Areas and Volumes of Revolution — Integration (9709 Pure Mathematics 1) — Mathematics

Introduction

Definite integration is the engine behind two of the most frequently examined geometric applications in 9709 Pure Mathematics 1: finding areas of regions bounded by curves and straight lines, and computing volumes of revolution when a region is rotated about a coordinate axis. These topics appear in virtually every examination sitting and typically carry 7–12 marks. Mastery requires careful attention to limits, signs, and the geometry of the region involved — errors in any of these are the most common source of lost marks.

Core Concept

Areas of Regions

The definite integral $\displaystyle\int_a^b f(x)\,dx$ gives the signed area between the curve $y = f(x)$ and the $x$ -axis over $[a, b]$ . For exam purposes, the following cases must all be handled:

Case 1 — Between a curve and the $x$ -axis:

If $f(x) \geq 0$ on $[a,b]$ , the area is simply $\displaystyle\int_a^b f(x)\,dx$ .

If the curve dips below the $x$ -axis on part of $[a,b]$ , split the integral at every root and take the absolute value of each part before summing.

Case 2 — Between a curve and a line parallel to the $y$ -axis (integrating with respect to $y$ ):

When a boundary is a horizontal line or the region is better described in terms of $y$ , express $x$ as a function of $y$ and use:

\text{Area} = \int_{y_1}^{y_2} x\,dy

Case 3 — Between a curve and a straight line, or between two curves:

If $f(x) \geq g(x)$ on $[a,b]$ (upper curve minus lower curve):

\text{Area} = \int_a^b \bigl[f(x) - g(x)\bigr]\,dx

The limits $a$ and $b$ are found by solving $f(x) = g(x)$ (or the given intersection conditions).

Volumes of Revolution

When a region is rotated through $2\pi$ radians (a full turn) about the $x$ -axis:

V = \pi \int_a^b y^2\,dx

When rotated about the $y$ -axis:

V = \pi \int_{y_1}^{y_2} x^2\,dy

For rotation of the region between two curves $y = f(x)$ (outer) and $y = g(x)$ (inner) about the $x$ -axis:

V = \pi \int_a^b \bigl[f(x)^2 - g(x)^2\bigr]\,dx

Key Formulae & Definitions

Area under a curve ( $y \geq 0$ ):

A = \int_a^b y\,dx

Area between two curves ( $f(x) \geq g(x)$ ):

A = \int_a^b \bigl[f(x) - g(x)\bigr]\,dx

Area integrating with respect to $y$ :

A = \int_{y_1}^{y_2} x\,dy

Volume of revolution about the $x$ -axis:

V = \pi\int_a^b y^2\,dx

Volume of revolution about the $y$ -axis:

V = \pi\int_{y_1}^{y_2} x^2\,dy

Washer/shell method (between two curves about the $x$ -axis):

V = \pi\int_a^b\bigl[\{f(x)\}^2 - \{g(x)\}^2\bigr]\,dx

Note: Volumes are always positive. If your answer is negative, check your limits or which curve is outer/inner.

Worked Examples

Example 1 — Area between a curve and a straight line

Find the area of the region enclosed by the curve $y = x^2 + 1$ and the line $y = 2x + 4$ .

Step 1 — Find intersection points.

Set $x^2 + 1 = 2x + 4$ :

x^2 - 2x - 3 = 0 \implies (x-3)(x+1) = 0 \implies x = -1,\; x = 3

Step 2 — Identify which is the upper curve on $(-1, 3)$ .

Test $x = 0$ : line gives $4$ , curve gives $1$ . So $y = 2x+4$ is above $y = x^2+1$ .

Step 3 — Set up and evaluate the integral.

