8.4 解答

(a) 点

と点

を結ぶ直線をパラメター化し，ベクトル表示すると

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

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

$\displaystyle \int_{C}xy^2 ds$	$\displaystyle =$	$\displaystyle \sqrt{13}\int_{0}^{1}(2t-1)(3t)^2 dt = \sqrt{13}\int_{0}^{1}(18t^3 - 9t^2)dt$
	$\displaystyle =$	$\displaystyle \sqrt{13}\left[\frac{9}{2}t^4 - 3t^3\right]_{0}^{1} = \sqrt{13}(\frac{9}{2} - 3) = \frac{3\sqrt{13}}{2}$

(b) 点

と点

を結ぶ直線をパラメター化し，ベクトル表示すると

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

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

$\displaystyle \int_{C}(x + y^2) ds$	$\displaystyle =$	$\displaystyle \sqrt{14}\int_{0}^{1}(t + 9t^2) dt = \sqrt{14}\left[\frac{t^2}{2} + 3t^3\right]_{0}^{1}$
	$\displaystyle =$	$\displaystyle \sqrt{14}(\frac{1}{2} + 3) = \frac{7\sqrt{14}}{2}$

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

$\displaystyle \int_{C}(-y^3 dx + x^2 dy)$	$\displaystyle =$	$\displaystyle \int_{0}^{\frac{\pi}{2}}(-\sin^{3}{t}(-\sin{t}) + \cos^{2}{t}\cos{t})dt$
	$\displaystyle =$	$\displaystyle \int_{0}^{\frac{\pi}{2}}\sin^{4}{t}dt + \int_{0}^{\frac{\pi}{2}}\cos^{3}{t}dt$
	$\displaystyle =$	$\displaystyle \frac{3\cdot1}{4\cdot2}\frac{\pi}{2} + \frac{2}{3\cdot1} = \frac{3\pi}{16} + \frac{2}{3}$

(d) 点

と点

を結ぶ直線

をパラメター化し，ベクトル表示すると

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

$\displaystyle \int_{C}(3x^2 + 6y, -14yz, 20xz^3) \cdot d{\bf r}$	$\displaystyle =$	$\displaystyle \int_{0}^{1}(3t^2 + 6t, -14t^2, 20t^4) \cdot (1,1,1)dt$
	$\displaystyle =$	$\displaystyle \int_{0}^{1}(3t^2 + 6t - 14t^2 + 20t^4)dt = \int_{0}^{1}(20t^4 - 11t^2 + 6t)dt$
	$\displaystyle =$	$\displaystyle \left[\frac{4t^5 - \frac{11}{3}t^3} + 3t^2\right]_{0}^{1}$
	$\displaystyle =$	$\displaystyle 4 - \frac{11}{3} + 3 = \frac{10}{3}$

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

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