Question

区域D：1/2<=|x|+|y|《=1 求∫∫ln(x^2+Y^2)dxdy的符号

区域D：1/2<=|x|+|y|《=1 求∫∫ln(x^2+Y^2)dxdy的符号请详细一点！不要极坐标法哦

Answer 1

由对称性，原式=4(∫<0,1/2>dx∫<1/2-x,1-x>+∫<1/2,1>dx∫<0,1-x>)ln(x^2+y^2)dy
=4∫<0,1/2>dx[yln(x^2+y^2)-2y+2xarctan(y/x)]|<1/2-x,1-x>
+4∫<1/2,1>dx[yln(x^2+y^2)-2y+2xarctan(y/x)]|<0,1-x>
=4∫<0,1/2>[(1-x)ln(2x^2-2x+1)-(1/2-x)ln(2x^2-x+1/4)-1+2xarctan(1/x-1)-2xarctan(0.5/x-1)]dx+4∫<1/2,1>[(1-x)ln(2x^2-2x+1)-2(1-x)+2xarctan(1/x-1)]dx,繁！
考虑第一象限，0<=x<=1/2时1/2-x<=y<=1-x,
∴x^2+y^2<=x^2+(1-x)^2=2(x-1/2)^2+1/2<=1;
1/2<x<1时0<y<1-x,
仿上，x^2+y^2<=1,
∴ln(x^2+y^2)<=0,
∴原式<0.