7.5 2重積分の応用

1. (a)

$\includegraphics[width=5cm]{CALCFIG1/cycloid.eps}$

まず，曲線 $r = 1 - \cos{\theta}$ の $r = 0$

の点を求める．すると， $1 - \cos{\theta} = 0$ より， $\theta = 0, 2\pi$ が求まる．これより，極座標を用いて領域を表すと，

$\displaystyle \Gamma = \left\{(r,\theta):0 \leq \theta \leq 2\pi, 0 \leq r \leq 1-\cos{\theta}\right\}$

よって，求める面積は

$\displaystyle I$	$\displaystyle =$	$\displaystyle \iint_{\Gamma}rdr d\theta$
	$\displaystyle =$	$\displaystyle \int_{0}^{2\pi}\int_{0}^{1-\cos{\theta}}rdr d\theta$
	$\displaystyle =$	$\displaystyle \int_{0}^{2\pi}\frac{r^2}{2}\mid_0^{1-\cos{\theta}}d\theta$
	$\displaystyle =$	$\displaystyle \int_{0}^{2\pi}\frac{(1-\cos{\theta})^2}{2}d\theta = \frac{1}{2}\int_{0}^{2\pi}(1-2\cos{\theta}+\cos^{2}{\theta})d\theta$
	$\displaystyle =$	$\displaystyle \frac{1}{2}\int_{0}^{2\pi}(1 - 2\cos{\theta} + \frac{1 + \cos{2\t... ...})d\theta = \frac{1}{2}\left[\frac{3\theta}{2}\right]_0^{2\pi} = \frac{3\pi}{2}$

(b)

$\includegraphics[width=5cm]{CALCFIG1/ellipse-right.eps}$

$r = 3\cos{\theta}$ と $r = \frac{3}{2}$ の交点を求めると， $3\cos{\theta} = \frac{3}{2}$ より， $\cos{\theta} = \frac{1}{2}$ .　したがって， $\theta = \pm \frac{\pi}{3}$ . これより求める領域は

$\displaystyle \Gamma = \{(r,\theta):-\frac{\pi}{3} \leq \theta \leq \frac{\pi}{3}, \frac{3}{2} \leq r \leq 3\cos{\theta}\}$

求める面積は

$\displaystyle I$	$\displaystyle =$	$\displaystyle \iint_{\Gamma}rdr d\theta$
	$\displaystyle =$	$\displaystyle 2\int_{0}^{\pi/3}\int_{\frac{3}{2}}^{3\cos{\theta}}rdrd\theta = \int_{0}^{\pi/3}r^2\mid_{\frac{3}{2}}^{3\cos{\theta}}d\theta$
	$\displaystyle =$	$\displaystyle \int_{0}^{\pi/3}(9\cos^{2}{\theta}-\frac{9}{4})d\theta = 9\int_{0}^{\pi/3}(\frac{1+\cos{2\theta}}{2} - \frac{1}{4})d\theta$
	$\displaystyle =$	$\displaystyle 9(\frac{1}{4}\theta + \frac{\sin{2\theta}}{4} = 9(\frac{\pi}{12}+\frac{\sqrt{3}}{8})$

(c)

$\includegraphics[width=5cm]{CALCFIG1/ellipse-ellipse.eps}$

$r = 3\cos{\theta}$ と $r = \cos{\theta}$ の交点を求めると， $3\cos{\theta} = \cos{\theta}$ より， $2\cos{\theta} = 0$ . よって， $theta = \pm \frac{\pi}{2}$ .　これより求める領域は

$\displaystyle \Gamma = \{(r,\theta):-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}, \cos{\theta} \leq r \leq 3\cos{\theta}\}$

