Question 117284
Given:

{{{diameter =d= 4 m}}}	

{{{height=h= 2 m}}}

the surface area {{{A= 2Pi*r[c]*h}}}

	
first find radius of curvature {{{r[c]}}}

{{{r[c]=(h^2 + (d/2)^2)/(2h)}}}………....where {{{d/2=2m}}}

{{{r[c]=((2m)^2 + (2m)^2)/(2*2m)}}}

	
{{{r[c]=(4m^2 + 4m^2)/(4m)}}}


{{{r[c]= 8m^2 /4m^1}}}

{{{r[c]= 2m^(2-1)}}}

{{{r[c]= 2m}}}


Then:


{{{A= 2Pi*r[c]*h}}}

{{{A= 2*(3.14)*2m*2m}}}

{{{A= 8*(3.14)m^2}}}

{{{A= 25.12m^2}}}