0=<x<=1,0=<y<=1,求e^(x^2+y^2)在这个范围内的二重积分 求解谢谢_百度...答:y=rsint x^2+y^2=r^2 r:[0,1]t:[0,pi/2]dxdy=rdr 还是四分之一,因为只有第一象限 ∫(0,1)∫(0,1)e^(x^2+y^2)dxdy =∫(0,pi/2)∫(0,1)e^(r^2)rdrdt =pi/2*(1/2)∫(0,1)e^(r^2)d(r^2)=pi/4*e^(r^2)|[0,1]=(pi/4)(e-1)=pi(e-1)/4 ...
用极坐标计算二重积分,∫∫e^(x^2+y^2)dxdy,其中D={(x,y)丨x^2+y^...答:D={(x,y)丨x^2+y^2≤4} 令x=pcosa,y=psina 0≤p≤2,0≤a≤2π ∫∫e^(x^2+y^2)dxdy =∫[0,2π]da∫[0,2]e^p^2*pdp =a[0,2π]*1/2e^p^2[0,2]=π*(e^4-1)