Tuesday, April 16, 2013
Why πr2 ?
Starting from:
y2+x2=r2y=√r2−x2 |
∫r−r2√r2−x2dx |
∫r−r2√r2−x2dx=∫π/2−π/22r⋅cosθ√r2−r2sin2θdθ=2r2∫π/2−π/2cos2θdθ=2r2∫π/2−π/21+cos2θ2dθ=r2∫π/2−π/2(1+cos2θ)dθ |
r2∫π/2−π/2(1+cos2θ)dθ=r22∫π−π(1+cosu)du=r22(u+sinu)|π−π=πr2 |