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