求める面積は

$\displaystyle I$	$\displaystyle =$	$\displaystyle \iint_{\Gamma}rdr d\theta$
	$\displaystyle =$	$\displaystyle 2\int_{0}^{\pi/2}\int_{\cos{\theta}}^{3\cos{\theta}}rdrd\theta = \int_{0}^{\pi/2}r^2 \mid_{\cos{\theta}}^{3\cos{\theta}} d\theta$
	$\displaystyle =$	$\displaystyle \int_{0}^{\pi/2}(9\cos^{2}{\theta} - \cos^{2}{\theta})d\theta = 8\int_{0}^{\pi/2}\cos^{2}{\theta}d\theta$
	$\displaystyle =$	$\displaystyle 4\int_{0}^{\pi/2}(1 + \cos{2\theta})d\theta = 4(\theta + \frac{\sin{2\theta}}{2}\mid_{0}^{\pi/2} = 2\pi$

2. (a)

$\includegraphics[width=5cm]{CALCFIG1/parabola.eps}$

の曲面積で， $0 \leq x^2 + y^2 \leq 1$ の部分であるから，Vertical simpleを用いて，

$\displaystyle \Omega = \left\{(x,y):-1 \leq x \leq 1, -\sqrt{1-x^2} \leq y \leq \sqrt{1-x^2}\right\}$

曲面積は $S = \iint_{\Omega}\sqrt{z_x^2 + z_y^2 + 1}dxdy$ で与えられるので， $z_x, z_y$

を求める． $z_x = \frac{x}{\sqrt{x^2 + y^2}}, z_y = \frac{y}{\sqrt{x^2 + y^2}}$ より，求める曲面積は

$\displaystyle S$	$\displaystyle =$	$\displaystyle \int_{-1}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\sqrt{2}dydx$
	$\displaystyle =$	$\displaystyle 4\sqrt{2}\int_{0}^{1}\sqrt{1-x^2}dx$

ここで， $x = \sin{t}, t = \sin^{-1}{x}$ とおくと, $\sqrt{1-x^2} = \cos{t}, dx = \cos{t}dt$ より，

$\displaystyle S$	$\displaystyle =$	$\displaystyle 4\sqrt{2}\int_{0}^{\pi/2}\cos^{2}{t}dt = 2\sqrt{2}\int_{0}^{\pi/2}(1+\cos{2t})dt$
	$\displaystyle =$	$\displaystyle 2\sqrt{2}(t + \frac{\sin{2t}}{2}\mid_{0}^{\pi/2} = \sqrt{2}\pi$

(b)

$x+y+z = 2$ の曲面積で， $x \geq 0, y \geq 0, z \geq 0$ の部分であるから， $z = 2-x-y \geq 0$ より， $x+y \leq 2$ となる．この領域をVertical simpleで求めると，

$\displaystyle \Omega = \left\{(x,y):0 \leq x \leq 2, 0 \leq y \leq 2-x\right\}$

曲面積は $S = \iint_{\Omega}\sqrt{z_x^2 + z_y^2 + 1}dxdy$ で与えられるので， $z_x, z_y$

を求める．

より，求める曲面積は

$\displaystyle S$	$\displaystyle =$	$\displaystyle \iint_{\Omega}\sqrt{3}dxdy$
	$\displaystyle =$	$\displaystyle \sqrt{3}\int_{0}^{2}\int_{0}^{2-x}dydx = \sqrt{3}\int_{0}^{2}y\mid_{0}^{2-x}dx$
	$\displaystyle =$	$\displaystyle \sqrt{3}\int_{0}^{2}(2-x)dx = \sqrt{3}(2x-\frac{x^2}{2}\mid_{0}^{2} = 2\sqrt{3}$

(c)

$z = xy$ の曲面積で， $x^2 + y^2 \leq a^2$ の部分であるから，この領域をVertical simpleで求めると，

$\displaystyle \Omega = \left\{(x,y):-a \leq x \leq a, -\sqrt{a^2 -x^2} \leq y \leq \sqrt{a^2 - x^2}\right\}$

曲面積は $S = \iint_{\Omega}\sqrt{z_x^2 + z_y^2 + 1}dxdy$ で与えられるので， $z_x, z_y$

を求める．

より，求める曲面積は $\iint_{\Omega}\sqrt{z_x^2 + z_y^2 + 1}dxdy$ . ここで，極座標変換を用いると， $x = r\cos{\theta}, y = r\sin{\theta}$ より，

