Volumes of Revolution (1.8.5) | CIE A-Level Maths Notes

In this section, we will focus on using definite integration to determine the volumes of solids of revolution. These solids are created by revolving a region bounded by curves around an axis (x-axis or y-axis). We'll particularly explore methods to calculate the volume of solids formed by revolving regions between curves, like the volume of a solid formed when revolving the region between $y = 9 - x^2$ and $y = 5$ around the x-axis.

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Volume of Revolution Between Two Curves

With Respect to x

Concept:

Finding the volume of revolution between two curves involves calculating the volumes formed by each curve around the x-axis and subtracting one from the other.

Procedure:

a. Ensure $y$ is the subject in the equations of the curves.

b. Apply the formula:

\pi \int_a^b (y_1^2 - y_2^2) \, \mathrm{dx}

\int_a^b \pi y_1^2 \, \mathrm{dx} - \int_a^b \pi y_2^2 \, \mathrm{dx}

With Respect to y

Procedure:

a. Make $x$ the subject in the equations of the curves.

b. Use the formula:

\pi \int_a^b (x_1^2 - x_2^2) \, \mathrm{dy}

\int_a^b \pi x_1^2 \, \mathrm{dy} - \int_a^b \pi x_2^2 \, \mathrm{dy}

Example Questions

Problem 1 :

Given: Curve $y = 2(3x - 1)^{\frac{1}{3}}$ and lines $x = \frac{2}{3}$ , $x = 3$ .

Task: Calculate the volume when the shaded region is rotated 360° about the x-axis.

Solution:

1. Formula: Volume of revolution formula:

\int_a^b \pi y^2 \, \mathrm{dx}

2. Integration:

\int{\frac{2}{3}}^3 \pi \left( 2(3x - 1)^{\frac{1}{3}} \right)^2 \, \mathrm{dx}

simplifies to

\int{\frac{2}{3}}^3 \pi \left( 4(3x - 1)^{\frac{2}{3}} \right) \, \mathrm{dx}

3. Final Calculation:

\left[ 4\pi(3x - 1)^{\frac{1}{3}} \right]_{\frac{2}{3}}^3 = 24.8\pi \, \text{(Volume of the solid)}

Problem 2:

Given: The region between the curves $y = x^2$ and $y = x$ in the first quadrant.

Task: Find the volume of the solid formed when this region is revolved about the y-axis.

Solution:

1. Preparation: Express $x$ as the subject. Here, $x = \sqrt{y}$ and $x = y$ .

2. Formula: Apply the formula for revolution about the y-axis:

\pi \int_a^b (x_1^2 - x_2^2) \, \mathrm{dy}

3. Integration:

\int_0^1 \pi \left( y^2 - (\sqrt{y})^2 \right) \, \mathrm{dy}

simplifies to

\pi \int_0^1 (y^2 - y) \, \mathrm{dy}

4. Final Calculation:

\pi \left[ \frac{y^3}{3} - \frac{y^2}{2} \right]_0^1 = \frac{\pi}{6} \, \text{(Volume of the solid)}

Written by:

Dr Rahil Sachak-Patwa

Oxford University - PhD Mathematics

Rahil spent ten years working as private tutor, teaching students for GCSEs, A-Levels, and university admissions. During his PhD he published papers on modelling infectious disease epidemics and was a tutor to undergraduate and masters students for mathematics courses.

Oxford University - PhD Mathematics

CIE A-Level Maths Study Notes

1.8.5 Volumes of Revolution

Volume of Revolution Between Two Curves

With Respect to x

Concept:

Procedure:

With Respect to y

Procedure:

Example Questions

Problem 1 :

Solution:

Problem 2:

Solution:

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