解答题
8.[2011年] 已知f(x，y)具有二阶连续偏导数，且f(1，y)=0，f(x，1)=0，
(x，y)dxdy=a，其中D={(x，y)｜0≤x≤1，0≤y≤1}，计算二重积分I=
(x，y)dxdy=a，其中D={(x，y)｜0≤x≤1，0≤y≤1}，计算二重积分I=
【正确答案】注意到f(x，y)的二阶导数连续，有f''xy(x，y)=f''yx(x，y)，故
f''xy(x，y)dy=f''yx(x，y)dy=df'x(x，y)， f'x(x，y)dx=df(x，y)，
有I=∫01xdx∫01yf''xy(x，y)dy=∫01xdx∫01ydf'x(x，y)
=∫01[yf'x(x，y)｜01—∫01f'x(x，y)dy]xdx
=∫01f'x(x，1)xdx—∫01xdx∫01f'x(x，y)dy
=0一∫01dy∫01xf'x(x，y)dx (因f(x，1)=0，故f'x(x，1)=0)
=一∫01dy∫01xdf(x，y)
=一∫01[xf(x，y)｜01—∫01f(x，y)dx]dy
=∫01∫01f(x，y)dxdy=
f''xy(x，y)dy=f''yx(x，y)dy=df'x(x，y)， f'x(x，y)dx=df(x，y)，
有I=∫01xdx∫01yf''xy(x，y)dy=∫01xdx∫01ydf'x(x，y)
=∫01[yf'x(x，y)｜01—∫01f'x(x，y)dy]xdx
=∫01f'x(x，1)xdx—∫01xdx∫01f'x(x，y)dy
=0一∫01dy∫01xf'x(x，y)dx (因f(x，1)=0，故f'x(x，1)=0)
=一∫01dy∫01xdf(x，y)
=一∫01[xf(x，y)｜01—∫01f(x，y)dx]dy
=∫01∫01f(x，y)dxdy=
【答案解析】