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2014-12-22T03:56:27+05:30

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See diagram.

Let us take the top half of the sphere of Radius R, located with center at origin.

Let us us consider a small ring of radius r on the surface of the sphere, such that, its plane is parallel to the x-y plane.  Its height above x-y plane be z.  Then its radius will be  r  =   √(R² - z²)          as,       r² + z²  = R²  
        equation of the circle (of the ring) with radius r :  x² + y² = r² = R² - z²
     Its circumference is 2 π r = 2 π √(R² - z²)

   Let the width of the ring along the surface of sphere be ds.
          ds  = approximated as = √ [(dz)² + (dr)² ]   -  by Pythagoras theorem.
          ds = √ [1 + (dr/dz)² ] * dz
    
              z² + r²  = R² ,            Differentiating we get,
           2 z + 2 r  dr/dz = 0      => dr/dz = -z/r
         ds = √(1 + z²/r²) * dz  = √ [ (r² + z²) / r² ] * dz = R / r * dz

     Area of the small infinitesimal ring =  2π r ds = 2π r R / r dz = 2 π R dz

     Total surface area of Sphere = 2 * surface area of hemisphere z = 0 to R

= 2*\int\limits^{R}_{z=0} {2\pi R} \, dz = 4\pi * R* [ z ]_0^R = 4 \pi R^2\\

Hence answer.


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