∫∫∑ 1 / √ (1 + 4x & # 178; + 4Y & # 178;) ds, ∑ surface z = x & # 178; + Y & # 178; (0 ≤ Z ≤ 1)

z=x²+y²
z'x=2x z'y=2y
ds=√（1+4x²+4y²）dxdy
∫∫Σ1/√（1+4x²+4y²）ds
=∫∫dxdy
=π

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