8.5 解答

8.5

1. スカラー場$f$の曲面$S$上での面積分は

$$\iint_{S}f(x,y,z)dS = \iint_{\Omega}f(x(u,v),y(u,v),z(u,v))\Vert{\bf r}_{x} \times {\bf r}_{y}\Vert du dv$$

曲面 $S$ 上のベクトル場 ${\bf F}$ の面積分は次のように2重積分で表わされます．

$$\iint_{S}{\bf F}\cdot \hat{\bf n}dS = \iint_{\Omega}{\bf F} \cdot \frac{{\bf r}_{u} \times {\bf r}_{v}}... ...f r}_{u} \times {\bf r}_{v}\Vert}\Vert{\bf r}_{u} \times {\bf r}_{v}\Vert du dv$$

$$= \iint_{\Omega}{\bf F}\cdot ({\bf r}_{u} \times {\bf r}_{v}) du dv$$

$$= \iint_{\Omega}\left\vert\begin{array}{ccc} F_{1} & F_{2} & F_{3}\... ...ial y}{\partial v} & \frac{\partial z}{\partial v} \end{array}\right\vert du dv$$

(a) ${\bf r} = (x,y,1-x-y)$より

$${\bf r}_{x} \times {\bf r}_{y} = (1,0,-1) \times (0,1,1-1) = (1,1,1)$$

よって面積素 $dS = \sqrt{3}dx dy$．また， $S : x + y +z = 1$の$xy$平面への射影$\Omega$はV-simpleを用いて表わすと

$$\Omega = \{(x,y) : 0 \leq x \leq 1, 0 \leq y \leq 1 - x\}$$

これより

$$\iint_{S}xy^2 dS = \sqrt{3}\iint_{\Omega}x y^2 dx dy$$

$$= \int_{0}^{1} \int_{0}^{1-x}xy^2 dy dx = \sqrt{3}\int_{0}^{1}\left[\frac{xy^3}{3}\right]_{0}^{1-x} dx$$

$$= \sqrt{3}\int_{0}^{1}\frac{x(1-x)^3}{3}dx = \frac{sqrt{3}}{3}\int_{0}^{1}x(1 - 3x + 3x^2 - x^3)dx$$

$$= \frac{\sqrt{3}}{3}\int_{0}^{1}(x - 3x^2 + 3x^3 - x^4)dx$$

$$= \frac{\sqrt{3}}{3}\left[\frac{x^2}{2} - x^3 + \frac{3x^4}{4} - \frac{x^5}{5}\right]_{0}^{1}$$

$$= \frac{\sqrt{3}}{3}\left(\frac{1}{2} - 1 + \frac{3}{4} - \frac{1}{5}\right)$$

$$= \frac{\sqrt{3}}{3}\left(\frac{10 -20 + 15 -4}{20}\right) = \frac{1}{20\sqrt{3}}$$

(b) $S : x^2 + y^2 = 9, x \geq 0, y \geq 0, 0 \leq z \leq 4$より正射影は$yz$平面または，$xz$平面となる．そこで，$yz$平面に正射影を行うとすると， $\Omega_{yz}$は

$$\Omega_{yz} = \{(y,z) : -3 \leq y \leq 3, 0 \leq z \leq 4\}$$

また， $x^2 = 9 - y^2$より， $x = \pm \sqrt{9 - y^2}$．$x \geq 0$より，

$${\bf r} = x{\bf i} + y{\bf j} + z{\bf k} = \sqrt{9-y^2}{\bf i} + y{\bf j} + z{\bf k}$$

$${\bf n} = \frac{{\bf r}_{y} \times {\bf r}_{z}}{\vert{\bf r}_{y} \times {\bf r}_{z}\vert}$$

したがって，

$$\iint_{S}(6z，3\sqrt{9-y^2}+2y-z,-\sqrt{9-y^2}) \cdot \hat{\bf n}dS = \iint_{S, x>0}(6z，3\sqrt{9-y^2}+2y-z,-\sqrt{9-y^2}) \cdot \hat{\bf n}dS$$

$$= \iint_{\Omega_{yz}}\left\vert\begin{array}{ccc} 6z & 3\sqrt{9-y^2... ...t\vert dx dy = \iint_{\Omega}(6z + 3y + \frac{y(2y - z)}{\sqrt{9 - y^2}}) dy dz$$

$$= \int_{y=0}^{3}\int_{0}^{4}(6z + 3y + \frac{y(2y - z)}{\sqrt{9 - y... ..._{0}^{3}\left[3z^2 + 3yz + \frac{2y^2z - yz^2/2}{\sqrt{9-y^2}}\right]_{0}^{4}dy$$

$$= \int_{0}^{3}(48 + 12y + \frac{8y^2 - 8y}{\sqrt{9-y^2}})dy$$

ここで，

$$int_{0}^{3}\frac{8y^2}{\sqrt{9-y^2}}dy = \int_{0}^{\pi/2}\frac{72\sin^2{t}}{3\cos{t}}(3\cos{t})dt = 72 \cdot \frac{\pi}{4} = 18\pi$$

に注意すると，

$$\iint_{S}(6z，3\sqrt{9-y^2}+2y-z,-\sqrt{9-y^2}) \cdot \hat{\bf n}dS = 156 + 18\pi$$

