问答题
设f(x，y)在全平面有连续偏导数，曲线积分∫
L
f(x，y)dx+xcosydy在全平面与路径无关，且∫
(0，0)
(t，t2)
f(x，y)dx+xcosydy=t
2
，求f(x，y)．
【正确答案】正确答案：①∫
L
(x，y)dx+xcosydy在全平面与路径无关
积分得f(x，y)=siny+C(x)． ②求f(x，y)转化为求C(x)． 因f(x，y)dx+xcosydy=sinydx+xcosydy+C(x)dx=sinydx+xdsiny+
积分得f(x，y)=siny+C(x)． ②求f(x，y)转化为求C(x)． 因f(x，y)dx+xcosydy=sinydx+xcosydy+C(x)dx=sinydx+xdsiny+
【答案解析】