令x=ρcosθ,y=ρsinθ，则D:{(x,y)|x^2+y^2<=2y}即为{(ρ,θ)|ρ<=2sinθ}
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第1个回答 2020-05-26
x^2+y^2 = 2y 化为极坐标是 r = 2Rsint， 0 ≤ t ≤ π
I = ∫∫<D>√(x^2+y^2)dxdy = ∫<0, π>dt∫<0, 2Rsint> r rdr
= (8/3)R^3∫<0, π>(sint)^3dt
= -(8/3)R^3∫<0, π>[1-(cost)^2]dcost
= -(8/3)R^3[cost-(1/3)(cost)^3]<0, π>
= (32/9)R^3本回答被提问者和网友采纳
I = ∫∫<D>√(x^2+y^2)dxdy = ∫<0, π>dt∫<0, 2Rsint> r rdr
= (8/3)R^3∫<0, π>(sint)^3dt
= -(8/3)R^3∫<0, π>[1-(cost)^2]dcost
= -(8/3)R^3[cost-(1/3)(cost)^3]<0, π>
= (32/9)R^3本回答被提问者和网友采纳
第2个回答 2020-08-29
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