设f(u)连续可导,且∫04f(u)du=2,L为半圆周y=,起点为原点,终点为B(2,0),则I=∫Lf(x2+y2)(xdx+ydy)=_______.

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问题 设f(u)连续可导,且∫04f(u)du=2,L为半圆周y=,起点为原点,终点为B(2,0),则I=∫Lf(x2+y2)(xdx+ydy)=_______.

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答案1

解析 P(x,y)=xf(x2+y2),Q(x,y)=yf(x2+y2),
因为=2xyf’(x2+y2),所以曲线积分与路径无关,
故I=∫Lf(x2+y2)(xdx+ydy)=1/2
(0,0)(2,0)f(x2+y2)d(x2+y2)=1/2∫01f(t)dt=1.
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