[math]2022年京都大学大学院理学研究科数学・数理解析学専攻数学系入試問題基礎科目1

math

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2022.10.09

Table of Contents

問題

$\mathbb{R}$の部分集合$D$を$$D = \left\{(x, y, z)\in \mathbb{R}\middle| \frac{1}{4}\leq x^2+y^2-2yz+4z^2\leq 1\right\}$$で定める。積分$$\iiint_{D}{\log{(x^2+y^2-2yz+4z^2)}dxdydz}$$の値を求めよ。

方針

変数変換してみる。

解答

$x = r\sin{\theta}\cos{\phi}, y-z = r\sin{\theta}\sin{\phi}, \sqrt{3}z = r\cos{\theta}\ \ (r\geq 0, 0\leq \theta\leq \pi, 0\leq \phi\leq 2\pi)$とする。このとき$x^2+y^2-2yz+4z^2 = r^2$である。$$\begin{eqnarray}J & = & \begin{vmatrix}\frac{dx}{dr} & \frac{dx}{d\theta} & \frac{dx}{d\phi}\\ \frac{dy}{dr} & \frac{dy}{d\theta} & \frac{dy}{d\phi}\\ \frac{dz}{dr} & \frac{dz}{d\theta} & \frac{dz}{d\phi}\end{vmatrix}\\ & = & \begin{vmatrix}\sin{\theta}\cos{\phi} & r\cos{\theta}\cos{\phi} & -r\sin{\theta}\sin{\phi}\\ \sin{\theta}\sin{\phi} + \frac{1}{\sqrt{3}}\cos{\theta} & r\cos{\theta}\sin{\phi} -\frac{r}{\sqrt{3}}\sin{\theta} & r\sin{\theta}\cos{\phi}\\ \frac{\cos{\theta}}{\sqrt{3}} & -\frac{r}{\sqrt{3}}\sin{\theta} & 0\end{vmatrix}\\ & = & \left|\frac{r^2}{\sqrt{3}}\sin{\theta}\cos^2{\theta}\cos^2{\phi}+\frac{r^2}{\sqrt{3}}\sin^2{\theta}\sin{\phi}\left(\sin{\theta}\sin{\phi}+\frac{1}{\sqrt{3}}\cos{\theta}\right)\right.\\ & & \left. + \frac{r}{\sqrt{3}}\sin{\theta}\cos{\theta}\sin{\phi}\left(r\cos{\theta}\sin{\phi}-\frac{r}{\sqrt{3}}\sin{\theta}\right)+\frac{r^2}{\sqrt{3}}\sin^3{\theta}\cos^2{\phi}\right|\\ & = & \left|\frac{r^2}{\sqrt{3}}\sin{\theta}\cos^2{\theta} + \frac{r^2}{\sqrt{3}}\sin^3{\theta}\right|\\ & = & \frac{r^2}{\sqrt{3}}\sin{\theta}\end{eqnarray}$$である。したがって、$$\begin{eqnarray}\iiint_{D}{\log{(x^2+y^2-2yz+4z^2)}dxdydz} & = & \iiint_{D}{\frac{r^2\sin{\theta}}{\sqrt{3}}\log{r^2}drd\theta d\phi}\\ & = & \frac{2}{\sqrt{3}}\iiint_{D}{r^2\sin{\theta}\log{r}drd\theta d\phi}\\ & = & \frac{2}{\sqrt{3}}\int_{0}^{\pi}{\sin{\theta}d\theta}\int_{0}^{2\pi}{d\phi}\int_{\frac{1}{2}}^{1}{r^2\log{r}dr}\\ & = & \frac{8\pi}{\sqrt{3}}\int_{\frac{1}{2}}^{1}{r^2\log{r}dr}\end{eqnarray}$$である。また、$$\begin{eqnarray}\int_{\frac{1}{2}}^{1}{r^2\log{r}dr} & = & \int_{\frac{1}{2}}^{1}{\left(\frac{r^3}{3}\right)^{\prime}\log{r}dr}\\ & = & \left[\frac{r^3}{3}\log{r}\right]_{\frac{1}{2}}^{1}-\frac{1}{3}\int_{\frac{1}{2}}^{1}{r^3\cdot \frac{1}{r}dr}\\ & = & -\frac{1}{3}\cdot \frac{1}{8}\log{\frac{1}{2}}-\frac{1}{3}\left[\frac{r^3}{3}\right]_{\frac{1}{2}}^{1}\\ & = & \frac{\log{2}}{24}-\frac{7}{72}\end{eqnarray}$$となる。よって、求める積分の値は$\displaystyle \frac{8\pi}{\sqrt{3}}\left(\frac{\log{2}}{24}-\frac{7}{72}\right) = \underline{\frac{\sqrt{3}\pi}{{27}}(3\log{2}-7)}$となる。