2020/07/13 スカラー場の線積分

本日のお題

曲線 \(\displaystyle \mbox{C}:\ \boldsymbol{r} = \left(\cos\theta,\ \sin\theta,\ \theta\right)\quad\left(0 \le \theta \le \frac{\pi}{2}\right)\) に沿って,スカラー場 \(\varphi(x,\ y,\ z) = x + z\) の

  曲線弧長 \(s\) に関する線積分 \(\displaystyle \int_C \varphi(x,\ y,\ z)\,ds\ ,\)

  \(x\) に関する線積分 \(\displaystyle \int_C \varphi(x,\ y,\ z)\,dx\ ,\)

  \(y\) に関する線積分 \(\displaystyle \int_C \varphi(x,\ y,\ z)\,dy\ ,\)

  \(z\) に関する線積分 \(\displaystyle \int_C \varphi(x,\ y,\ z)\,dz\)

をそれぞれ求めよ

はい,かしこまりました

\(\hspace{1em} \displaystyle \int_C \varphi(x,\ y,\ z)\,ds \begin{array}[t]{l} \displaystyle  = \int_0^{\frac{\pi}{2}} (x + z)\sqrt{\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2 + \left(\frac{dz}{d\theta}\right)^2} \, d\theta \\ \displaystyle = \int_0^{\frac{\pi}{2}} (\cos\theta + \theta) \sqrt{\sin^2 \theta + \cos^2 \theta + 1}\,d\theta \\ \displaystyle = \sqrt{2}\,\int_0^{\frac{\pi}{2}} (\cos\theta + \theta) \,d\theta \\ \displaystyle = \sqrt{2} \left[\sin\theta + \frac{1}{2}\theta^2\right]_0^{\frac{\pi}{2}} \\ \displaystyle = \sqrt{2} \left(1 - \frac{1}{8} \pi^2 \right) \end{array}\)


\(\hspace{1em} \displaystyle \int_C \varphi(x,\ y,\ z)\,dx  \begin{array}[t]{l} \displaystyle = \int_0^{\frac{\pi}{2}} (x + z)\,\frac{dx}{d\theta} \, d\theta  \\ \displaystyle = \int_0^{\frac{\pi}{2}} (\cos\theta + \theta )(-\sin\theta)\,d\theta \\ \displaystyle = \int_0^{\frac{\pi}{2}} (-\sin\theta \, \cos\theta - \theta\,\sin\theta)\,d\theta\\ \displaystyle = - \frac{1}{2} \int_0^{\frac{\pi}{2}} \sin 2\theta \,d\theta + \int_0^{\frac{\pi}{2}} \theta(\cos\theta)'\,d\theta \\ \displaystyle = \frac{1}{4}\Big[\cos 2\theta\Big]_0^{\frac{\pi}{2}} + \Big[\theta\,\cos\theta\Big]_{0}^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}} \cos\theta\,d\theta \\ \displaystyle = \frac{1}{4}\cdot(-2) + 0 - 1 = - \frac{3}{2} \end{array}\)

途中の \(\displaystyle \int_0^{\frac{\pi}{2}} \theta\,\sin\theta\,d\theta\) については,部分積分を施しています

\(u = \theta,\quad v' = \sin\theta\) とすれば \(u' = 1,\quad v = -\cos\theta\) です

したがって,\(\displaystyle \int_0^{\frac{\pi}{2}} \theta\,\sin\theta\,d\theta = \int_0^{\frac{\pi}{2}}u\,v'\,d\theta = \Big[uv\Big]_0^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}} u'\,v\,d\theta = -\Big[\theta\,\cos\theta\Big]_0^{\frac{\pi}{2}} + \int_0^{\frac{\pi}{2}} 1\cdot \cos\theta \, d\theta\)

次の \(y\) に関する積分に登場する \(\displaystyle \int_0^{\frac{\pi}{2}} \theta \cos\theta\,d\theta\) についても同様です

\(\hspace{1em} \displaystyle \int_C \varphi(x,\ y,\ z)\,dy  \begin{array}[t]{l} \displaystyle = \int_0^{\frac{\pi}{2}} (x + z)\,\frac{dy}{d\theta} \, d\theta  \\ \displaystyle = \int_0^{\frac{\pi}{2}} (\cos\theta + \theta ) \,\cos\theta \,d\theta \\ \displaystyle = \int_0^{\frac{\pi}{2}} (\cos^2 \theta + \theta\,\cos\theta)\,d\theta\\ \displaystyle = \int_0^{\frac{\pi}{2}} \frac{1 + \cos 2\theta}{2} \,d\theta + \int_0^{\frac{\pi}{2}} \theta\,(\sin\theta)'\,d\theta \\ \displaystyle = \left[\frac{1}{2}\theta + \frac{1}{4}\,\sin 2\theta\right]_0^{\frac{\pi}{2}} + \Big[\theta \,\sin\theta \Big]_0^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}} \sin\theta\,d\theta \\ \displaystyle = \frac{\pi}{4} + \frac{\pi}{2} - 1 = \frac{3}{4}\pi - 1 \end{array}\)


\(\hspace{1em} \displaystyle \int_C \varphi(x,\ y,\ z)\,dz \begin{array}[t]{l} \displaystyle = \int_0^{\frac{\pi}{2}} (x + z)\,\frac{dz}{d\theta} \,d\theta \\ \displaystyle = \int_0^{\frac{\pi}{2}} (\cos\theta + \theta)\,d\theta \\ \displaystyle = \left[\sin\theta + \frac{1}{2}\theta^2\right]_0^{\frac{\pi}{2}} \\ \displaystyle = 1 + \frac{\pi^2}{8} \end{array}\)

Last modified: Friday, 5 March 2021, 5:30 PM