其中d是由圆周x2+y2=1及坐标轴所围成的在第一象限内的闭区域；

Question

如题所述

Answer 1

极坐标
∫∫（D）ln(1+x²+y²)dxdy
=∫∫（D）rln(1+r²)drdθ
=∫[0→2π]dθ∫[0→1] rln(1+r²)dr
=2π∫[0→1] rln(1+r²)dr
=π∫[0→1] ln(1+r²)d(r²)
=πr²ln(1+r²)-2π∫[0→1] r³/(1+r²)dr
=πr²ln(1+r²)-2π∫[0→1] (r³+r-r)/(1+r²)dr
=πr²ln(1+r²)-2π∫[0→1] rdr+2π∫[0→1] r/(1+r²)dr
=πr²ln(1+r²)-πr²+π∫[0→1] 1/(1+r²)d(r²)
=πr²ln(1+r²)-πr²+πln(1+r²) |[0→1]
=πln2-π+πln2
=π(2ln2-1)
做错了,当作整圆做的了.结果再除以4本回答被网友采纳