设D为单位圆x^2+y^2≤1在第一象限部分,则：∫∫De^(x^2+y^2)/2dxdy=?注：∫∫的下面是D

Question

设D为单位圆x^2+y^2≤1在第一象限部分,则：∫∫De^(x^2+y^2)/2dxdy=?
注：∫∫的下面是D

Answer 1

用极坐标吧
D为单位圆x^2+y^2≤1在第一象限部分
x=pcosa,y=psina,0≤p≤1,0≤a≤π/2,dxdy=pdadp
∫∫De^(x^2+y^2)/2dxdy
=∫[0,π/2]∫[0,1]e^(p^2/2)*pdpda
=∫[0,π/2]da∫[0,1]e^(p^2/2)*pdp
=a[0,π/2]*e^(p^2/2)[0,1]
=π/2*(e^(1/2)-1)