第1个回答 2011-05-10
∫∫(x^2-y^2)dσ
=∫∫(x^2-y^2)dxdy
=∫[0,π]dx∫[0,sinx](x^2-y^2)dy
=∫[0,π]dx*(y*x^2-y^3/3)|[0,sinx]
=∫[0,π][x^2*sinx-(sinx)^3/3]dx
=∫[0,π]x^2*sinx*dx-1/3*∫[0,π](sinx)^3*dx
=(-x^2*cosx+2xsinx+2cosx)|[0,π]-2/3*∫[0,π/2](sinx)^3*dx
=(π^2-2)-2-2/3*2/3
=π^2-4-4/9
=π^2-40/9本回答被提问者采纳
=∫∫(x^2-y^2)dxdy
=∫[0,π]dx∫[0,sinx](x^2-y^2)dy
=∫[0,π]dx*(y*x^2-y^3/3)|[0,sinx]
=∫[0,π][x^2*sinx-(sinx)^3/3]dx
=∫[0,π]x^2*sinx*dx-1/3*∫[0,π](sinx)^3*dx
=(-x^2*cosx+2xsinx+2cosx)|[0,π]-2/3*∫[0,π/2](sinx)^3*dx
=(π^2-2)-2-2/3*2/3
=π^2-4-4/9
=π^2-40/9本回答被提问者采纳
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