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Topic 5.11 · HL only

Area under/between a curve, volumes of revolution: notes and practice questions

Summary
  • Integration is anti-differentiation.
  • Definite integral calculates exact accumulation with specific boundary values (limits aa and bb).
  • Constant of integration (+c+c) is not needed for definite integrals as it cancels out.
  • For physical area, use modulus (absolute value) if the definite integral evaluates to a negative number (area below x-axis or left of y-axis).
  • Volume of Revolution (HL): 3D solid formed by rotating a 2D area 2π2\pi radians around an axis.
  • Fundamental Theorem of Calculus: ∫abf(x)dx=F(b)−F(a) \int_{a}^{b} f(x) dx = F(b) - F(a)
  • Area under a curve (x-axis): A=∫ab∣y∣dx A = \int_{a}^{b} |y| dx
  • Area between a curve and the y-axis (HL): A=∫ab∣x∣dy A = \int_{a}^{b} |x| dy (Rearrange y=f(x)y=f(x) to x=g(y)x=g(y)).
  • Volume of Revolution around the x-axis (HL): V=π∫aby2dx V = \pi \int_{a}^{b} y^2 dx
  • Volume of Revolution around the y-axis (HL): V=π∫abx2dy V = \pi \int_{a}^{b} x^2 dy (Rearrange y=f(x)y=f(x) to x=g(y)x=g(y)).
  • Always sketch the graph to identify limits and potential negative areas.
  • If limits are not given, find axis intercepts (roots).
  • For area against y-axis or volume around y-axis, rearrange y=f(x)y=f(x) to make xx the subject (x=g(y)x=g(y)).
  • For area between curve and line: Find intersection points, determine if areas need to be added or subtracted.
  • GDC 'Variable Dummy Trick': When integrating with respect to yy (e.g., ∫x2dy\int x^2 dy), type 'x' as the dummy variable on the GDC (e.g., ∫x2dx\int x^2 dx).
  • Use GDC's absolute value function ('Abs' or ∣…∣| \dots |) for areas that cross an axis.
  • For exact volumes with π\pi, evaluate the integral without π\pi on GDC, convert to fraction, then multiply by π\pi.
  • SL: Area under curve (x-axis), area between curves/lines.
  • HL: Area between curve and y-axis, all Volumes of Revolution.
  • Do not include the constant of integration (+c+c) in definite integrals.
  • Always sketch to avoid errors from positive and negative areas cancelling out.

How it is examined

The rotation axis is the detail that decides which variable the integral runs over, and a question that rotates about the y-axis expects the curve rearranged as xx in terms of yy first. Region-enclosed-by-a-curve questions that dip below the axis need the interval split at the root and each piece's absolute value summed, or the negative and positive parts cancel and understate the true area.

Given in the booklet

V=∫abπy2 dxV = \displaystyle\int_a^b \pi y^2\,dx and V=∫abπx2 dyV = \displaystyle\int_a^b \pi x^2\,dy.

Key ideas
  • Find the area of the region enclosed by a curve and the x-axis or y-axis over a given interval, including where the integral is negative.
  • Find volumes of revolution about the x-axis or y-axis, V=∫abπy2 dxV = \displaystyle\int_a^b \pi y^2\,dx or V=∫abπx2 dyV = \displaystyle\int_a^b \pi x^2\,dy.

Linking questions

  • Other contexts: industrial design, architecture.
  • International-mindedness: Liu Hui's calculation of the volume of a cylinder, the use of infinitesimals by Greek geometers, and Ibn al-Haytham's integration of a function to find the volume of a paraboloid.

Practice questions

40 questions · 25 medium · 15 hard
Showing 20 of 20

Question 1

MediumPaper 1 · calculator8 marks
(a)(i)

(a) An architect is designing the entrance to a new tunnel. The shape of the entrance is modelled by a quadratic curve, with the base of the tunnel on the x-axis. The curve has end points (0, 6) and (10, 6), and its vertex is (5, 9). Distances are measured in metres.

The quadratic curve can be expressed in the form y=ax2+bx+cy = ax^2 + bx + c for 0≤x≤100 \leq x \leq 10.

(a.i) Write down the value of cc.

[1]
(a)(ii)

(a.ii) Hence, form two equations in terms of aa and bb.

[2]
(a)(iii)

(a.iii) Hence, find the equation of the quadratic curve.

[2]
(b)

(b) Calculate the area of the tunnel entrance.

