高等数学提高曲线曲面积分计算题查看解析已掌握收入错题本题讨论不看本题#9562设 LLL 是从点 A(1,−1)A\left({1, - 1}\right)A(1,−1) 沿曲线 x2+y2=−2y(y≥−1){x}^{2} + {y}^{2} = - {2y}\left({y \geq - 1}\right)x2+y2=−2y(y≥−1) 到点 B(−1,−1)B\left({-1, - 1}\right)B(−1,−1) 的有向曲线， f(x)f\left(x\right)f(x) 是连续函数,计算 I=∫Lx[f(x)+1]dy−y2[f(x)+1]+2yf(x)1−x2 dxI = {\int }_{L}x\left\lbrack {f\left(x\right) + 1}\right\rbrack \mathrm{d}y - \frac{{y}^{2}\left\lbrack {f\left(x\right) + 1}\right\rbrack + {2yf}\left(x\right) }{\sqrt{1 - {x}^{2}}}\,\mathrm{d}xI=∫Lx[f(x)+1]dy−1−x2y2[f(x)+1]+2yf(x)dx.
