∫∫sin(1+x2+y2)dxdy,其中D为x2+y2≤1(x≥0,y≥0),那俩积分号下面有个D
问题描述：
∫∫sin(1+x2+y2)dxdy,其中D为x2+y2≤1(x≥0,y≥0),那俩积分号下面有个D
答
设x=rcosθ,y=rsinθ (极坐标代换)原式=∫(0,π/2)dθ∫(0,1)sin(1+r²)rdr(符号∫(0,π/2)表示从0到π/2积分,其它类同) =(2π)/2∫(0,1)sin(1+r²)d(1+r²) =[π(-1...