Edexcel IAL - Pure Maths 4- 6.1 Volumes of Revolution- Study notes - New syllabus
Edexcel IAL – Pure Maths 4- 6.1 Volumes of Revolution -Study notes- New syllabus
Edexcel IAL – Pure Maths 4- 6.1 Volumes of Revolution -Study notes -Edexcel A level Physics – per latest Syllabus.
Key Concepts:
- 6.1 Volumes of Revolution
Volume of Revolution
When a curve is rotated about the \( x \)-axis, it sweeps out a three-dimensional solid. The volume of this solid can be found using integration.
Only the formula involving integration with respect to \( x \) is required at IAL level.
Standard Formula
If a curve is given by \( y = f(x) \) and is rotated about the \( x \)-axis between \( x = a \) and \( x = b \), then the volume \( V \) is:
\( V = \pi \displaystyle\int_{a}^{b} y^2 \, dx \)
The formula \( \pi \int x^2 \, dy \) is not required.
Using Parametric Equations
If the curve is given parametrically as:
\( x = g(t),\quad y = h(t) \)
then:
\( dx = \dfrac{dx}{dt}\,dt \)
The volume becomes:
\( V = \pi \displaystyle\int y^2 \dfrac{dx}{dt}\,dt \)
The limits of integration must be changed to the corresponding values of the parameter.
Key Points
- The axis of rotation here is the \( x \)-axis.
- The curve must be written in terms of \( y \).
- Always square \( y \) inside the integral.
- Use parametric differentiation if needed.
Example
Find the volume generated when the curve \( y = x \) is rotated about the \( x \)-axis from \( x = 0 \) to \( x = 2 \).
▶️ Answer / Explanation
Use:
\( V = \pi \displaystyle\int_{0}^{2} y^2 \, dx \)
Since \( y = x \):
\( V = \pi \displaystyle\int_{0}^{2} x^2 \, dx \)
\( = \pi \left[ \dfrac{x^3}{3} \right]_{0}^{2} \)
\( = \pi \left( \dfrac{8}{3} \right) = \dfrac{8\pi}{3} \)
Example
Find the volume when the curve \( y = \sqrt{x} \) is rotated about the \( x \)-axis from \( x = 0 \) to \( x = 4 \).
▶️ Answer / Explanation
\( y^2 = x \)
\( V = \pi \displaystyle\int_{0}^{4} x \, dx \)
\( = \pi \left[ \dfrac{x^2}{2} \right]_{0}^{4} \)
\( = \pi \cdot \dfrac{16}{2} = 8\pi \)
Example
The curve is given by \( x = t^2,\quad y = 2t \), where \( 0 \le t \le 2 \). Find the volume generated when the curve is rotated about the \( x \)-axis.
▶️ Answer / Explanation
First find \( dx \):
\( x = t^2 \Rightarrow \dfrac{dx}{dt} = 2t \)
So \( dx = 2t \, dt \)
Now use:
\( V = \pi \displaystyle\int y^2 \, dx \)
\( y = 2t \Rightarrow y^2 = 4t^2 \)
\( V = \pi \displaystyle\int_{0}^{2} 4t^2 \cdot 2t \, dt \)
\( = \pi \displaystyle\int_{0}^{2} 8t^3 \, dt \)
\( = \pi \left[ 2t^4 \right]_{0}^{2} \)
\( = \pi (2 \cdot 16) = 32\pi \)