求大神看看这道不定积分题是怎么化的?
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2∫<0, π/2>dθ∫<0, Rcosθ>√(R^2-r^2)rdr
= - ∫<0, π/2>dθ∫<0, Rcosθ>√(R^2-r^2)d(R^2-r^2)
= - (2/3)∫<0, π/2>dθ[(R^2-r^2)^(3/2)]<0, Rcosθ>
= (2/3)R^3∫<0, π/2>[1-(sinθ)^3]dθ
= (2/3)R^3 [ π/2 - ∫<0, π/2>(sinθ)^3dθ]
= (2/3)R^3 { π/2 + ∫<0, π/2>[1-(cosθ)^2]dcosθ }
= (2/3)R^3 { π/2 + [cosθ-(1/3)(cosθ)^3]<0, π/2>}
= (2/3)R^3(π/2-2/3) = (1/3)R^3(π-4/3)
= - ∫<0, π/2>dθ∫<0, Rcosθ>√(R^2-r^2)d(R^2-r^2)
= - (2/3)∫<0, π/2>dθ[(R^2-r^2)^(3/2)]<0, Rcosθ>
= (2/3)R^3∫<0, π/2>[1-(sinθ)^3]dθ
= (2/3)R^3 [ π/2 - ∫<0, π/2>(sinθ)^3dθ]
= (2/3)R^3 { π/2 + ∫<0, π/2>[1-(cosθ)^2]dcosθ }
= (2/3)R^3 { π/2 + [cosθ-(1/3)(cosθ)^3]<0, π/2>}
= (2/3)R^3(π/2-2/3) = (1/3)R^3(π-4/3)
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