7.3 解答

1. 積分の領域が円，楕円，ひし形などの場合は，変数変換を用いる．

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

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

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

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

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

$\displaystyle I$	$\displaystyle =$	$\displaystyle \iint_{\Omega} \log(x^2 + y^2) dxdy$
	$\displaystyle =$	$\displaystyle \iint_{\Gamma}\log(r^2) \vert J(r,\theta)\vert dr d\theta ({\rm where},\vert J(r,theta)\vert = r)$
	$\displaystyle =$	$\displaystyle \int_{0}^{2\pi}\int_{1}^{2}r\log{r^2} dr d\theta {\rm where} \... ...r^2 \log{r} - \int{r}dr\\ = r^2 \log{r} - \frac{r^2}{2} + c \end{array}\right.$
	$\displaystyle =$	$\displaystyle 2\pi[r^2\log{r} - \frac{r^2}{2}]_{1}^{2} = 2\pi(4\log{2} - 2 + \frac{1}{2})$
	$\displaystyle =$	$\displaystyle 2\pi(4\log{2} - \frac{3}{2}) = 8\pi\log{2} - 3\pi$

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

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

$\displaystyle I$	$\displaystyle =$	$\displaystyle \iint_{\Omega} e^{(y-x)/(y+x)} dxdy$
	$\displaystyle =$	$\displaystyle \iint_{\Gamma} e^{-v/u} \vert J(u,v)\vert du dv {\rm ただし}, \lef... ...\right\vert \vert\\ &= \vert-\frac{1}{2}\vert = \frac{1}{2} \end{array}\right.$
	$\displaystyle =$	$\displaystyle \frac{1}{2}\int_{0}^{1}\int_{-u}^{u}e^{-v/u}dvdu$
	$\displaystyle =$	$\displaystyle \frac{1}{2}\int_{0}^{1}[-ue^{-v/u}]_{-u}^{u} du$
	$\displaystyle =$	$\displaystyle \frac{1}{2}\int_{0}^{1}(-ue^{-1} + ue)du$
	$\displaystyle =$	$\displaystyle \frac{1}{2}[-\frac{u^2}{2e} + \frac{u^2 e}{2}]_{0}^{1}$
	$\displaystyle =$	$\displaystyle \frac{1}{4}(-\frac{1}{e} + e)$

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

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

$\displaystyle I$	$\displaystyle =$	$\displaystyle \iint_{\Omega} e^{x^2 + y^2} dxdy$
	$\displaystyle =$	$\displaystyle \iint_{\Gamma} e^{r^2} \vert J(r,\theta)\vert dr d\theta ({\rm where}, \vert J(r,theta)\vert = r)$
	$\displaystyle =$	$\displaystyle \int_{0}^{2\pi}\int_{1}^{2}re^{r^2} dr d\theta {\rm where} \lef... ...e^{t}}dt\\ = \frac{1}{e^{t}} + c\\ = \frac{1}{e^{r^2}} + c \end{array}\right.$
	$\displaystyle =$	$\displaystyle (\int_{0}^{2\pi} d\theta)[\frac{1}{2}{e^{r^2}}]_{1}^{2}$
	$\displaystyle =$	$\displaystyle 2\pi(\frac{1}{2}(e^4 - e)) = \pi(e^4 - e)$

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

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

$\displaystyle I$	$\displaystyle =$	$\displaystyle \iint_{\Omega} \sqrt{1 - x^2 - y^2} dxdy$
	$\displaystyle =$	$\displaystyle \iint_{\Gamma} \sqrt{1 - r^2} \vert J(r,\theta)\vert dr d\theta ({\rm where}, \vert J(r,theta)\vert = r)$
	$\displaystyle =$	$\displaystyle \int_{0}^{2\pi}d\theta \int_{0}^{1}\sqrt{1 - r^2}r dr {\rm wher... ...t^{\frac{3}{2}}\\ = -\frac{1}{3}(1 - r^2)^{\frac{3}{2}} + c \end{array}\right.$
	$\displaystyle =$	$\displaystyle 2\pi[-\frac{1}{3}(1 -r^2)^{\frac{3}{2}}]_{0}^{1}$
	$\displaystyle =$	$\displaystyle 2\pi(\frac{1}{3}) = \frac{2\pi}{3}$

