A rectangular sheet of paper of …

Length of sheet = 33 cm 

Breadth of sheet = 22 cm 

We can fold the sheet in two ways, first along its lenght and second along its breadth. 

So, if we fold the sheet along its breadth, than its height = 33 cm

And the circumference = 22 cm

                        {tex}\eqalign{ & 2\,\pi \,r\, = \,22 \cr & 2\, \times \,{{22} \over 7}\, \times \,r\, = \,22 \cr & r\,\, = \,\,{{22\,\, \times \,\,7} \over {2\,\, \times \,\,22}}\,\, = \,\,{7 \over 2} cm \cr & \,First\,Volume\, = \,\pi \,{r^2}\,h\, = \,{{22} \over 7}\,\, \times \,\,{7 \over 2}\,\, \times \,\,{7 \over 2}\,\, \times \,33\,\, = \,\,1270.5\,c{m^3} \cr} {/tex}

Now, if we fold sheet along its length, than its height = 22 cm

And circumference = 33 cm

                {tex}\eqalign{ & 2\,\pi \,R\,\, = \,\,33 \cr & 2\,\, \times \,\,{{22} \over 7}\,\, \times \,\,R\,\, = \,\,33 \cr & R\,\, = \,\,{{33\,\, \times \,\,7} \over {2\,\, \times \,\,22}}\,\, = \,\,{{21} \over 4} \cr & \,Second\,Volume\, = \,\pi \,{R^2}\,h\, = \,{{22} \over 7}\,\, \times \,\,{{21} \over 4}\,\, \times \,\,{{21} \over 4}\,\, \times \,\,22\,\, = \,\,1905.75\,c{m^3} \cr} {/tex}

Hence, difference in the cylinder formed = 1905.75 - 1270.5 = 635.25 cm³