=x3/2e−x/2y = x^{3/2} e^{-x/2} and the xx-axis for x≥0x \ge 0. The region RR is rotated by 2π2\pi radians about the xx-axis. Find the volume of the solid generated.

[5]
(d)

(d) Show that lim⁡x→∞xne−x=0\lim_{x \to \infty} x^n e^{-x} = 0 for any n∈Nn \in \mathbb{N}.

[3]
(e)(i)

Consider the function h(x)=xe−xh(x) = x e^{-x}.

(e) (i) Find the Maclaurin series for h(x)h(x) up to and including the term in x4x^4.

[3]
(e)(ii)

(ii) Hence, find the value of the fourth derivative of h(x)h(x) at x=0x=0, i.e. h(4)(0)h^{(4)}(0).

[2]

Question 4

MediumPaper 1 · no calculator6 marks

The function ff is defined as f(x)=arctan⁡x1+x2f(x) = \sqrt{\frac{\arctan x}{1+x^2}}, where x≥0x \ge 0.

Consider the shaded region R enclosed by the graph of ff, the xx-axis and the line x=1x = 1, as shown in the following diagram.

Graph of y=f(x) with shaded region R

The shaded region R is rotated by 2π2\pi radians about the xx-axis to form a solid.

Show that the volume of the solid is π332\frac{\pi^3}{32}.

Question 5

HardPaper 2 · calculator20 marks
(a)

A designer is creating a decorative glass container shaped like a dome. The outer profile of the container can be modelled by the function f(x)=9−x2f(x) = \sqrt{9-x^2}, where 0≤x≤30 \le x \le 3 and xx and yy are measured in metres.

Sketch the curve y=f(x)y = f(x), clearly indicating the coordinates of the endpoints.

[2]
(b)(i)

Show that the inverse function of ff is given by f−1(x)=9−x2f^{-1}(x) = \sqrt{9-x^2}.

[3]
(b)(ii)

State the domain and range of f−1f^{-1}.

[2]
(c)(i)

The container is formed by rotating the curve y=f(x)y = f(x) by 2π2\pi about the y-axis. Show that the volume, V m3V \text{ m}^3, of liquid in the container when it is filled to a height of hh metres is given by V=π(9h−13h3)V = \pi \left( 9h - \frac{1}{3}h^3 \right).

[3]
(c)(ii)

Hence, determine the maximum volume of the container.

[2]
(d)

At t=0t = 0, the container is empty. Liquid is then added to the container at a constant rate of 0.5 m3s−10.5 \text{ m}^3\text{s}^{-1}.

Find the time it takes to fill the container to its maximum volume.

[2]
(e)

Find the rate of change of the height of the liquid when the container is filled to half its maximum volume.

[6]

Question 6

MediumPaper 1 · no calculator7 marks
(a)

Consider the function g(x)=csc⁡(x+π3)g(x) = \csc\left(x+\frac{\pi}{3}\right), for 0≤x≤π30\le x\le \frac{\pi}{3}.

Determine the range of gg.

[3]
(b)

The region bounded by the graph of y=g(x)y = g(x), the xx-axis and the lines x=0x = 0 and x=π3x = \frac{\pi}{3} is rotated 2π2\pi radians about the xx-axis.

Find the volume of the solid generated.

[4]

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What does Volume of revolutions + HL area problems cover in IB Maths AA?

Calculate the area between curves by integrating the difference between the upper and lower functions with respect to x or y. If limits are not provided, find them by solving for intersection points. For volumes of revolution, integrate π times the square of the function(s) with respect to the axis of rotation.

Is Volume of revolutions + HL area problems SL or HL?

Volume of revolutions + HL area problems is HL only. SL students are not examined on it.

How do I revise Volume of revolutions + HL area problems for IB Maths AA?

Start from the core idea: calculate the area between curves by integrating the difference between the upper and lower functions with respect to x or y. In the exam: rotating about the y-axis means writing x in terms of y and changing the limits to y values, and doing neither is the common failure. The π gets dropped often enough that it is worth a marking point of its own. Then practise exam-style questions, easiest first, writing out every step of your working before you check it.

How does FourtyFive help me practise Volume of revolutions + HL area problems?

FourtyFive has 6 Volume of revolutions + HL area problems questions. Every answer you write is marked mark by mark, IB-style, and you see where each mark was won or lost. Every part has a hint, the AI tutor helps you through the step you are stuck on, and your Study Profile picks what to practise next.

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