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

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

$\displaystyle I$	$\displaystyle =$	$\displaystyle \iint_{\Omega} (1 - x - 2y) dxdy$
	$\displaystyle =$	$\displaystyle \iint_{\Gamma} (1 - r\cos{\theta} - 2r\sin{\theta}) \vert J(r,\theta)\vert dr d\theta ({\rm ただし}, \vert J(r,\theta)\vert = r)$
	$\displaystyle =$	$\displaystyle \int_{0}^{\frac{\pi}{2}} \int_{0}^{1}(1 - r\cos{\theta} - 2r\sin{\theta})r dr$
	$\displaystyle =$	$\displaystyle \int_{0}^{\frac{\pi}{2}}\int_{0}^{1}(r - r^2 \cos{\theta} - 2r^2\sin{\theta})dr d\theta$
	$\displaystyle =$	$\displaystyle \int_{0}^{\frac{\pi}{2}}([\frac{r^2}{2} - \frac{r^3 \cos{\theta}}{3} - \frac{2r^3 \sin{\theta}}{3}]_{0}^{1}) d\theta$
	$\displaystyle =$	$\displaystyle \int_{0}^{\frac{\pi}{2}}(\frac{1}{2} - \frac{\cos{\theta}}{3} - \frac{2\sin{\theta}}{3})d\theta$
	$\displaystyle =$	$\displaystyle [\frac{\theta}{2} - \frac{\sin{\theta}}{3} + \frac{2\cos{\theta}}{3}]_{0}^{\frac{\pi}{2}}$
	$\displaystyle =$	$\displaystyle \frac{\pi}{4} - \frac{1}{3}$

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

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

$\displaystyle x^2 + y^2 = (\frac{u+v}{2})^2 + (\frac{u-v}{2})^2 = \frac{1}{2}(u^2 + v^2)$

$\displaystyle I$	$\displaystyle =$	$\displaystyle \iint_{\Omega} (x^2 + y^2)e^{-x+y} dxdy$
	$\displaystyle =$	$\displaystyle \frac{1}{2}\iint_{\Gamma} (u^2 + v^2)e^{-v} \vert J(u,v)\vert du ... ...{array}\right\vert \vert\\ &= \vert-\frac{1}{2}\vert = \frac{1}{2} \end{array}$
	$\displaystyle =$	$\displaystyle \frac{1}{4}\int_{v=-1}^{1}\int_{u=-1}^{1}(u^2 + v^2)e^{-v}dudv$
	$\displaystyle =$	$\displaystyle \frac{1}{4}\int_{-1}^{1}[\frac{u^3}{3}e^{-v} + uv^2 e^{-v}]_{-1}^{1} dv$
	$\displaystyle =$	$\displaystyle \frac{2}{4}\int_{-1}^{1}(\frac{1}{3}e^{-v} + v^2e^{-v})dv$

$\displaystyle \int{v^2 e^{-v}}dv$	$\displaystyle =$	$\displaystyle \int{v^2 (-e^{-v})'}dv$
	$\displaystyle =$	$\displaystyle -v^2 e^{-v} + \int{e^{-v}(2v)}dv$
	$\displaystyle =$	$\displaystyle -v^2e^{-v} + 2\int{(-e^{-v})' v} dv$
	$\displaystyle =$	$\displaystyle -v^2 e^{-v} + 2[-e^{-v}v + \int{e^{-v}}dv]$
	$\displaystyle =$	$\displaystyle -v^2 e^{-v} - 2v e^{-v} -2e^{-v} + c$

$\displaystyle I$	$\displaystyle =$	$\displaystyle \frac{1}{2}[-\frac{1}{3} e^{-v} - v^2 e^{-v} - 2v e^{-v} -2e^{-v}]_{-1}^{1}$
	$\displaystyle =$	$\displaystyle \frac{1}{2}(-\frac{e^{-1}}{3} - e^{-1} -2e^{-1} -2e^{-1} -(-\frac{e}{3} - e + 2e - 2e))$
	$\displaystyle =$	$\displaystyle \frac{1}{2}(\frac{4e}{3} - \frac{16}{3e}) = \frac{2e}{3} - \frac{8}{3e}$