8.6 解答

8.6

1. (Greenの定理) 区分的に滑らかな閉曲線 $C$ を境界に持つ $xy$ 平面上の有界閉領域を $\Omega$ とする．また，ベクトル場 ${\bf F} = (P(x,y),Q(x,y))$ は $C^{1}$ 級とする．このとき

$\displaystyle \oint_{C}{\bf F}\cdot d{\bf r} = \iint_{\Omega}(\nabla \times {\bf F}\cdot {\hat{\bf k}} dx dy$

$\displaystyle \oint_{C}Pdx + Qdy = \iint_{\Omega}\left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)dx dy$

が成り立つ．

(Stokes の定理)

$\displaystyle \oint_{\partial S}{\bf F}\cdot d{\bf r} = \iint_{S}(\nabla \times {\bf F}\cdot {\hat{\bf n}}dS$

(a) Greenの定理より

$\displaystyle \oint_{C}(x^2 - xy^3)dx + (y^2 - 2xy) dy$	$\displaystyle =$	$\displaystyle \iint_{\Omega}(-2y - (-2xy))dx dy$
	$\displaystyle =$	$\displaystyle \int_{0}^{2}\int_{0}^{2}(-2y + 2xy)dx dy = \int_{0}^{2}\left[-2xy + x^2 y\right]_{0}^{2} dy$
	$\displaystyle =$	$\displaystyle \int_{0}^{2}(-4y + 4y)dy = 0$

(b)

$\displaystyle \nabla \times {\bf F}$	$\displaystyle =$	$\displaystyle \left\vert\begin{array}{ccc} {\hat{\bf i}} & {\hat{\bf j}} & {\ha... ...z}\\ 2xy^3 - y^2 \cos{x} & 1 - 2y\sin{x} + 3x^2 y^2 & 0 \end{array}\right\vert$
	$\displaystyle =$	$\displaystyle {\hat{\bf k}}(-2y\cos{x} + 6xy^2 - (6xy^2 - 2y\cos{x})) = {\bf0}$

よって，Fは保存場となり， $\oint_{C}{\bf F}\cdot d{\bf r} = 0$

$\displaystyle \nabla \times {\bf F} = \left\vert\begin{array}{ccc} {\hat{\bf i... ...{\partial}{\partial z}\\ -z^2 & xy^2 & z \end{array}\right\vert = (0,2z,y^2)$

また， ${\bf r} = (x,y,z) = (x,y,\sqrt{1 - (x^2 + y^2)})$ より法線ベクトルを求めると

$\displaystyle {\bf r}_{x} \times {\bf r}_{y} = \left\vert\begin{array}{ccc} {\... ...\\ 0 & 1 & -\frac{y}{z} \end{array}\right\vert = (\frac{x}{z},\frac{y}{z},1)$

となる．よって

$\displaystyle \oint_{\partial S}-z^2dx + xy^2 dy + zdz$	$\displaystyle =$	$\displaystyle \iint_{\Omega}(0,2z,y^2) \cdot {\bf r}_{x} \times {\bf r}_{y} dx dy$
	$\displaystyle =$	$\displaystyle \int_{\Omega} (0,2z,y^2) \cdot (\frac{x}{z},\frac{y}{z},1) dx dy$
	$\displaystyle =$	$\displaystyle \int_{\Omega} (2y + y^2)dx dy ここで \Omega = \{(x,y):x^2 + y^2 =1\}$
	$\displaystyle =$	$\displaystyle \int_{0}^{2\pi}\int_{0}^{1}(2r\sin{\theta} + r^2 \sin^{2}{\theta})r dr d\theta$
	$\displaystyle =$	$\displaystyle \int_{0}^{2\pi}\int_{0}^{1}(2r^2 \sin{\theta} + r^3 \sin^{2}{\theta})dr d\theta$
	$\displaystyle =$	$\displaystyle \int_{0}^{2\pi}\left[\frac{2}{3}r^3 \sin{\theta} + \frac{r^4}{4} \sin^{2}{\theta}\right]_{0}^{1}d\theta$
	$\displaystyle =$	$\displaystyle \int_{0}^{2\pi}\left(\frac{2}{3}\sin{\theta} + \frac{1}{4} \sin^{2}{\theta}\right)d\theta$
	$\displaystyle =$	$\displaystyle \frac{1}{4}$

