解：原式=∫<0,π>dx∫<0,sinx>(x²-y²)dy
=∫<0,π>(x²sinx-sin³x/3)dx
=∫<0,π>x²sinxdx-∫<0,π>sin³x/3dx
=(-x²cosx+2xsinx+2cosx)│<0,π>-(cos³x/3-cosx)│<0,π>
=(π²-2-2)-(-1/3+1-1/3+1)
=π²-16/3。
=∫<0,π>(x²sinx-sin³x/3)dx
=∫<0,π>x²sinxdx-∫<0,π>sin³x/3dx
=(-x²cosx+2xsinx+2cosx)│<0,π>-(cos³x/3-cosx)│<0,π>
=(π²-2-2)-(-1/3+1-1/3+1)
=π²-16/3。
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第1个回答 2012-05-19
∫∫(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
=∫∫(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
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