7.3 変数変換

(a)

$\displaystyle \Omega = \{(x,y) : x^2 + y^2 \leq 4\}$

より極座標を用いる. $x = r\cos{\theta}, y = r\sin{\theta}$ より $x^2 + y^2 = r^2 \leq 4$ ．よって $\Omega$ は

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

に移るので，

$\displaystyle \iint_{\Omega} (x^2 + y^2) dxdy$	$\displaystyle =$	$\displaystyle \iint_{\Gamma}r^2 \vert J(r,\theta)\vert dr d\theta ({\rm where}, \vert J(r,theta)\vert = r)$
	$\displaystyle =$	$\displaystyle \int_{0}^{2\pi}\int_{0}^{2}r^3 dr d\theta = (\int_{0}^{2\pi}d\theta) (\int_{0}^{2}r^{3}dr)$
	$\displaystyle =$	$\displaystyle 2\pi ([\frac{r^4}{4}]_{0}^{2} = 2\pi \cdot 4 = 8\pi$

(b)

$\displaystyle \Omega = \{(x,y) : 1 \leq x^2 + y^2 \leq 4 y \geq 0\}$

より極座標を用いる． $x = r\cos{\theta}, y = r\sin{\theta}$ より $1 \leq r^2 \leq 4$ ．また， $y = r\cos{\theta} \geq 0$ より， $0 \leq \theta \leq \pi$ . よって $\Omega$ は

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

に移るので，

$\displaystyle I$	$\displaystyle =$	$\displaystyle \iint_{\Omega} \frac{1}{x^2 + y^2} dxdy$
	$\displaystyle =$	$\displaystyle \iint_{\Gamma}\frac{1}{r^2} \vert J(r,\theta)\vert dr d\theta ({\rm where},\vert J(r,theta)\vert = r)$
	$\displaystyle =$	$\displaystyle \int_{0}^{\pi}\int_{1}^{2}\frac{1}{r} dr d\theta$
	$\displaystyle =$	$\displaystyle \int_{0}^{\pi} d\theta [\log{r}]_{1}^{2} = \pi \log{2}$

(c)

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

より極座標を用いる. $x = r\cos{\theta}, y = r\sin{\theta}$ より $x^2 + y^2 = r^2 \leq 1$ ．よって $\Omega$ は

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

に移るので，

$\displaystyle \iint_{\Omega} y^2 dxdy$	$\displaystyle =$	$\displaystyle \iint_{\Gamma}r^2 \sin^{2}{\theta} \vert J(r,\theta)\vert dr d\theta ({\rm where}, \vert J(r,\theta)\vert = r)$
	$\displaystyle =$	$\displaystyle \int_{0}^{2\pi}\int_{0}^{1}r^3 \sin^{2}{\theta} dr d\theta$
	$\displaystyle =$	$\displaystyle \int_{0}^{2\pi}\left[\frac{r^4}{4} \sin^{2}{\theta}\right]_{0}^{1}\; d\theta$
	$\displaystyle =$	$\displaystyle \int_{0}^{2\pi}\frac{1}{4}\sin^{2}{\theta}\; d\theta$
	$\displaystyle =$	$\displaystyle \frac{1}{4}\int_{0}^{2\pi}\frac{1 - \cos{2\theta}}{2}d\theta = \frac{1}{4}\left[\frac{\theta}{2} - \frac{\sin{2\theta}}{4}\right]_{0}^{2\pi}$
	$\displaystyle =$	$\displaystyle \frac{1}{4}\cdot \frac{2\pi}{2} = \frac{\pi}{4}$

(d)

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

は， $y \geq -x, y \leq -x + 1, y \geq x - 1, y \leq x - 1$ で囲まれた領域である．

$\begin{figure}\includegraphics[width=5cm]{CALCFIG/ren7-3-1d.eps} \end{figure}$

ここで， $u = x + y$ , $v = x - y$ とおくと， $0 \leq x + y \leq 1$ より $0 \leq u \leq 1$ に移り， $-1 \leq x-y \leq 1$ より $-1 \leq v \leq 1$ に移る．したがって， $\Omega$ は

$\displaystyle \Gamma = \{(u,v) : 0 \leq u \leq 1, -1 \leq v \leq 1\}$

に移る．また，

と

を

と

について解くと，

$\displaystyle x = \frac{u+v}{2}, y = \frac{u - v}{2}$

これよりJacobianは，

$\displaystyle J(u,v) = \frac{\partial(x,y)}{\partial(u,v)} = \left\vert\begin{a... ...ac{1}{2}\\ \frac{1}{2} & -\frac{1}{2} \end{array}\right\vert = -\frac{1}{2}$

よって，

$\displaystyle \iint_{\Omega} （x＋y)\; dxdy$	$\displaystyle =$	$\displaystyle \iint_{\Gamma} u \vert J(u,v)\vert \; du dv$
	$\displaystyle =$	$\displaystyle \frac{1}{2}\int_{0}^{1}\int_{-1}^{1}udvdu$
	$\displaystyle =$	$\displaystyle \frac{1}{2}\int_{0}^{1}\left[\frac{u^2}{2}\right]_{-1}^{1} du$
	$\displaystyle =$	$\displaystyle \frac{1}{2}\int_{0}^{1}du = \frac{1}{2}$

(e)