2. [Gaussの発散定理] ベクトル場 ${\bf F}(x,y,z) = (F_{1},F_{2},F_{3})$ において，区分的に滑らかな閉曲面 $S$ で囲まれた空間の領域を $V$ とし， $S$ の内部から外部に向かう法線ベクトルを $\hat{\bf n}$ とすると，

$\displaystyle \iint_{S}{\bf F} \cdot \hat{\bf n}dS = \iiint_{V} {\rm div}{\bf F} dV$

$\displaystyle \iint_{S}(F_{1}dydz + F_{2}dzdx + F_{3}dxdy) = \iiint_{V} \left(\... ...artial F_{2}}{\partial y} + \frac{\partial F_{3}}{\partial z} \right) dx dy dz$

が成り立つ．

(a) ${\bf r} = (x,y,z)$

$\displaystyle dS$	$\displaystyle =$	$\displaystyle \Vert{\bf r}_{x} \times {\bf r}_{y}\Vert dx dy$
	$\displaystyle =$	$\displaystyle \Vert\left\vert\begin{array}{ccc} {\hat{\bf i}} & {\hat{\bf j}} &... ...t {\bf k}}\\ 1 & 0 & z_{x}\\ 0 & 1 & z_{y} \end{array}\right\vert \Vert dx dy$
	$\displaystyle =$	$\displaystyle \Vert(-z_{x},-z_{y},1)\Vert dx dy = \sqrt{z_{x}^2 + z_{y}^2 + 1} dx dy$
	$\displaystyle =$	$\displaystyle \sqrt{(\frac{3x}{z})^2 + (\frac{3y}{z})^2 + 1}dx dy$
	$\displaystyle =$	$\displaystyle \sqrt{\frac{9x^2 + 9y^2 + z^2}{z^2}} dx dy = 2\sqrt{x^2 + y^2}$

よって

$\displaystyle \iint_{S}(x^2 + y^2)dS$	$\displaystyle =$	$\displaystyle 2\iint_{S}(x^2 + y^2)^{3/2} dx dy$
	$\displaystyle =$	$\displaystyle 2\int_{0}^{2\pi}\int_{0}^{\sqrt{3}}r^3 r dr d\theta$
	$\displaystyle =$	$\displaystyle 2\int_{0}^{2\pi}d\theta \int_{0}^{\sqrt{3}}r^4 dr = 2(2\pi)(\frac{9\sqrt{3}}{5})$
	$\displaystyle =$	$\displaystyle \frac{36\sqrt{3}\pi}{5}$

(b)

$\displaystyle {\rm curl}{\bf F} = \left\vert\begin{array}{ccc} {\hat{\bf i}} &... ...\partial}{\partial z}\\ x^2 - x & -xy & 3z \end{array}\right\vert = (0,0,-y)$

$\displaystyle {\hat{\bf n}} = \frac{{\bf r}_{x} \times {\bf r}_{y}}{\Vert{\bf r}_{x} \times {\bf r}_{y}\Vert} = (1,0,z_{x}) \times (0,1,z_{y})$

$\displaystyle dS = \Vert{\bf r}_{x} \times {\bf r}_{y}\Vert dx dy$

より

$\displaystyle \iint_{S}{\rm curl}{\bf F} \cdot {\hat {\bf n}}dS$	$\displaystyle =$	$\displaystyle \iint_{S}(0,0,-y) \cdot (1,0,z_{x}) \times (0,1,z_{y}) dx dy$
	$\displaystyle =$	$\displaystyle \iint_{\Omega}\left\vert\begin{array}{ccc} 0& 0 & -y\\ 1 & 0 & z_{x}\\ 0 & 1 & z_{y} \end{array}\right\vert dx dy$
	$\displaystyle =$	$\displaystyle \iint_{\Omega}-y dx dy$
	$\displaystyle =$	$\displaystyle \int_{0}^{2\pi}\int_{0}^{2}-r\sin{\theta} r dr d\theta$
	$\displaystyle =$	$\displaystyle \int_{0}^{2\pi}\sin{\theta}d\theta \int_{0}^{2}-r^2 dr = 0$