[3]

Question 2

HardPaper 1 · calculator10 marks
(a)

A mathematical model for a physical phenomenon involves the expression (1u+2)2\left(\frac{1}{u} + 2\right)^2.

(a) Expand (1u+2)2\left(\frac{1}{u} + 2\right)^2.

[2]
(b)

The rate of change of a certain quantity is given by f′(x)=1(x+1)2+4x+1+4f'(x) = \frac{1}{(x+1)^2} + \frac{4}{x+1} + 4.

(b) Find the indefinite integral ∫(1(x+1)2+4x+1+4)dx\int \left( \frac{1}{(x+1)^2} + \frac{4}{x+1} + 4 \right) dx.

[4]
(c)

A designer is creating a custom-shaped container. The cross-sectional profile of the container is defined by the curve y=1x+1+2y = \frac{1}{x+1} + 2 for 1≤x≤31 \le x \le 3. The container is formed by rotating this region 2π2\pi about the x-axis.

(c) Calculate the volume of the solid formed. Give your answer in the form π4(a+bln⁡(c))\frac{\pi}{4}(a + b \ln(c) ), where a,b,c∈Za, b, c \in \mathbb{Z}.

[4]

Question 3

MediumPaper 1 · calculator7 marks
(a)

The internal shape of a specialized chemical reactor vessel is formed by rotating the curve y=8ln⁡xy = 8 \ln x, for 0≤y≤120 \le y \le 12, about the yy-axis, where xx and yy are measured in centimetres. The vessel contains a liquid to a height of hh cm.

Show that the volume of the liquid, VV, in terms of hh is V=4π(eh4−1)V = 4\pi(e^{\frac{h}{4}} - 1).

[5]
(b)

Hence find the maximum capacity of the reactor vessel in cm3^3. Give your answer to three significant figures.

[2]

Question 4

HardPaper 1 · calculator9 marks
(a)(i)

A landscape architect is designing a decorative water channel. The cross-section of the channel is modelled by the function f(x)=x3−4xf(x) = x^3 - 4x, for −2≤x≤2-2 \leq x \leq 2. The shaded region, RR, represents the cross-sectional area of the channel, bounded by the graph of y=f(x)y = f(x) and the xx-axis.

Graph of y=f(x) = x^3-4x from -2 to 2, with shaded region R between the curve and x-axis. The curve is above the x-axis for x in [-2,0 and below for x in [0,2].]

Write down an integral that represents the area of RR.

[2]
(a)(ii)

Find the area of RR.

[2]
(b)

The architect considers a modified design, where the cross-section is given by g(x)=0.5f(x+1)g(x) = 0.5f(x+1).

On the following set of axes, the graph of y=f(x)y = f(x) has been drawn. On the same set of axes, sketch the graph of y=g(x)y = g(x).

Axes with graph of y=f(x)=x^3-4x drawn from -3 to 3, ready for sketch of y=g(x). The graph of f(x) has roots at -2, 0, 2 and local max/min at approx (-1.15, 3.08) and (1.15, -3.08).
[2]
(c)

The region RR (the original cross-section) is rotated through 2π2\pi radians about the xx-axis to form a three-dimensional decorative element. Find the volume of this element.

[3]

Question 5

MediumPaper 1 · calculator4 marks

A rocket engineer is designing a new nozzle. The internal shape of the nozzle is formed by rotating a curve about the central axis (y-axis).

The nozzle is 12 cm long. The internal radius of the nozzle is measured at 3 cm intervals along its length:

Length from base (cm)Radius (cm)
05
34
63
92
121.5

Use the trapezoidal rule to estimate the internal volume of the nozzle.

Question 6

HardPaper 1 · calculator7 marks

(a) A designer is creating a unique component for a specialized optical instrument. The cross-section of the component's profile can be modelled by the curve y=xcos⁡(x)y = \sqrt{x \cos(x)} for x∈[π6,π2]x \in \left[\frac{\pi}{6}, \frac{\pi}{2}\right]. The component is formed by rotating this curve about the xx-axis.

Calculate the exact volume of this solid of revolution. Give your answer in the form Aπ2+BπC+DEA\pi^2 + B\pi\sqrt{C} + D\sqrt{E} or a similar exact form, and then to three significant figures.

Question 7

MediumPaper 1 · calculator10 marks
(a)

A landscape architect is designing a section of a park. The boundary of a planned pathway can be modelled by the line y=−x+8y = -x + 8 and a decorative flower bed by the curve y=0.5x2−4x+10.5y = 0.5x^2 - 4x + 10.5. These two features intersect at points (1,7)(1, 7) and (5,3)(5, 3), as shown in the following diagrams.

