TutorChase logo
Decorative notebook illustration
CIE A-Level Maths Study Notes

1.8.5 Volumes of Revolution

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=9x2y = 9 - x^2 and y=5y = 5 around the x-axis.

volumes of revolution

Image courtesy of Nagwa

Volume of Revolution Between Two Curves

With Respect to x


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.


a. Ensure yy is the subject in the equations of the curves.

b. Apply the formula:

πab(y12y22)dx\pi \int_a^b (y_1^2 - y_2^2) \, \mathrm{dx}abπy12dxabπy22dx \int_a^b \pi y_1^2 \, \mathrm{dx} - \int_a^b \pi y_2^2 \, \mathrm{dx}

With Respect to y


a. Make xx the subject in the equations of the curves.

b. Use the formula:

πab(x12x22)dy\pi \int_a^b (x_1^2 - x_2^2) \, \mathrm{dy}abπx12dyabπx22dy \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(3x1)13y = 2(3x - 1)^{\frac{1}{3}}and lines x=23x = \frac{2}{3}, x=3x = 3.

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


1. Formula: Volume of revolution formula:

abπy2dx\int_a^b \pi y^2 \, \mathrm{dx}

2. Integration:

233π(2(3x1)13)2dx\int{\frac{2}{3}}^3 \pi \left( 2(3x - 1)^{\frac{1}{3}} \right)^2 \, \mathrm{dx}

simplifies to

233π(4(3x1)23)dx\int{\frac{2}{3}}^3 \pi \left( 4(3x - 1)^{\frac{2}{3}} \right) \, \mathrm{dx}

3. Final Calculation:

[4π(3x1)13]233=24.8π(Volume of the solid)\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=x2y = x^2 and y=xy = x in the first quadrant.

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


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

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

πab(x12x22)dy\pi \int_a^b (x_1^2 - x_2^2) \, \mathrm{dy}

3. Integration:

01π(y2(y)2)dy\int_0^1 \pi \left( y^2 - (\sqrt{y})^2 \right) \, \mathrm{dy}

simplifies to

π01(y2y)dy\pi \int_0^1 (y^2 - y) \, \mathrm{dy}

4. Final Calculation:

π[y33y22]01=π6(Volume of the solid)\pi \left[ \frac{y^3}{3} - \frac{y^2}{2} \right]_0^1 = \frac{\pi}{6} \, \text{(Volume of the solid)}
Dr Rahil Sachak-Patwa avatar
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.

Hire a tutor

Please fill out the form and we'll find a tutor for you.

1/2 About yourself
Still have questions?
Let's get in touch.