求二次积分∫(0~2)dx∫(0~√4-x^2)√(4-x^2-y^2)dy的解!∫∫,∫4-x的

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求二次积分∫(0~2)dx∫(0~√4-x^2)√(4-x^2-y^2)dy的解!

∫,∫4-x的

∫dx∫√(4-x²-y²)dy=∫dθ∫√(4-r²)rdr (做极坐标变换)
=(π/2)(-1/2)∫√(4-r²)d(4-r²)
=(-π/4)(2/3)(4-r²)^(3/2)│
=(-π/6)[(4-2²)^(3/2)-(4-0²)^(3/2)]
=(-π/6)(-8)
=4π/3.