设D:x^2+y^2

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设D:x^2+y^2

设D:x^2+y^2
设D:x^2+y^2<=a^2,计算二重积分∫∫√(a^2+x^2+y^2)dxdy

设D:x^2+y^2
极坐标
∫∫√(a²+x²+y²)dxdy
=∫∫r√(a²+r²)drdθ
=∫[0→2π]dθ∫[0→a] r√(a²+r²)dr
=2π∫[0→a] r√(a²+r²)dr
=π∫[0→a] √(a²+r²)d(r²)
=π(2/3)(a²+r²)^(3/2) |[0→a]
=(2π/3)[(2a²)^(3/2)-a³]
=(2πa³/3)(2√2-1)