Volumes of Revolution (Edexcel IGCSE Further Maths)

Volumes of Revolution About the x-axis

Spec point — 9C

What is a volume of revolution about the x-axis?

• A solid of revolution is formed

  • when an area bounded by a function $y=f(x)$ and the lines $x=a$ and $x=b$
  • is rotated $2π$ radians $(360^{\circ})$ about the $x$-axis

• The volume of revolution is the volume of this solid

• Be careful – the ’front’ and ‘back’ of this solid are flat

  • they were created from straight (vertical) lines
  • 3D sketches can be misleading!

What is the formula for a volume of revolution about the x-axis?

The volume of revolution of a solid rotated $2π$ radians ($360^{\circ}$) about the $x$-axis between $x=a$ and $x=b$ is given by:

• $V = \pi \int_{a}^{b} y^{2} d x$

• This is not given on the exam formula sheet, so you need to remember it

  • Note that $πy^{2}$ is the area of the circular cross-section of the solid at any value of $x$
  • That might help you remember the form of the volume integral

• $y$ is a function of$x$

  • i.e. $y = \mathrm{f} ( x )$

• $x=a$ and $x=b$ are the equations of the (vertical) lines bounding the area

  • $a<b$ ($a$ is the 'left boundary' and $b$ is the 'right boundary')
  • $x=a$ and $x=b$ may be given in the question
  • one boundary may be the $y$-axis ($x=0$)
  • the $x$-axis intercepts of $y=f(x)$ may also be boundaries

How do I calculate the volume of revolution about the x-axis?

• STEP 1 Identify the limits $a$ and $b$

  • These may be given in the question

    • or be indicated on a graph in the question

  • Sketching the graph of $y=f(x)$ can help if the graph is not provided

• STEP 2 Square the function $y = \mathrm{f} ( x )$

  • e.g. $y = x^{2} + 1 \Leftrightarrow y^{2} = ( x^{2} + 1 )^{2} = x^{4} + 2 x^{2} + 1$
  • or $y = \sqrt{4 - x} \Leftrightarrow y^{2} = ( \sqrt{4 - x} )^{2} = 4 - x$

• STEP 3 Evaluate the integral in the volume formula $V = \pi \int_{a}^{b} y^{2} d x$

  • An answer may be required in exact form

    • i.e. as a multiple of $π$

Exam Hint

• Don't panic if $y = \mathrm{f} ( x )$ involves a square root

  • The square root will disappear when you find $y^{2}$

• Don't forget to bring $π$ back in after working out the integral

  • In my experience that is a very common student error

Worked Example Find the volume of the solid of revolution formed by rotating the region bounded by the graph of $y=\sqrt{3x^{2}+2}$, the coordinate axes and the line $x=3$ by $2π$ radians about the $x$-axis. Give your answer as an exact value.

Start by finding the values of $a$ and $b$ for the formula

'Bounded by the coordinate axes' tells us that the $y$-axis ($x=0$) is one boundary The question tells us that $x=3$ is the other one So $a=0$ and $b=3$

If in doubt, drawing a sketch can help

Now square the function $y = \mathrm{f} ( x )$

$y^{2} = ( \sqrt{3 x^{2} + 2} )^{2} = 3 x^{2} + 2$

Substitute everything into $V = \pi \int_{a}^{b} y^{2} \mathrm{d} x$

$V = \pi \int_{0}^{3} ( 3 x^{2} + 2 ) \mathrm{d} x$

Work out the definite integral

$\begin{gathered} \int_{0}^{3} ( 3 x^{2} + 2 ) \mathrm{d} x & = [ 3 ( \frac{x^{2 + 1} }{2 + 1} ) + 2 x ]_{0}^{3} \\ = [ x^{3} + 2 x ]_{0}^{3} \\ = ( ( 3 )^{3} + 2 ( 3 ) ) - ( ( 0 )^{3} + 2 ( 0 ) ) \\ = 33 - 0 \\ = 33 \end{gathered}$

Put that value back into the volume formula

Don't forget to include the $π$ !

$V=33π$

The question asks for an exact value answer, so leave the answer in terms of $π$

Final answer: Volume$=33π\mathrm{units}^{3}$