Question

求二重积分∫∫D （1-x^2-y^2)^(1/2)dδ=?,其中D={(x,y)|x^2+y^2<=1}

∫∫D （1-x^2-y^2)^(1/2)dδ=?,其中D={(x,y)|x^2+y^2<=1}

Answer 1

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直接极坐标换元，x^2+y^2=r^2，区域D是0<=θ<=2π，0<=r<=1
原积分=∫(0,2π)dθ∫(0,1)r√(1-r^2)dr
=π∫(0,1)√(1-r^2)dr^2
=-2π/3(1-r^2)^(3/2)|(0,1)
=2π/3