# 曲线积分的计算

Eufisky posted @ 2017年9月19日 16:05 in 数学分析 with tags 积分计算 , 825 阅读

\begin{align*}U(a,b)&=R\int_0^{2\pi}\ln \frac 1{\sqrt{(R\cos\theta-a)^2+(R\sin\theta-b)^2}}d\theta\\&=R\int_0^{2\pi}\ln \frac 1{\sqrt{R^2+a^2+b^2-2aR\cos\theta-2bR\sin\theta}}d\theta\\&=-\frac R2\int_0^{2\pi}\ln \left(R^2+a^2+b^2-2R\sqrt{a^2+b^2}\sin(\theta+\varphi)\right)d\theta,\quad \text{其中}\, \tan\varphi=\frac ab\\&=-\frac R2\int_0^{2\pi}\ln \left(R^2+a^2+b^2-2R\sqrt{a^2+b^2}\cos\theta\right)d\theta\\&=-\frac R2\int_0^{\pi}\ln \left(R^2+a^2+b^2-2R\sqrt{a^2+b^2}\cos\theta\right)d\theta-\frac R2\int_\pi^{2\pi}\ln \left(R^2+a^2+b^2-2R\sqrt{a^2+b^2}\cos\theta\right)d\theta\\&=-\frac R2\int_0^{\pi}\ln \left(R^2+a^2+b^2-2R\sqrt{a^2+b^2}\cos\theta\right)d\theta-\frac R2\int_0^\pi\ln \left(R^2+a^2+b^2+2R\sqrt{a^2+b^2}\cos\theta\right)d\theta.\end{align*}

$$U(a,b)=\begin{cases}-2\pi |R|\ln |R|,& a^2+b^2<R^2\\-\pi |R|\ln \left(a^2+b^2\right),& a^2+b^2>R^2\end{cases}$$

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AP SSC Sanskrit Ques 说:
2022年9月25日 22:23

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