Question 1043641
Without loss of generality, let the semicircle of radius 6 be represented by the equation

{{{y = sqrt(36-x^2)}}}.

Then the surface area is obtained by rotating a rectangle with radius {{{y = sqrt(36-x^2)}}} and width dx around the x-axis.  
The element of surface area is then given by 

{{{dA = 2*pi*sqrt(36-x^2)dx}}}.

The surface area is gotten by adding this infinite number of infinitesimal elements of surface area--


{{{A = int(1,dA,-6,6) = 2*int(2*pi*sqrt(36-x^2),dx,0,6) = 4*pi*int(sqrt(36-x^2),dx,0,6)}}}