In diagram 1, the region enclosed by the line y=−x+8y = -x + 8, x=1x = 1, x=5x = 5 and the xx-axis has been shaded.

Diagram 1 showing a shaded region under the line y=-x+8 from x=1 to x=5, with intersection points (1,7) and (5,3).

Calculate the area of the shaded region in diagram 1.

[2]
(b)(i)

In diagram 2, the region enclosed by the curve y=0.5x2−4x+10.5y = 0.5x^2 - 4x + 10.5, and the lines x=1x = 1, x=5x = 5 and the xx-axis has been shaded.

Diagram 2 showing a shaded region under the curve y=0.5x^2-4x+10.5 from x=1 to x=5, with intersection points (1,7) and (5,3).

Write down an integral for the area of the shaded region in diagram 2.

[3]
(b)(ii)

Calculate the area of this region.

[3]
(c)

Hence, determine the area enclosed between the line y=−x+8y = -x + 8 and the curve y=0.5x2−4x+10.5y = 0.5x^2 - 4x + 10.5.

[2]

Question 8

HardPaper 1 · calculator10 marks
(a)

Alex is preparing for a calculus exam and encounters a series of integrals. For each integral, identify whether it can be evaluated analytically using standard techniques (and if so, state the appropriate analytical method) or if it requires numerical approximation using technology (e.g., a GDC).

(a) ∫cos⁡(3x+2) dx\int \cos(3x + 2)\,dx

[2]
(b)

(b) ∫2xex2 dx\int 2x e^{x^2}\,dx

[2]
(c)

(c) ∫xsin⁡(x) dx\int x \sin(x)\,dx

[2]
(d)

(d) ∫e2x2 dx\int e^{2x^2}\,dx

[1]
(e)

(e) ∫124x2x3+3 dx\int_{1}^{2} \frac{4x^2}{x^3 + 3}\,dx

[2]
(f)

(f) ∫144sin⁡(x2) dx\int_{1}^{4} 4\sin(x^2)\,dx

[1]

Question 9

MediumPaper 1 · calculator6 marks

A landscape architect designs a decorative water feature for a public park. The cross-section of the feature, when viewed from the side, is a region bounded by the vertices (2,1)(2, 1), (2,5)(2, 5), (4,5)(4, 5), (4,3)(4, 3), (6,3)(6, 3) and (6,1)(6, 1). This region is rotated about the y-axis to form a solid water basin.

2D cross-section of the water basin with vertices labeled

(a) Find the volume of this water basin.

Question 10

HardPaper 1 · calculator11 marks
(a)

(a) A designer is creating a prototype for a decorative vase. The cross-section of the vase can be modelled by the function f(x)=x4−x2f(x) = x\sqrt{4-x^2}, for −2≤x≤2-2 \le x \le 2.

Sketch the graph of y=f(x)y = f(x) on the following pair of axes.

graph of y=f(x) on axes from -3 to 3 for x and -3 to 3 for y. The curve passes through the origin, has a maximum in the first quadrant and a minimum in the third quadrant. The curve is symmetric about the origin. The endpoints are at x=-2 and x=2. The maximum is at x=sqrt(2) and y=2, and the minimum is at x=-sqrt(2) and y=-2. The curve is smooth. The x-axis is labelled from -3 to 3 and the y-axis is labelled from -3 to 3.
[2]
(b)(i)

(b) The region enclosed by the graph of y=f(x)y = f(x) and the x-axis is rotated 360∘360^\circ about the x-axis to form the body of the vase.

(i) Write down an integral that represents the volume of this vase.

[2]
(b)(ii)

(ii) Calculate the value of this integral.

[4]
(c)

(c) The designer decides to create a new, larger version of the vase, y=g(x)y = g(x), by applying the following transformations to the original cross-section y=f(x)y = f(x):

  • A horizontal stretch by a scale factor of 3, parallel to the x-axis.
  • A vertical stretch by a scale factor of 0.75, parallel to the y-axis.

Find the volume of this new vase.