(c) $S : 2x+3y+6z=12, x \geq 0, y \geq 0, z \geq 0$ より ${\bf r} = (x,y,z) = (x, y , \frac{12 - 2x - 3y}{6})$

$$\Omega = \{(x,y) : 0 \leq x \leq 6, 0 \leq y \leq 4 - \frac{2}{3}x\}$$

$$\iint_{S}(6z，-4x,y) \cdot \hat{\bf n}dS = \iint_{\Omega}\left\vert\begin{array}{ccc} 6z & -4x & y\\ 1 & 0 ... ...& -\frac{1}{2} \end{array}\right\vert dx dy = \iint_{\Omega}(2z - 2x + y) dx dy$$

$$= \frac{1}{3}\int_{0}^{6}\int_{0}^{4-\frac{2}{3}x}(6z - 6x + 3y)dy dx$$

$$= \frac{1}{3}\int_{0}^{6}\int_{0}^{4-\frac{2}{3}x}(12 -2x - 3y - 6x + 3y)dy dx$$

$$= \frac{1}{3}\int_{0}^{6}\int_{0}^{4-\frac{2}{3}x}(12 - 8x)dy dx$$

$$= \frac{1}{3}\int_{0}^{6}\left[12y - 4x^2\right]_{0}^{4 - \frac{2}{3}x} dx$$

$$= \frac{1}{3}\int_{0}^{6}\left(48 - 8x - 4(\frac{(12- 2x)^2}{9})\right) dx$$

$$= \frac{1}{3}\int_{0}^{6}\left(48 - 8x - \frac{16}{9}(36 - 12x + x^2)\right)dx$$

$$= \frac{1}{3}\left[-\frac{16x^3}{27} + \frac{40x^2}{6} - 16x\right]_{0}^{6}$$

$$= \frac{1}{3}(-128 + 240 - 96) = \frac{16}{3}$$

(d) $S : x^2 + y^2 + z^2 = a^2, x \geq 0, y \geq 0, z \geq 0$ より ${\bf r} = (x,y,z(x,y))$

$$\Omega = \{(t,\theta) : 0 \leq \theta \leq \frac{\pi}{2}, 0 \leq r \leq a\}$$

$$\iint_{S}(6z，-4x,y) \cdot \hat{\bf n}dS = \iint_{\Omega}\left\vert\begin{array}{ccc} 6z & -4x & y\\ 1 & 0 ... ...{z} \end{array}\right\vert dx dy = \iint_{\Omega}(6x - \frac{4xy}{z} + y) dx dy$$

$$= \int_{0}^{\frac{\pi}{2}}\int_{0}^{a}(6r\cos{\theta} - \frac{4r^2 \sin{\theta}\cos{\theta}}{\sqrt{a^2 - r^2}} + r\sin{\theta})r dr d\theta$$

$$= \int_{0}^{\frac{\pi}{2}}\int_{0}^{a}(6r^2 \cos{\theta} - \frac{4r^3 \sin{\theta}\cos{\theta}}{\sqrt{a^2 - r^2}} + r^2\sin{\theta}) dr d\theta$$

$$= \int_{0}^{\frac{\pi}{2}}\left[2r^3 \cos{\theta} + \frac{r^3}{3}\s... ...}(\frac{2}{3}(a^2 - r^2)^{3/2} - 2a^2 (a^2 - r^2)^{1/2})\right]_{0}^{a} d\theta$$

$$= \int_{0}^{\frac{\pi}{2}}\left(2a^3\cos{\theta} + \frac{a^3}{3}\sin{\theta} + 2\sin{\theta}\cos{\theta}(\frac{2}{3}a^3 - 2a^3)\right)d\theta$$

$$= 2a^3 + \frac{a^3}{3} + \frac{2}{3}a^3 - 2a^3 = a^3$$

(e) $S : z = x^2 + y^2 , x^2 + y^2 \leq 1$より$\Omega$は

$$\Omega = \{(r,\theta) : 0 \leq \theta \leq \pi, 0 \leq r \leq 1\}$$

また， ${\bf r} = (x,y,x^2 + y^2)$．

$$\iint_{S}(xy,-2y,z-x) \cdot \hat{\bf n}dS = \iint_{\Omega}\left\vert\begin{array}{ccc} xy & -2y & z-x\\ 1 & ... ...& 2y \end{array}\right\vert dx dy = \iint_{\Omega}(-2x^2 y + 4y^2 + z -x) dx dy$$

$$= \iint_{\Omega}(-2x^2 y + 4y^2 + x^2 + y^2 -x) dx dy = \iint_{\Omega}(-2x^2 y + 5y^2 + x^2 - x) dx dy$$

$$= \int_{0}^{\pi}\int_{0}^{1}(-2r^3 \cos^{2}{\theta}\sin{\theta} + 5r^2 \sin^{2}{\theta} + r^2 \cos^{2}{\theta} - r\cos{\theta}) rdr d\theta$$

$$= \int_{0}^{\pi}\int_{0}^{1}(-2r^4 \cos^{2}{\theta}\sin{\theta} + 5r^3 \sin^{2}{\theta} + r^3 \cos^{2}{\theta} - r^2\cos{\theta}) dr d\theta$$

$$= \int_{0}^{\pi}\left[\frac{-2}{5}r^5 \cos^{2}{\theta}\sin{\theta} ... ...rac{r^4}{4} \cos^{2}{\theta} - \frac{r^3}{3}\cos{\theta}\right]_{0}^{1} d\theta$$

$$= \int_{0}^{\pi}\left(\frac{-2}{5} \cos^{2}{\theta}\sin{\theta} + \... ...+ \frac{1}{4} \cos^{2}{\theta} - \frac{1}{3}\cos{\theta}\right]_{0}^{1} d\theta$$

ここで，

$$\int_{0}^{2\pi}\cos^{2}{\theta}\sin{\theta}d\theta = 0$$

$$\int_{0}^{2\pi}\cos^{2}{\theta} d\theta = \int_{0}^{2\pi}\sin^{2}{\theta}d\theta = \pi$$

に注意すると

$$\iint_{S}(xy,-2y,z-x) \cdot \hat{\bf n}dS = \frac{5\pi}{4} + \frac{\pi}{4} = \frac{3\pi}{2}$$