Volumes of Revolution Revision Notes for AQA A-Level Further Maths

Volumes of Revolution

Introduction

When you rotate a two-dimensional region around an axis, you create a three-dimensional solid. This is called a solid of revolution, and calculus allows us to calculate its exact volume. This technique is particularly useful for finding volumes of shapes like cones, spheres, and more complex curved solids.

The method works by imagining the solid as being made up of many thin circular discs stacked together. Each disc has a small thickness, and we can find the volume of each disc and then add them all together using integration.

Consider a curve given by $y = f(x)$ . When this curve is rotated 360° (a full turn) around the x-axis between $x = a$ and $x = b$ , it creates a solid. Since the rotation is through 360°, any cross-section perpendicular to the axis of rotation will be a circle with radius $y$ .

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The key insight of the disc method is that we can approximate the entire solid using many thin circular discs. Each disc has:

Radius: $y$ (the distance from the axis to the curve)
Thickness: $\delta x$ (an infinitesimally small width)
Volume: $\delta V = \pi y^2 \delta x$

As we make $\delta x$ smaller and smaller, our approximation becomes exact, and summation becomes integration.

Now imagine dividing this solid into many thin slices, each with thickness $\delta x$ . Each slice is approximately a cylinder (or disc) with radius $y$ and thickness $\delta x$ , giving volume $\delta V = \pi y^2 \delta x$ .

The total volume is found by summing all these thin discs and taking the limit as $\delta x \to 0$ . This gives us an integral.

Rotation about the x-axis

When a curve is rotated about the x-axis, each cross-section perpendicular to the x-axis forms a circular disc with radius equal to the y-value at that point.

Key formula: The volume of the solid formed by rotating the curve $y = f(x)$ between $x = a$ and $x = b$ a full turn (360° or 2π radians) around the x-axis is:

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

chatImportant

Since this is an integral with respect to $x$ , you must express $y^2$ as a function of $x$ only. Simply square your function and substitute it into the formula.

Remember: Rotate about x → square the y

Worked example 1: Rotation about the x-axis

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Worked Example: Finding Volume by Rotation About the x-axis

Question: Find the volume formed when the area enclosed by the curve $y = x^3 + 1$ , the coordinate axes, and the line $x = 2$ is rotated:

a) 360° around the x-axis

b) 180° around the x-axis

Solution for part a:

First, identify that we're rotating about the x-axis, so we use the formula $V = \int_a^b \pi y^2 \, dx$ where $y = x^3 + 1$ .

Step 1: Substitute $y = x^3 + 1$ into the formula.

$V = \pi \int_0^2 (x^3 + 1)^2 \, dx$

Step 2: Expand the squared term before integrating.

$(x^3 + 1)^2 = x^6 + 2x^3 + 1$

Step 3: Write out the integral with the expanded form.

$V = \pi \int_0^2 (x^6 + 2x^3 + 1) \, dx$

Step 4: Integrate each term.

$V = \pi \left[ \frac{x^7}{7} + \frac{x^4}{2} + x \right]_0^2$

Step 5: Evaluate the definite integral.

$V = \pi \left( \frac{128}{7} + 8 + 2 - 0 \right)$

$V = \pi \left( \frac{128}{7} + 10 \right)$

$V = \frac{198}{7}\pi \text{ cubic units}$

Solution for part b:

Rotating through 180° produces exactly half of the shape obtained by rotating through 360°.

Therefore, the volume is:

$V = \frac{1}{2} \times \frac{198}{7}\pi = \frac{99}{7}\pi \text{ cubic units}$

chatImportant

Common Mistake to Avoid:

Always expand brackets fully before attempting to integrate. Don't try to integrate $(x^3 + 1)^2$ directly - this is a very common error that leads to incorrect answers.

Rotation about the y-axis

When rotating a region around the y-axis instead of the x-axis, the roles of $x$ and $y$ are swapped. The cross-sections are still circular, but now they are perpendicular to the y-axis, with radius equal to the x-value.

You need the function in the form $x = f(y)$ , and the limits must be in terms of $y$ values (from $y = a$ to $y = b$ ).

Key formula: The volume of the solid formed by rotating the curve $x = f(y)$ between $y = a$ and $y = b$ a full turn around the y-axis is:

$V = \int_a^b \pi x^2 \, dy$

chatImportant

You must rearrange your equation into the form $x = f(y)$ before using this formula. This means making $x$ the subject.

Remember: Rotate about y → square the x

Also remember: Integrate along what you rotate around - when rotating about the y-axis, integrate with respect to $y$ .

Worked example 2: Rotation about the y-axis

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Worked Example: Finding Volume by Rotation About the y-axis

Question: The region A is enclosed by the curve $y = x^5$ , the y-axis, and the lines $y = 0.5$ and $y = 1$ . Calculate the volume when region A is rotated 2π radians around the y-axis.

