设D={(x,y)|x^2+y^2≤a^2},当a=?时,∫∫√a^2-x^2-y^2dxdy=π

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设D={(x,y)|x^2+y^2≤a^2},当a=?时,∫∫√a^2-x^2-y^2dxdy=π设D={(x,y)|x^2+y^2≤a^2},当a=?时,∫∫√a^2-x^2-y^2dxdy=π设D

设D={(x,y)|x^2+y^2≤a^2},当a=?时,∫∫√a^2-x^2-y^2dxdy=π
设D={(x,y)|x^2+y^2≤a^2},当a=?时,∫∫√a^2-x^2-y^2dxdy=π