$\displaystyle \iint_{S}{\bf F}\cdot{\hat{\bf n}}dS$	$\displaystyle =$	$\displaystyle \iiint_{V}(5-y^2)dV$
	$\displaystyle =$	$\displaystyle \int_{0}^{3}\int_{-2}^{2}\int_{z=0}^{4-y^2}(5-y^2)dz dy dx$
$\displaystyle \int_{0}^{3}dx \int_{-2}^{2}(5(4-y^2) - y^2 (4-y^2))dy$
	$\displaystyle =$	$\displaystyle 6\left[20y - 3y^3 + \frac{y^5}{5}\right]_{0}^{2}$
	$\displaystyle =$	$\displaystyle 6(40 - 24 + \frac{32}{5}) = \frac{672}{5}$

(d) ${\bf F} = (x,y,z)$ より ${\rm div}{\bf F} = 3$ ．よって

$\displaystyle \iint_{S}xdydz+ ydzdx+ zdxdy = \iiint_{V}3dV$

ここで，

は円柱

と平面

で囲まれた領域なので，その体積は $\pi r^2 h = 27 \pi$ ．これより

$\displaystyle \iint_{S}xdydz+ ydzdx+ zdxdy = \iiint_{V}3dV = 81\pi$

3. $A$ を閉曲線 $C$ で囲まれた領域とすると，Greenの定理で $P = -y, Q = x$ とおくと

$\displaystyle \oint_{C}xdy - ydx = \iint_{\Omega}\left(\frac{\partial x}{\partial x} - \frac{\partial (-y)}{\partial y}\right)dx dy = \iint_{\Omega}2 dx dy = 2A$

よって， $A = \frac{1}{2}\oint_{C}xdy - ydx$

$\displaystyle A$	$\displaystyle =$	$\displaystyle \frac{1}{2}\oint_{C}xdy - y dx$
	$\displaystyle =$	$\displaystyle \frac{1}{2}\int_{0}^{2\pi}(a\cos{\theta})(b\cos{\theta}d\theta - (b\sin{\theta})(-a\sin{\theta})d\theta$
	$\displaystyle =$	$\displaystyle \frac{1}{2}\int_{0}^{2\pi}ab(\cos^{2}{\theta} + \sin^{2}{\theta})d\theta$
	$\displaystyle =$	$\displaystyle \frac{1}{2}\int_{0}^{2\pi}ab d\theta = \pi ab$

5. labelenshU:8-6-5 発散定理において ${\bf F} = f {\rm grad}g$ とおくと

$\displaystyle {\bf } \cdot {\hat{\bf n}} = f ({\rm grad}g \cdot {\hat{\bf n}} = f\frac{\partial g}{\partial n}$

また，

$\displaystyle {\rm div}{\bf F}$	$\displaystyle =$	$\displaystyle {\rm div}(f {\rm grad}g)$
	$\displaystyle =$	$\displaystyle \nabla \cdot (f \nabla g) = (\nabla f)\cdot \nabla g + f\nabla \cdot \nabla g$
	$\displaystyle =$	$\displaystyle {\rm grad}f \cdot {\rm grad}g + f \nabla^2 g$

よって

$\displaystyle \iint_{S}f \frac{\partial g}{\partial n} dS = \iiint_{V}(f \nabla^2 g + {\rm grad}f \cdot {\rm grad}g)dV$

定理 10..1 (積分公式) ～
$(1) \int x^{\alpha} dx = \frac{x^{\alpha + 1}}{\alpha + 1} + c (\alpha \neq -1)$
$(2) \int \frac{1}{x} dx = \log{\vert x\vert} + c$
$(3) \int e^{x} dx = e^{x} + c$
$(4) \int \sin{x}dx = -\cos{x} + c$
$(5) \int \cos{x} dx = \sin{x} + c$
$(6) \int \tan{x} dx = \log{\vert\sec{x}\vert} + c$
$(7) \int \sec^{2}{x}dx = \int \frac{1}{\cos^{2}{x}} = \tan{x} + c$
$(8) \int \csc^{2}{x}dx = \int \frac{1}{\sin^{2}{x}} = -\cot{x} +c$
$(9) \int \frac{dx}{x^{2}-a^{2}} = \frac{1}{2a}\log{\vert\frac{x-a}{x+a}\vert} + c$
$(10) \int \frac{dx}{x^{2}+a^{2}} = \frac{1}{a}\tan^{-1}{\frac{x}{a}} + c$
$(11) \int \frac{dx}{\sqrt{a^{2}-x^{2}}} = \sin^{-1}{\frac{x}{a}} + c$
$(12) \int \frac{dx}{\sqrt{x^{2} + A}} = \log{\vert x + \sqrt{x^{2} + A}\vert} + c$