$\displaystyle \Omega = \{(x,y) : x^2 + y^2 \leq 2x\}$

は，中心が

で半径1の円より，極座標を用いる． $x = r\cos{\theta}, y = r\sin{\theta}$ より $r^2 \leq 2r\cos{\theta}$ ．よって， $r \leq 2\cos{\theta}$ ．次に， $\theta$ の範囲を求める． $r = 2\cos{\theta}$ において， $r = 0$

となる $\theta$ を求めると， $\cos{\theta} = 0$ より， $\theta = -\pi/2, \pi/2$ となる．よって $\Omega$ は

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

に移るので，

$\displaystyle I$	$\displaystyle =$	$\displaystyle \iint_{\Omega} \sqrt{4 - x^2 - y^2} dxdy$
	$\displaystyle =$	$\displaystyle \iint_{\Gamma} \sqrt{4 - r^2} \vert J(r,\theta)\vert dr d\theta$
	$\displaystyle =$	$\displaystyle \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}d\theta \int_{0}^{2\cos{\theta}}\sqrt{4 - r^2}r dr$

ここで， $t = 4 - r^2, dt = -2rdr$ より， $\begin{displaymath}\begin{array}{l\vert l} r & 0 \rightarrow 2\cos{\theta}\ \hline t & 4 \rightarrow 4 - 4\cos^{2}{\theta} \end{array}\end{displaymath}$

したがって，

$\displaystyle I$	$\displaystyle =$	$\displaystyle \int_{-\pi/2}^{\pi/2} \int_{r=0}^{2\cos{\theta}} \sqrt{4 - r^2} r\; dr d\theta$
	$\displaystyle =$	$\displaystyle \int_{-\pi/2}^{\pi/2} \int_{t = 4}^{4\sin^{2}{\theta}} \sqrt{t} (-\frac{dt}{2})\;d\theta$
	$\displaystyle =$	$\displaystyle \int_{-\pi/2}^{\pi/2} -\frac{1}{2}\cdot \frac{2}{3}[t^3/2]_{t=4}^{4\sin^{2}{\theta}}\;d\theta$
	$\displaystyle =$	$\displaystyle -\frac{1}{3}\int_{-\pi/2}^{\pi/2}(8\sin^{3}{\theta} - 8)\; d\theta$
	$\displaystyle =$	$\displaystyle \frac{8}{3}\int_{-\pi/2}^{\pi/2}d\theta = \frac{16}{3}\cdot \frac{\pi}{2} = \frac{8\pi}{3}$

2. $u = x - y, v = x + y$ とおくと，領域 $\Omega$ は領域

$\displaystyle \Gamma = \{(u,v): 0 \leq u \leq 1, 0 \leq v \leq 1\}$

に移る．また， $x = \frac{u+v}{2}, y = \frac{v - u}{2}$ より，

$\displaystyle J(u,v) = \left\vert\begin{array}{cc} \frac{1}{2} & \frac{1}{2}\\ -\frac{1}{2} & \frac{1}{2} \end{array}\right\vert = \frac{1}{2}$

したがって，

$\displaystyle \iint_{\Omega}(2x+3y)\;dxdy$	$\displaystyle =$	$\displaystyle \iint_{\Gamma}(\frac{-u+5v}{2}(\frac{1}{2})\;dv du$
	$\displaystyle =$	$\displaystyle \int_{u=0}^{1}\int_{v=0}^{1}\left(\frac{-u+5v}{4}\right)\; dvdu$
	$\displaystyle =$	$\displaystyle \frac{1}{4}\left[-uv + \frac{5v^2}{2}\right]_{0}^{1}du$
	$\displaystyle =$	$\displaystyle \frac{1}{4}\int_{0}^{1}(-u + \frac{5}{2})du = \frac{1}{4}\left[-\frac{u^2}{2} + \frac{5u}{2}\right]_{0}^{1}$
	$\displaystyle =$	$\displaystyle \frac{1}{4}(-\frac{1}{2} + \frac{5}{2}) = \frac{1}{4}(\frac{4}{2}) = \frac{1}{2}$

3. $u = x-y, v = x + 2y$ とおくと，領域 $\Omega$ は領域

$\displaystyle \Gamma = \{(u,v): 0 \leq u \leq 1, 0 \leq v \leq 1\}$

に移る．また， $x = \frac{2u+v}{3}, y = \frac{v - u}{3}$ より，

$\displaystyle J(u,v) = \left\vert\begin{array}{cc} \frac{2}{3} & \frac{1}{3}\\ -\frac{1}{3} & \frac{1}{3} \end{array}\right\vert = \frac{1}{3}$

したがって，

$\displaystyle \iint_{\Omega}2x\;dxdy$	$\displaystyle =$	$\displaystyle \iint_{\Gamma}(\frac{4u + 2v}{3}(\frac{1}{3})\;dv du$
	$\displaystyle =$	$\displaystyle \int_{u=0}^{1}\int_{v=0}^{1}\left(\frac{4u+2v}{9}\right)\; dvdu$
	$\displaystyle =$	$\displaystyle \frac{2}{9}\left[2uv + \frac{v^2}{2}\right]_{0}^{1}du$
	$\displaystyle =$	$\displaystyle \frac{2}{9}\int_{0}^{1}(\frac{1}{2} + 2u)du = \frac{2}{9}\left[\frac{u}{2} + u^2 \right]_{0}^{1}$
	$\displaystyle =$	$\displaystyle \frac{2}{9}(\frac{1}{2} + 1) = \frac{1}{3}$