$\displaystyle \Gamma = \left\{(r,\theta):0 \leq \theta \leq 2\pi, 0 \leq r \leq a\right\}$

これより，

$\displaystyle S$	$\displaystyle =$	$\displaystyle \int_{0}^{2\pi}\int_{0}^{a}\sqrt{r^2 + 1}rdrd\theta$
	$\displaystyle =$	$\displaystyle \frac{1}{3}\int_{0}^{2\pi}(r^2 + 1)^{\frac{3}{2}}\mid_{0}^{a} d\theta = \frac{1}{3}\int_{0}^{2\pi}(a^2 + 1)^{\frac{3}{2}} - 1)d\theta$
	$\displaystyle =$	$\displaystyle \frac{2\pi}{3}((a^2+1)^{\frac{3}{2}}-1)$

3. (a) $z = x^2 + y^2$ と $z = 2y$ で囲まれた部分の体積を求めるには，まず，交線を求める必要がある． $z = x^2 + y^2 = 2y$ より， $x^2 + (y-1)^2 = 1$ . これより，縦線集合を用いると

$\displaystyle \Omega = \left\{(x,y):-1 \leq x \leq 1, 1-\sqrt{1-x^2} \leq y \leq 1+\sqrt{1-x^2}\right\}$

ここで，極座標変換を用いると， $x^2 + y^2 = 2y \Rightarrow r^2 = 2r\sin{\theta}$ より，

$\displaystyle \Gamma = \left\{(r,\theta):0 \leq \theta \leq \pi, 0 \leq y \leq 2\sin{\theta}\right\}$

これより，求める体積は

$\displaystyle V$	$\displaystyle =$	$\displaystyle \iint_{\Omega}(2y - (x^2 + y^2))dxdy$
	$\displaystyle =$	$\displaystyle \iint_{\Gamma}(2r\sin{\theta} - r^2)rdrd\theta = 2\int_{0}^{\pi/2}\int_{r=0}^{2\sin{\theta}}(2r^2 \sin{\theta}-r^3)dr d\theta$
	$\displaystyle =$	$\displaystyle 2\int_{0}^{\pi/2}(\frac{2r^3}{3}\sin{\theta} - \frac{r^4}{4}\mid_... ...{0}^{\pi/2}(\frac{16\sin^{4}{\theta}}{3} - \frac{16\sin^{4}{\theta}}{4})d\theta$
	$\displaystyle =$	$\displaystyle \frac{8}{3}\int_{0}^{\pi/2}\sin^{4}{\theta}d\theta = \frac{8}{3}\frac{3 \cdot 1}{4 \cdot 2}\frac{\pi}{2} = \frac{\pi}{2}$

(b) 放物面 $z = x^2 + y^2$ と円柱 $x^2 + y^2 \leq 1$ で囲まれた部分の体積を求めるには，まず，交線を求める必要がある． $z = x^2 + y^2 = 1$ より， $x^2 + y^2 = 1$ . これより，縦線集合を用いると

$\displaystyle \Omega = \left\{(x,y):-1 \leq x \leq 1, -\sqrt{1-x^2} \leq y \leq \sqrt{1-x^2}\right\}$

ここで，極座標変換を用いると， $x^2 + y^2 = 1 \Rightarrow r^2 = 1$ より，

$\displaystyle \Gamma = \left\{(r,\theta):0 \leq \theta \leq 2\pi, 0 \leq y \leq 1\right\}$

これより，求める体積は

$\displaystyle V$	$\displaystyle =$	$\displaystyle \iint_{\Omega} (x^2 + y^2) dxdy$
	$\displaystyle =$	$\displaystyle \iint_{\Gamma}r^2rdrd\theta = \int_{0}^{2\pi}\int_{0}^{1}r^3 dr d\theta$
	$\displaystyle =$	$\displaystyle \int_{0}^{2\pi}\frac{r^4}{4}\mid_{0}^{1}d\theta = \frac{1}{4}\theta\mid_{0}^{2\pi} = \frac{\pi}{2}$