A = \int_{-1}^{3} \bigl[(2x+4) - (x^2+1)\bigr]\,dx = \int_{-1}^{3} \bigl(-x^2 + 2x + 3\bigr)\,dx

= \left[-\frac{x^3}{3} + x^2 + 3x\right]_{-1}^{3}

At $x = 3$ : $-9 + 9 + 9 = 9$

At $x = -1$ : $\dfrac{1}{3} + 1 - 3 = -\dfrac{5}{3}$

A = 9 - \left(-\frac{5}{3}\right) = 9 + \frac{5}{3} = \frac{32}{3}

\boxed{A = \tfrac{32}{3} \text{ square units}}

Example 2 — Volume of revolution about the $x$ -axis

The region bounded by the curve $y = \sqrt{x+1}$ , the $x$ -axis, $x = 0$ and $x = 3$ is rotated through $2\pi$ radians about the $x$ -axis. Find the exact volume of the solid formed.

Step 1 — Write down the formula.

V = \pi \int_0^3 y^2\,dx

Step 2 — Substitute $y^2$ .

Since $y = \sqrt{x+1}$ , we have $y^2 = x + 1$ .

V = \pi \int_0^3 (x+1)\,dx

Step 3 — Integrate.

V = \pi\left[\frac{x^2}{2} + x\right]_0^3 = \pi\left(\frac{9}{2} + 3 - 0\right) = \pi \cdot \frac{15}{2}

\boxed{V = \frac{15\pi}{2} \text{ cubic units}}

Example 3 — Volume of revolution about the $y$ -axis

The region enclosed by $y = x^2$ , $y = 0$ , and $y = 4$ is rotated fully about the $y$ -axis. Find the volume generated.

Step 1 — Express $x^2$ in terms of $y$ .

From $y = x^2$ : $x^2 = y$ .

Step 2 — Apply the formula for rotation about the $y$ -axis.

V = \pi \int_0^4 x^2\,dy = \pi \int_0^4 y\,dy

Step 3 — Integrate.

V = \pi\left[\frac{y^2}{2}\right]_0^4 = \pi \cdot \frac{16}{2} = 8\pi

\boxed{V = 8\pi \text{ cubic units}}

Common Mistakes & Examiner Pitfalls

Mistake	What goes wrong	How to avoid it
Wrong sign for area below $x$ -axis	Negative integral treated as area	Always sketch; use $\lvert\int\rvert$ for portions below axis
Forgetting $\pi$ in volume formula	Volume answer is off by a factor of $\pi$	Write $V = \pi\int\ldots$ from the outset
Using $y$ instead of $y^2$ in the volume integral	Computes area, not volume	Remember: rotation gives $\pi y^2\,dx$
Swapping upper and lower curve	Area comes out negative	Always verify which curve is greater with a test point
Incorrect limits when integrating w.r.t. $y$	Limits are $y$ -values, not $x$ -values	Check that limits match the variable of integration
Not squaring correctly when using washer method	E.g. writing $f(x)^2 - g(x)^2 = (f-g)^2$	Expand each term separately before subtracting
Omitting absolute value for a region split by the $x$ -axis	Positive and negative parts cancel	Split the integral at the roots and add magnitudes

Practice Questions

Q1. Find the area enclosed between the curve $y = x^3 - 4x$ and the $x$ -axis.

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The curve meets the $x$ -axis where $x^3 - 4x = 0 \Rightarrow x(x^2-4)=0 \Rightarrow x = -2, 0, 2$ .

On $(-2, 0)$ : $y > 0$ ; on $(0,2)$ : $y < 0$ . By symmetry the two parts have equal magnitude.

A_1 = \int_{-2}^{0}(x^3-4x)\,dx = \left[\frac{x^4}{4}-2x^2\right]_{-2}^{0} = 0-(4-8) = 4

A_2 = \left|\int_0^2(x^3-4x)\,dx\right| = \left|\left[\frac{x^4}{4}-2x^2\right]_0^2\right| = |4-8| = 4

\text{Total area} = 4 + 4 = \boxed{8 \text{ square units}}

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Q2. The region $R$ is bounded by the curve $y = 4 - x^2$ and the line $y = x + 2$ . Find the area of $R$ .

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Intersections: $4-x^2 = x+2 \Rightarrow x^2+x-2=0 \Rightarrow (x+2)(x-1)=0$ , so $x=-2,\,x=1$ .

