关于曲面积分～∫∫s(xy+yz+zx)ds,其中 - 知识问答

关于曲面积分～∫∫s(xy+yz+zx)ds,其中s为锥面z＝(x^2+y^2)^(1/2)被柱面x^2+y^2＝2ax

1个回答

∵αz/αx=x/√(x²+y²),αz/αy=y/√(x²+y²)
∴ds=√[1+(αz/αx)²+(αz/αy)²]dxdy
=√2dxdy
故原式=∫∫(xy+yz+zx)√2dxdy
=√2∫dθ∫(r²sinθcosθ+r²sinθ+r²cosθ)rdr (做极坐标变换)
=4√2a^4∫(sinθcosθ+sinθ+cosθ)(cosθ)^4dθ
=4√2a^4∫[((cosθ)^5+(cosθ)^4)sinθ+(1-2sin²θ+(sinθ)^4)cosθ]dθ
=(4√2a^4)*[2(1-2/3+1/5)]
=64√2a^4/15.

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