Areas and Volumes of Revolution

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.

Show answer

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}}

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

Show answer

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}}

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.

Show answer

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}}

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.

Show answer

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}}

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.

Show answer

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}}

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.

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