[3]

Question 11

MediumPaper 1 · calculator8 marks
(a)

The rate of profit, P(t)P(t), in thousands of dollars per month, for a new product is modelled by the piecewise function

P(t)={P1(t),0≤t≤TP2(t),t≥TP(t) = \begin{cases} P_1(t), & 0 \leq t \leq T \\ P_2(t), & t \geq T \end{cases}

where P1(t)=−t2+6tP_1(t) = -t^2 + 6t and P2(t)=−2t+16P_2(t) = -2t + 16. For a smooth transition in the profit rate, it is required that P1(T)=P2(T)P_1(T) = P_2(T).

Find the value of TT.

[2]
(b)

Show that P1′(T)=P2′(T)P_1'(T) = P_2'(T).

[2]
(c)

The total profit from the product at time t=0t=0 is zero.

Find the time when the total profit returns to its initial position.

[4]

Question 12

HardPaper 1 · calculator7 marks
(a)

A population of microorganisms grows according to the function P(t)=t3−9t2+23t−15P(t) = t^3 - 9t^2 + 23t - 15, where P(t)P(t) is the population in thousands and tt is the time in hours, for 0≤t≤50 \le t \le 5 hours.

The population is zero when t=1t=1, t=5t=5 and t=at=a hours.

Find the value of aa.

[2]
(b)

Calculate the total population growth (area enclosed by the curve and the tt-axis) from t=1t=1 hour to t=at=a hours.

[2]
(c)

Another species of microorganism has its population modelled by Q(t)=−2t2+10t−8Q(t) = -2t^2 + 10t - 8 (in thousands). The total population growth of species PP from t=1t=1 to t=bt=b hours is equal to the total population growth of species QQ from t=1t=1 to t=bt=b hours.

Find the value of bb, where 1<b<31 < b < 3.

[3]

Question 13

MediumPaper 1 · calculator7 marks
(a)

The rate of change of a certain quantity, QQ, with respect to time, tt (in minutes), is modelled by the function dQdt=105t−2\frac{dQ}{dt} = \frac{10}{5t-2}.

(a) Find an expression for Q(t)Q(t) in terms of tt, assuming an arbitrary constant of integration.

[3]
(b)

(b) Given that the model is valid for t>25t > \frac{2}{5}, find the exact total change in the quantity QQ from t=1t=1 minute to t=4t=4 minutes. Give your answer in the form aln⁡ba \ln b, where a,b∈Na, b \in \mathbb{N}.

[4]

Question 14

HardPaper 2 · calculator9 marks
(a)

The cross-section of a decorative garden bed is modelled by the curve y=2(x−1)(x−3)2y = 2(x-1)(x-3)^2, where xx and yy are measured in metres. The garden bed is built on flat ground, represented by the xx-axis.

(a) Write down the xx-intercepts of this curve.

[2]
(b)

(b) Write down a definite integral that represents the area of the cross-section of the garden bed.

[2]
(c)

(c) Find the value of this area.

[5]

Question 15

MediumPaper 1 · calculator6 marks

A landscape architect is designing a new feature for a botanical garden. The boundary of a particular flower bed is defined by two curves. The upper boundary can be modelled by the equation y=−x2+7x−6y = -x^2 + 7x - 6 and the lower boundary by y=x+2y = x + 2, where xx and yy are measured in metres.

(a) Find the area of the flower bed.

Question 16

HardPaper 1 · calculator8 marks
(a)(i)

A landscape architect is designing a large decorative fountain for a new public park. The outer structure of the fountain consists of a cylindrical base topped by a conical section. The inner part of the fountain, which holds the water, is a hollow space created by rotating a specific curve around the vertical axis.

The shape of the inner hollow is based on a transformation of the graph y=−x3y = -x^3. The curve that defines the profile of the inner hollow is given by y=40−0.064x3y = 40 - 0.064x^3. This transformation involves a vertical translation of aa units and a stretch parallel to the x-axis with a scale factor of bb.

(a.i) Write down the value of aa.

[1]
(a)(ii)

(a.ii) Find the value of bb.

[2]
(b)

The cylindrical base of the fountain has a radius of 1515 m and a height of 55 m. The conical section on top has the same base radius of 1515 m and a height of 88 m. The inner hollow, described by the curve y=40−0.064x3y = 40 - 0.064x^3, extends from the base of the fountain (y=0y=0) up to the total height of the outer structure.

Find the volume of the solid material that makes up the fountain (i.e., the volume of the outer structure minus the volume of the inner hollow).

