设f(x,y)连续,且f(x,y)=x(y^2)+∫∫f(x,y)dxdy,其中D是由x=1,y=0,y=x^2所围,求f(x,y)

Question

设f(x,y)连续,且f(x,y)=x(y^2)+∫∫f(x,y)dxdy,其中D是由x=1,y=0,y=x^2所围,求f(x,y)

Answer 1

∫∫f(x,y)dxdy的结果为一常数,设a=∫∫f(x,y)dxdy
则f(x,y)=xy²+a
两边作二重积分
∫∫f(x,y)dxdy=∫∫(xy²+a)dxdy
=∫[0---->1] dx∫[0---->x²] (xy²+a)dy
=∫[0---->1] (1/3xy³+ay) |[0---->x²] dx
=∫[0---->1] (1/3x⁷+ax²) dx
=(1/24)x⁸+(a/3)x³ |[0---->1]
=1/24+a/3
则：1/24+a/3=a
因此解得：a=1/16
因此f(x,y)=xy²+1/16