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Tuesday, April 16, 2013

 

Why πr2 ?

Starting from:
  y2+x2=r2y=r2x2
The area of the circle is therefore:
  rr2r2x2dx
Let x=rsinθ, so dx=rcosθdθ:
  rr2r2x2dx=π/2π/22rcosθr2r2sin2θdθ=2r2π/2π/2cos2θdθ=2r2π/2π/21+cos2θ2dθ=r2π/2π/2(1+cos2θ)dθ
Let u=2θ, so du=2dθ:
  r2π/2π/2(1+cos2θ)dθ=r22ππ(1+cosu)du=r22(u+sinu)|ππ=πr2


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