∫∫√(1-x^2-y^2),定义域D是圆域x^2+y^2≤1 ,

题目

∫∫√(1-x^2-y^2),定义域D是圆域x^2+y^2≤1 ,

答案

不是“定义域D”,而是“积分域D".
∫∫√(1-x^2-y^2)dxdy
= ∫dt∫√(1-r^2)rdr
= -π∫√(1-r^2)d(1-r^2)
= (-2π/3)[(1-r^2)^(3/2)]
= 2π/3