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【小题】 玩玩儿!
x-r=rsinθ,dx=rcosθ,
∫√[r²-(x-r)²]dx=∫r²cos²θdθ
=(r²/2)θ+(r²/4)sin2θ+C
=(r²/2)arcsin[(x-r)/r]+(r²/2)(x-r)√[x(2r-x)]/r²+C
∫[(2-√3)r/2,r/2]√[r²-(x-r)²]dx
=(r²/2)arcsin(-1/2)-√3r²/8-(r²/2)arcsin(-√3/2)+√3r²/8
=πr²/12
[(√3-1)r/2](r/2)=(√3-1)r²/4
4×[πr²/12-(√3-1)r²/4]=(π/3-√3+1)r²=0.315r² |
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