8.4 解答

8.4

1.

(a)$(-1,0,2)$と点$(1,3,2)$を結ぶ直線をパラメター化し,ベクトル表示すると

$\displaystyle {\bf r}(t) = (-1,0,2) + t(2,3,0) = (2t-1,3t,2)  (0 \leq t \leq 1)$

この直線$C$の弧長$s$

$\displaystyle s(t) = \int_{0}^{t}\Vert\frac{d{\bf r}}{dt}\Vert dt = \int_{0}^{t}\Vert(2,3,0)\Vert dt = \sqrt{13}t$

これより $ds = \sqrt{13}$となり,求める線積分は
$\displaystyle \int_{C}xy^2 ds$ $\displaystyle =$ $\displaystyle \sqrt{13}\int_{0}^{1}(2t-1)(3t)^2 dt = \sqrt{13}\int_{0}^{1}(18t^3 - 9t^2)dt$  
  $\displaystyle =$ $\displaystyle \sqrt{13}\left[\frac{9}{2}t^4 - 3t^3\right]_{0}^{1} = \sqrt{13}(\frac{9}{2} - 3) = \frac{3\sqrt{13}}{2}$  

(b)$(0,0,0)$と点$(1,3,2)$を結ぶ直線をパラメター化し,ベクトル表示すると

$\displaystyle {\bf r}(t) = t(1,3,2) = (t,3t,2t)  (0 \leq t \leq 1)$

この直線$C$の弧長$s$

$\displaystyle s(t) = \int_{0}^{t}\Vert\frac{d{\bf r}}{dt}\Vert dt = \int_{0}^{t}\Vert(1,3,2)\Vert dt = \sqrt{14}t$

これより $ds = \sqrt{14}$となり,求める線積分は
$\displaystyle \int_{C}(x + y^2) ds$ $\displaystyle =$ $\displaystyle \sqrt{14}\int_{0}^{1}(t + 9t^2) dt = \sqrt{14}\left[\frac{t^2}{2} + 3t^3\right]_{0}^{1}$  
  $\displaystyle =$ $\displaystyle \sqrt{14}(\frac{1}{2} + 3) = \frac{7\sqrt{14}}{2}$  

(c) 曲線$C$をパラメター化し,ベクトル表示すると

$\displaystyle {\bf r}(t) = (\cos{t},\sin{t})  (0 \leq t \leq \frac{\pi}{2})$

これより $x = \cos{t}, y = \sin{t}$となり 求める線積分は
$\displaystyle \int_{C}(-y^3 dx + x^2 dy)$ $\displaystyle =$ $\displaystyle \int_{0}^{\frac{\pi}{2}}(-\sin^{3}{t}(-\sin{t}) + \cos^{2}{t}\cos{t})dt$  
  $\displaystyle =$ $\displaystyle \int_{0}^{\frac{\pi}{2}}\sin^{4}{t}dt + \int_{0}^{\frac{\pi}{2}}\cos^{3}{t}dt$  
  $\displaystyle =$ $\displaystyle \frac{3\cdot1}{4\cdot2}\frac{\pi}{2} + \frac{2}{3\cdot1} = \frac{3\pi}{16} + \frac{2}{3}$  

(d)$(0,0,0)$と点$(1,1,1)$を結ぶ直線$C$をパラメター化し,ベクトル表示すると

$\displaystyle {\bf r}(t) = t(1,1,1) = (t,t,t)  (0 \leq t \leq 1)$

これより $x = t, y = t, z = t$.また, $d{\bf r} = (1,1,1)dt$となり 求める線積分は
$\displaystyle \int_{C}(3x^2 + 6y, -14yz, 20xz^3) \cdot d{\bf r}$ $\displaystyle =$ $\displaystyle \int_{0}^{1}(3t^2 + 6t, -14t^2, 20t^4) \cdot (1,1,1)dt$  
  $\displaystyle =$ $\displaystyle \int_{0}^{1}(3t^2 + 6t - 14t^2 + 20t^4)dt = \int_{0}^{1}(20t^4 - 11t^2 + 6t)dt$  
  $\displaystyle =$ $\displaystyle \left[\frac{4t^5 - \frac{11}{3}t^3} + 3t^2\right]_{0}^{1}$  
  $\displaystyle =$ $\displaystyle 4 - \frac{11}{3} + 3 = \frac{10}{3}$  

(e) $f(x,y,z) = 3x^2 - yz$より 点$(0,0,0)$と点$(1,1,1)$を結ぶ曲線$C$をベクトル表示すると

$\displaystyle {\bf r}(t) = (t,t^2,t^3)  (0 \leq t \leq 1)$

これより $d{\bf r} = (1,2t,3t^2)dt$となり 求める線積分は
$\displaystyle \int_{C}(3x^2 + 6y, -14yz, 20xz^3) \cdot d{\bf r}$ $\displaystyle =$ $\displaystyle \int_{0}^{1}(3t^2 + 6t^2, -14t^5, 20t^10) \cdot (1,t,t^2)dt$  
  $\displaystyle =$ $\displaystyle \int_{0}^{1}(9t^2 -14t^6 + 20t^12)dt$  
  $\displaystyle =$ $\displaystyle \left[3t^3 - 2t^7 + \frac{20}{13}{t^{13}}\right]_{0}^{1}$  
  $\displaystyle =$ $\displaystyle 3 - 2 + \frac{20}{13} = \frac{33}{13}$