[5]

Question 17

MediumPaper 1 · calculator10 marks
(a)

(a) A landscape architect is designing a curved retaining wall for a garden. The height of the wall, hh metres, is measured at 11 m intervals along its horizontal length, xx metres, from one end. The measurements are given in the table below.

xx (m)001122334455667788
hh (m)003.53.5667.57.5887.57.5663.53.500

Use the trapezium rule to estimate the cross-sectional area of the wall, giving your answer correct to 22 decimal places.

[3]
(b)

(b) It is later discovered that the equation connecting xx and hh is h(x)=0.5x(8−x)h(x) = 0.5x(8-x).

Calculate the true value of the cross-sectional area of the wall.

[4]
(c)

(c) Find the percentage error in your estimation of the area you made in part (a).

[2]
(d)

(d) Explain why, in this case, the trapezium rule underestimates the true value.

[1]

Question 18

HardPaper 2 · calculator14 marks
(a)

(a) An architect is designing a decorative archway for a park entrance. The cross-section of one half of the archway is modeled. The archway is symmetrical about the y-axis.

The architect models the base section of the archway as a straight line passing through the points (0,2)(0, 2) and (2,4)(2, 4), where all units are in metres.

Find the equation of the line passing through these two points.

[2]
(b)(i)

(b) The architect initially models the curved upper section of the archway using the following measured points:

(2,4)(2, 4), (4,5)(4, 5), (5.5,3)(5.5, 3), and (7,0)(7, 0).

(i) Find the equation of the least squares regression quadratic curve for these four points.

[2]
(b)(ii)

(ii) By considering the gradient of this curve when x=2x = 2, explain why it may not be a good model for the archway.

[1]
(c)

(c) The architect decides that a better model for the curved section would be a quadratic curve with a maximum point at (4.5,5.5)(4.5, 5.5) and that passes through the endpoint (7,0)(7, 0).

Find the equation of this new quadratic model.

[4]
(d)(i)

(d) Believing this to be a better model for the archway, the architect wants to estimate the volume of the solid generated by rotating this half-archway about the x-axis.

(i) Write down an expression for this estimate of the volume as a sum of two integrals.

[4]
(d)(ii)

(ii) Find the value of this estimate.

[1]

Question 19

MediumPaper 2 · calculator7 marks
(a)

(a) A designer is creating a decorative glass orb. The profile of the orb's cross-section is modelled by the curve f(x)=R2−x2f(x) = \sqrt{R^2 - x^2} for x∈[−R,R]x \in [-R, R], where RR is the radius of the orb.

The designer rotates the curve f(x)f(x) around the xx-axis by 2π2\pi radians. State the geometrical name of the solid generated.

[1]
(b)

(b) Write down a definite integral, including limits, that represents the volume of this solid.

[2]
(c)

(c) If the radius of the glass orb is R=6R = 6 cm, calculate the exact volume of the orb.

[4]

Question 20

HardPaper 2 · calculator15 marks
(a)

A drone is launched vertically upwards from a platform. Its vertical velocity, v ms−1v \text{ ms}^{-1}, at time tt seconds, is given by the function:

v=−3t2+18t−15v = -3t^2 + 18t - 15, for t≥0t \ge 0.

Find the times when the drone is momentarily at rest.

[2]
(b)

Find the magnitude of the drone's vertical acceleration at t=5t = 5 seconds.

[4]
(c)

Find the greatest speed of the drone in the interval 0≤t≤50 \le t \le 5.

[2]
(d)

The drone starts from an initial height of 1010 metres above the ground. Find an expression for the height of the drone, hh metres, above the ground at time tt seconds.

[4]
(e)

Find the total distance travelled by the drone in the interval 0≤t≤40 \le t \le 4.

[3]

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What does Area under/between a curve, volumes of revolution cover in IB Maths AI?

Integration is anti-differentiation. Definite integral calculates exact accumulation with specific boundary values (limits a and b). Constant of integration (+c) is not needed for definite integrals as it cancels out.

Is Area under/between a curve, volumes of revolution SL or HL?

Area under/between a curve, volumes of revolution is HL only. SL students are not examined on it.

How do I revise Area under/between a curve, volumes of revolution for IB Maths AI?

Start from the core idea: integration is anti-differentiation. In the exam: the rotation axis is the detail that decides which variable the integral runs over, and a question that rotates about the y-axis expects the curve rearranged as x in terms of y first. Region-enclosed-by-a-curve questions that dip below the axis need the interval split at the root and each piece's absolute value summed, or the negative and positive parts cancel and understate the true area. Then practise exam-style questions, easiest first, writing out every step of your working before you check it.

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