A hemispherical piece of wood is to be mounted at the end of a cylindrical wooden log with the same base
If the length of the log is four times its width, then what is the percentage increase in its total surface area?

Answer :

Given : A hemispherical piece of wood is to be mounted at the end of a cylindrical wooden log with the same base , So our diagram in as :

And h  =  4 ( Width )                            ( we know width of cylinder  =  diameter of cylinder )
So,
h  =  4 ( d )
h =  4 ( 2 r )
h = 8r
So,
r  = $\frac{h}{8}$

Total surface area of cylinder without hemispherical end  = $2\mathrm{\pi rh} +\hspace{0.17em}2{\mathrm{\pi r}}^2$
So,
Total surface area of cylinder without hemispherical end  = $2\times \pi \times \frac{h}{8}\times h +\hspace{0.17em}2\times \pi \times {\left(\frac{h}{8}\right)}^2 = \frac{\mathrm{\pi }{h}^2}{4} +\hspace{0.17em}2\times \pi \times \frac{h^2}{64} = \frac{\mathrm{\pi }{h}^2}{4} +\frac{\mathrm{\pi }{h}^2}{32} = \frac{8\mathrm{\pi }{h}^2 + \mathrm{\pi }{h}^2}{32} = \frac{9\mathrm{\pi }{h}^2}{32}$
And
Total surface area of cylinder with hemispherical end = Total surface area of cylinder without hemispherical end  - Surface area of top + Lateral surface area of  hemispherical end
So,
Total surface area of cylinder with hemispherical end = $\frac{9\mathrm{\pi }{h}^2}{32} - {\mathrm{\pi r}}^2 + 2{\mathrm{\pi r}}^2 = \frac{9\mathrm{\pi }{\mathrm{h}}^2}{32} + {\mathrm{\pi r}}^{2 } = \frac{9\mathrm{\pi }{\mathrm{h}}^2}{32} + \mathrm{\pi }\times {\left(\frac{\mathrm{h}}{8}\right)}^{2 } =\frac{9\mathrm{\pi }{\mathrm{h}}^2}{32} + {\frac{\mathrm{\pi h}}{64}}^{2 } = \frac{18\mathrm{\pi }{\mathrm{h}}^2 + \mathrm{\pi }{\mathrm{h}}^2}{64} = \frac{19\mathrm{\pi }{\mathrm{h}}^2}{64}$  
So,
Total surface area increased  = $\frac{19\mathrm{\pi }{\mathrm{h}}^2}{64}- \frac{9\mathrm{\pi }{\mathrm{h}}^2}{32}= \frac{19\mathrm{\pi }{\mathrm{h}}^2 - 18\mathrm{\pi }{\mathrm{h}}^2}{64} = \frac{\mathrm{\pi }{\mathrm{h}}^2}{64}$  
So,
Percentage increased in total surface area  = $\frac{\frac{\mathrm{\pi }{\mathrm{h}}^2}{64}}{{\displaystyle \frac{9{\mathrm{\pi h}}^2}{32}}} \times 100 = \frac{{\mathrm{\pi h}}^2 \times 32}{{\displaystyle 64 \times 9{\mathrm{\pi h}}^2}} \times 100 = \frac{1}{2 \times 9}\times 100 = \frac{100}{18} = \mathbf{5}\mathbf{.}\mathbf{5}\mathbf{\%}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{(}\mathbf{ }\mathbf{Ans}\mathbf{ }\mathbf{)}$