麻烦帮忙解答一下,求过程!谢谢啦!
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发布时间:2024-10-03 12:48
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时间:2024-12-13 00:58
S:z=√(1-x^2-y^2) Dxy:x^2+y^2<=1
dz/dx=-x/√(1-x^2-y^2) dz/dy=-y/√(1-x^2-y^2)
dS=√[1+(dz/dx)^2+(dz/dy)^2]dxdy=dxdy/√(1-x^2-y^2)
所以,原式=∫∫(Dxy) √(1-x^2-y^2)*dxdy/√(1-x^2-y^2)
=∫∫(Dxy) dxdy
=∫(0,2π)dΘ∫(0,1)rdr
=π