Volume of Solids of Revolution

Summary

  • Rotating about the x axis, Volume of Revolution:

v\quad =\quad \int _{ a }^{ b }{ \pi { y }^{ 2 } } dx

  • Rotating about the y axis, Volume of Revolution:

v\quad =\quad \int _{ a }^{ b }{ \pi { x }^{ 2 } } dy

 

We know that definite integrals can help us figure out areas underneath the curves.

Recalling the formula for definite integrals from the article ”Integration” :

\int _{ a }^{ b }{ f^{ ' }\left( x \right) } dx\quad =\quad f\left( b \right) \quad -\quad f\left( a \right)

Similarly, pretty much using the same principle we can work out volumes of rotating solids.

If the part of a curve y = f(x) between the giving limits x = a & x = b is rotated about the x axis or y axis through 360°, then the solid formed is called a ‘Solid of Revolution’.

Such a solid is always symmetrical about the axis of rotation. Examples of such solids are, cone, cylinder and sphere etc.

Consider an element with radius y and height  \Delta x.

Volume of the element   =\quad \pi { y }^{ 2 }\Delta x

Volume of the entire solid   =\quad \sum _{ a }^{ b }{ \pi { y }^{ 2 }\Delta x }

Note: The smaller the  \Delta x,  the approximation is closer to the actual value.

When  \Delta x\quad \rightarrow \quad 0 :

v\quad =\quad \int _{ a }^{ b }{ \pi { y }^{ 2 } } dx

Hence, we can deduce that:

The volume of revolution of y = f(x) between x = a & x = b about the axis of x is:

v\quad =\quad \int _{ a }^{ b }{ \pi { y }^{ 2 } } dx

Let’s look at some examples now.

Example 1

Rotating about the x axis

Q. The region between the curve  y\quad =\quad f\left( x \right) \quad =\quad { x }^{ 2 },  the x axis and the line x = 1 and x = 3 is rotated through 360° about the x axis.

Find the volume of revolution which is formed.

Solution:

Using the formula v\quad =\quad \int _{ a }^{ b }{ \pi { y }^{ 2 } } dx

Replacing y with { x }^{ 2 },

v\quad =\quad \int _{ 1 }^{ 3 }{ \pi { ({ x }^{ 2 }) }^{ 2 } } dx

=\quad\pi \int _{ 1 }^{ 3 }{ { x }^{ 4 } } dx

=\quad \pi { \left[ \frac { { x }^{ 5 } }{ 5 } \right] }_{ 1 }^{ 3 }

=\quad \frac { \pi }{ 5 } (243\quad -\quad 1)

Ans:         =\quad \frac { 242\pi }{ 5 } cubic units

Example 2

Rotating about the y axis

Q. The region between the curve  y\quad =\quad f\left( x \right) \quad =\quad { x }^{ 2 },  the y axis and the line y = 2 and y = 5 is rotated 360° around the y axis.

Find the volume of revolution obtained.

Solution:

Using the formula  v\quad =\quad \int _{ a }^{ b }{ \pi { x }^{ 2 } } dy

where  { y }^{ 2 }  is now replaced by  { x }^{ 2 }

Hence:

v\quad =\quad \int _{ 2 }^{ 5 }{ { \pi y } } dy  as   y\quad =\quad { x }^{ 2 }

=\quad\pi \int _{ 2 }^{ 5 }{ { y } } dy

=\quad \pi { \left[ \frac { { y }^{ 2 } }{ 2 } \right] }_{ 2 }^{ 5 }

=\quad \pi (\frac { 25 }{ 2 } \quad -\quad \frac { 4 }{ 2 } )

Ans:    =\quad \frac { 21\pi }{ 2 }   cubic units

Reference
  1. http://amsi.org.au/teacher_modules/Cones_Pyramids_and_Spheres.html
  2. MEI A Level Further Mathematics Year 2 4th Edition