Solution:

Step 1: Recognise that 2π radians equals 360°, which is a full turn. We're rotating about the y-axis, so we need to use the formula $V = \int_a^b \pi x^2 \, dy$ .

Step 2: Rearrange the equation $y = x^5$ into the form $x = f(y)$ .

$y = x^5$

$x^5 = y$

$x = y^{1/5}$

Step 3: Substitute $x = y^{1/5}$ into the volume formula.

$V = \int_{0.5}^{1} \pi \left(y^{1/5}\right)^2 \, dy$

$V = \pi \int_{0.5}^{1} y^{2/5} \, dy$

Step 4: Integrate using the power rule.

$V = \pi \left[ \frac{5}{7}y^{7/5} \right]_{0.5}^{1}$

Step 5: Evaluate the definite integral.

$V = \pi \left( \frac{5}{7}(1)^{7/5} - \frac{5}{7}(0.5)^{7/5} \right)$

$V = \pi \left( \frac{5}{7} - 0.2707 \right)$

$V = 1.39 \text{ cubic units (to 3 s.f.)}$

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Checking Your Work:

Always verify that you're rotating about the correct axis. A good way to remember:

If you see $dx$ in the integral, you're rotating about the x-axis
If you see $dy$ in the integral, you're rotating about the y-axis

The axis of rotation determines which variable you integrate with respect to.

Partial rotations

You don't always need to rotate through a full 360°. The volume is directly proportional to the angle of rotation, which means the relationship is linear and predictable.

For 180° rotation (π radians): The volume is exactly half the volume of a full 360° rotation.

$V_{180°} = \frac{1}{2} \times V_{360°}$

For 90° rotation (π/2 radians): The volume is exactly one quarter the volume of a full 360° rotation.

$V_{90°} = \frac{1}{4} \times V_{360°}$

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This proportional relationship holds because the volume swept out during rotation is directly proportional to the angle through which you rotate. If you rotate through half the angle, you get half the volume; if you rotate through a quarter of the angle, you get a quarter of the volume.

This property can save you significant calculation time in exams!

Composite shapes and problem-solving

Sometimes the solid of revolution you need to find is not a simple shape. You may need to subtract one volume from another, or add volumes together. This typically occurs when the region you're rotating has boundaries formed by multiple curves or lines.

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Problem-solving strategy:

Always sketch first, calculate second. Follow these steps:

Sketch a graph - Draw the region carefully, noting which axis you're rotating about. Mark all intersection points and boundaries clearly.
Choose the correct formula - Decide whether to use $V = \int \pi y^2 \, dx$ or $V = \int \pi x^2 \, dy$ based on the axis of rotation.
Identify composite shapes - Look for standard 3D shapes like cones, cylinders, or spheres within your solid of revolution.
Add or subtract volumes where necessary - If the region creates a composite shape, you may need to subtract unwanted volumes (like cones) or add separate volumes together.

Worked example 3: Composite shape with cone

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Worked Example: Composite Shape Requiring Subtraction

Question: The region R is enclosed by the curve $y = \sqrt{x}$ , the line $y = x - 6$ , the line $x = 9$ , and $y = 0$ . Calculate the volume of the solid formed when R is rotated 360° around the x-axis.

Solution:

Step 1: Sketch the region. The curve $y = \sqrt{x}$ meets the line $y = x - 6$ at $(9, 3)$ . The region R lies between the x-axis and the curve.

Step 2: Calculate the volume formed by rotating the curve $y = \sqrt{x}$ from $x = 0$ to $x = 9$ about the x-axis.

$V_{\text{curve}} = \int_0^9 \pi (\sqrt{x})^2 \, dx$

$= \pi \int_0^9 x \, dx$

$= \pi \left[ \frac{x^2}{2} \right]_0^9$

$= \pi \left( \frac{81}{2} - 0 \right)$

$= \frac{81\pi}{2}$

Step 3: The line $y = x - 6$ from $x = 6$ to $x = 9$ forms a cone when rotated. The cone has:

Base radius: $r = 3$ (when $x = 9$ , $y = 3$ )
Height: $h = 3$ (from $x = 6$ to $x = 9$ )

Using the cone volume formula:

$V_{\text{cone}} = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (3)^2 (3) = 9\pi$

Step 4: Subtract the cone volume from the volume under the curve.

$V_{\text{required}} = \frac{81\pi}{2} - 9\pi = \frac{81\pi - 18\pi}{2} = \frac{63\pi}{2}$

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Time-saving Tip:

When dealing with composite shapes, always identify standard shapes (cones, cylinders, spheres) that can be calculated using known formulas:

Cone: $V = \frac{1}{3}\pi r^2 h$
Cylinder: $V = \pi r^2 h$
Sphere: $V = \frac{4}{3}\pi r^3$

This can save significant time compared to setting up complex integrals for these standard shapes.

