高数第二题求解
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z=√(x^2+y^2) z'x=x/√(x^2+y^2) z'y=y/√(x^2+y^2)
dS=√[1+x^2/(x^2+y^2)+y^2/(x^2+y^2)]dxdy=√2dxdy
原式=√2∫∫[Dxy][xy+(y+x)√(x^2+y^2)]dxdy
=√2∫[-π/2,π/2]dθ∫[0,2acosθ][ρ^3cosθsinθ+ρ^3(cosθ+sinθ)]dρ
=√2∫[-π/2,π/2][cosθsinθ+(cosθ+sinθ)]ρ^4/4|[0,2acosθ]dθ
=4√2a^4∫[-π/2,π/2][cos^5θsinθ+cos^5θ+sinθcos^4θ]dθ
=8√2a^4∫[0,π/2]cos^5θdθ
=8√2a^4*4*2/5*3
=64/15√2a^4
dS=√[1+x^2/(x^2+y^2)+y^2/(x^2+y^2)]dxdy=√2dxdy
原式=√2∫∫[Dxy][xy+(y+x)√(x^2+y^2)]dxdy
=√2∫[-π/2,π/2]dθ∫[0,2acosθ][ρ^3cosθsinθ+ρ^3(cosθ+sinθ)]dρ
=√2∫[-π/2,π/2][cosθsinθ+(cosθ+sinθ)]ρ^4/4|[0,2acosθ]dθ
=4√2a^4∫[-π/2,π/2][cos^5θsinθ+cos^5θ+sinθcos^4θ]dθ
=8√2a^4∫[0,π/2]cos^5θdθ
=8√2a^4*4*2/5*3
=64/15√2a^4
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