Hollow Circular Cylinder

The example of hollow cylinders is pipes, circular buildings and bearing bushes, etc. If R is the outside radius of the cylinder and r is the inside radius of the cylinder, then


            (i)         V = \pi {R^2}h - \pi {r^2}h = \pi \left( {{R^2} -  {r^2}} \right)h
            (ii)        V = \frac{\pi  }{4}\left( {{D^2} - {d^2}} \right)

            D and d being outer and inner diameter, Vis volume.


A well with 10m inside diameter dug 14m deep. Earth taken out of it is spread all round to a width of 5m to form an embankment. Find the height of the embankment.

            Volume of the dug out            = \pi {r^2}h
                                                             = \frac{{22}}{7} \times 5 \times 5 \times 14



            Area of the embankment (shaded)

                                                             = \pi \left( {{R^2} - {r^2}} \right)
                                                             = \pi \left( {{{10}^2} - {5^2}} \right)            = 75 \times \frac{{22}}{7}\,{\text{sq}}{\text{.m}}

\therefore            Height of the embankment

                                                             = \frac{{{\text{Volume of the earth dug  out}}}}{{{\text{Area of the embankment}}}}

                                                             = \frac{{1100}}{{75 \times \frac{{22}}{7}}}\,\, =  \frac{{14}}{3}\,\,\, = 4\frac{2}{3}\,\,{\text{m}}


A hollow cylinder copper pipe is 21dm long. Its outer and inner diameters are 10cm and 6cm respectively. Find the volume of copper used in making the pipe.


Given that:
            Height of cylindrical pipe, h = 21{\text{dm}}\,\,\, = \,210{\text{cm}}

\therefore            External radius, R = \frac{{10}}{2}\,\, = \,5{\text{cm}}
            Internal radius, R = \frac{6}{2}\,\, = \,3{\text{cm}}
            Volume of the copper used in making the pipe
             = {\text{ Volume of external cylinder }} - {\text{  volume of internal cylinder}}
             = \pi {R^2}h - \pi {r^2}h\,\,\,\,\,\, = \,\pi  \left( {{R^2} - {r^2}} \right)h\,\,\,\, = \frac{{22}}{7}\left[ {{5^2} - {3^2}}  \right] \times 210
             = \frac{{22}}{7} \times 16 \times 210\,\,\,\, = 22  \times 16 \times 30\,\,\,\, = 10560\,{\text{cu}}{\text{.cm}}