Proving standard formulas

The volume of revolution method is not just a calculation tool - it's also a powerful technique for deriving formulas for standard 3D shapes. This demonstrates the deep connection between calculus and geometry.

Worked example 4: Proving the sphere volume formula

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Worked Example: Deriving the Volume of a Sphere

Question: By rotating the curve $y = \sqrt{r^2 - x^2}$ between $-r$ and $r$ around the x-axis, show that the volume of a sphere is $\frac{4}{3}\pi r^3$ .

Solution:

Step 1: Recognise that $y = \sqrt{r^2 - x^2}$ is the upper semicircle of a circle with equation $x^2 + y^2 = r^2$ . Rotating this semicircle around the x-axis creates a complete sphere of radius $r$ .

Step 2: Use the volume of revolution formula for rotation about the x-axis.

$V = \int_{-r}^{r} \pi y^2 \, dx$

Step 3: Substitute $y^2 = r^2 - x^2$ (from rearranging $y = \sqrt{r^2 - x^2}$ ).

$V = \pi \int_{-r}^{r} (r^2 - x^2) \, dx$

Step 4: Integrate with respect to $x$ . Remember that $r$ is just a constant.

$V = \pi \left[ r^2x - \frac{x^3}{3} \right]_{-r}^{r}$

Step 5: Evaluate the definite integral.

$V = \pi \left[ \left(r^2(r) - \frac{r^3}{3}\right) - \left(r^2(-r) - \frac{(-r)^3}{3}\right) \right]$

$= \pi \left[ r^3 - \frac{r^3}{3} + r^3 - \frac{r^3}{3} \right]$

$= \pi \left[ 2r^3 - \frac{2r^3}{3} \right]$

$= \pi \left[ \frac{6r^3 - 2r^3}{3} \right]$

$= \frac{4\pi r^3}{3}$

This proves the standard formula for the volume of a sphere: $V = \frac{4}{3}\pi r^3$ .

chatImportant

Key Point About Constants:

When proving standard formulas, remember that constants like $r$ are treated as constants during integration, not variables. Only the variable of integration (in this case, $x$ ) changes during the integration process.

Exam tips and common pitfalls

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Common Mistakes to Avoid:

Forgetting to square the function - The formula is $\pi y^2$ or $\pi x^2$ , not $\pi y$ or $\pi x$ . This is one of the most frequent errors.
Using the wrong variable - When rotating about the x-axis, integrate with respect to $x$ and use $y^2$ . When rotating about the y-axis, integrate with respect to $y$ and use $x^2$ .
Not rearranging for y-axis rotation - You must rearrange to get $x = f(y)$ before rotating about the y-axis. Don't skip this crucial step.
Forgetting the π - The formula includes $\pi$ from the circular cross-sections. This comes from the area formula $\pi r^2$ for a circle.
Expanding brackets incorrectly - Always expand brackets like $(x^3 + 1)^2$ fully before integrating. Show your working step-by-step.
Missing composite shapes - Always sketch the region. Look for triangular, conical, or other standard shapes that need to be subtracted or added.

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Exam Technique Tips:

Always sketch the region before attempting any calculation
Label your axes clearly and mark key points (intersections, boundaries)
State explicitly which axis you're rotating about
Write down the formula you're using before substituting values
Show all steps of your integration clearly - examiners award marks for method
Give answers to the required accuracy (often 3 significant figures or in terms of π)
Check units - volumes are always in cubic units
If the answer seems unreasonable, check your working for errors

Remember!

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Key Points to Remember:

Formulas:

Rotation about x-axis: Use $V = \int_a^b \pi y^2 \, dx$ where $y$ is expressed as a function of $x$ .
Rotation about y-axis: Use $V = \int_a^b \pi x^2 \, dy$ where $x$ is expressed as a function of $y$ .

Essential Rules:

2π radians = 360° = full turn - These all mean the same complete rotation.
For partial rotations, the volume is directly proportional to the angle.
180° rotation = half the volume of 360°
90° rotation = quarter the volume of 360°

Memory Aids:

Rotate about x → square the y
Rotate about y → square the x
Integrate along what you rotate around

Problem-solving Steps:

Always sketch first - A clear diagram helps you identify the correct limits, the axis of rotation, and any composite shapes.
Rearrange when needed - For y-axis rotation, you must rearrange to get $x = f(y)$ form.
Identify standard shapes - Look for cones, cylinders, or spheres that can be calculated using standard formulas.
Add or subtract volumes - Composite shapes require careful addition or subtraction of volumes.

Volumes of Revolution (AQA A-Level Further Maths): Revision Notes