Curve is above line on $(-2,1)$ :

A = \int_{-2}^{1}\bigl[(4-x^2)-(x+2)\bigr]\,dx = \int_{-2}^{1}(2-x-x^2)\,dx

=\left[2x-\frac{x^2}{2}-\frac{x^3}{3}\right]_{-2}^{1}

At $x=1$ : $2-\tfrac{1}{2}-\tfrac{1}{3}=\tfrac{7}{6}$

At $x=-2$ : $-4-2+\tfrac{8}{3}=-\tfrac{10}{3}$

A = \frac{7}{6}+\frac{10}{3} = \frac{7}{6}+\frac{20}{6} = \boxed{\frac{27}{6} = \frac{9}{2} \text{ square units}}

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Q3. The curve $y = 3\sqrt{x}$ and the lines $x = 0$ , $x = 4$ bound a region which is rotated fully about the $x$ -axis. Find the exact volume of revolution.

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y^2 = 9x

V = \pi\int_0^4 9x\,dx = 9\pi\left[\frac{x^2}{2}\right]_0^4 = 9\pi\cdot 8 = \boxed{72\pi \text{ cubic units}}

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Q4. The region bounded by $y = x^2$ and $y = 2x$ is rotated $2\pi$ radians about the $x$ -axis. Find the volume of the solid formed.

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Intersections: $x^2 = 2x \Rightarrow x = 0$ or $x = 2$ .

On $(0,2)$ : $2x \geq x^2$ , so $y = 2x$ is the outer curve.

V = \pi\int_0^2\bigl[(2x)^2-(x^2)^2\bigr]\,dx = \pi\int_0^2(4x^2-x^4)\,dx

= \pi\left[\frac{4x^3}{3}-\frac{x^5}{5}\right]_0^2 = \pi\left(\frac{32}{3}-\frac{32}{5}\right) = \pi\cdot\frac{160-96}{15} = \boxed{\frac{64\pi}{15} \text{ cubic units}}

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Q5. Find the volume generated when the region enclosed by $y = \dfrac{1}{x}$ , $y = 0$ , $x = 1$ and $x = 3$ is rotated completely about the $x$ -axis.

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y^2 = \frac{1}{x^2}

V = \pi\int_1^3\frac{1}{x^2}\,dx = \pi\left[-\frac{1}{x}\right]_1^3 = \pi\!\left(-\frac{1}{3}+1\right) = \boxed{\frac{2\pi}{3} \text{ cubic units}}

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Connections

Prerequisite knowledge relied upon:

Definite Integrals — evaluating $\int_a^b f(x)\,dx$ using the Fundamental Theorem of Calculus; handling polynomial, power, and simple composite integrands.
Coordinate Geometry — finding intersection points of lines and curves by solving simultaneous equations.
Curve Sketching — identifying which curve lies above the other and locating roots, essential for setting up correct limits.

Natural next steps within 9709:

Integration by substitution and integration by parts (A2 Pure) — needed for more complex integrands in area/volume problems.
Differential Equations — another major application of integration techniques.
Further volumes of revolution at A2 level may involve parametric equations as limits or integrands.

Areas and Volumes of Revolution — Integration (9709 Pure Mathematics 1)

Introduction

Core Concept

Areas of Regions

Volumes of Revolution

Key Formulae & Definitions

Worked Examples

Example 1 — Area between a curve and a straight line

Example 2 — Volume of revolution about the $x$ -axis

Example 3 — Volume of revolution about the $y$ -axis

Common Mistakes & Examiner Pitfalls

Practice Questions

Connections

Figures

Keep learning

Introduction

Core Concept

Areas of Regions

Volumes of Revolution

Key Formulae & Definitions

Worked Examples

Example 1 — Area between a curve and a straight line

Example 2 — Volume of revolution about the xxx-axis

Example 3 — Volume of revolution about the yyy-axis

Common Mistakes & Examiner Pitfalls

Practice Questions

Connections

Figures

Keep learning

Example 2 — Volume of revolution about the $x$ -axis

Example 3 — Volume of revolution about the $y$ -axis