第10章 10.12

(1)

$y'=2ax$ $\qquad y'=\dfrac{1}{x}$ 傾きが等しいので

$2ax=\dfrac{1}{x}$ $\qquad a=\dfrac{1}{2x^2}$ これを元の式に代入して $\qquad y=\dfrac{1}{2}$ これより $\qquad x=\sqrt{e}$

よって $\qquad a=\dfrac{1}{2e}$ $\qquad (x,y)=\left(\sqrt{e},\dfrac{1}{2}\right)$

(2)

$S=\int_0^{\sqrt{e}}\frac{x^2}{2e}\,dx-\int_1^{\sqrt{e}}\log x\,dx=\left[\frac{1}{6e}x^3\right]_0^{\sqrt{e}}-[x\log x-x]_1^{\sqrt{e}}=\frac{\sqrt{e}}{6}-\frac{\sqrt{e}}{2}+\sqrt{e}-1=\frac{2}{3}\sqrt{e}-1$

(3)

$V=\pi\int_0^{\sqrt{e}}\left(\frac{x^2}{2e}\right)^2dx-\pi\int_1^{\sqrt{e}}(\log x)^2dx=\pi\left[\frac{1}{20e^2}x^5\right]_0^{\sqrt{e}}-\pi\left\{[\log x(x\log x-x)]_1^{\sqrt{e}}-\int_1^{\sqrt{e}}\frac{1}{x}(x\log x-x)dx\right\}$

$=\pi\frac{\sqrt{e}}{20}-\pi\left\{\frac{1}{2}\left(\frac{\sqrt{e}}{2}-\sqrt{e}\right)-\int_1^{\sqrt{e}}(\log x-1)dx\right\}=\pi\frac{\sqrt{e}}{20}-\pi\left(-\frac{\sqrt{e}}{4}-[x\log x-x-x]_1^{\sqrt{e}}\right)$

$=\pi\frac{\sqrt{e}}{20}+\pi\frac{\sqrt{e}}{4}+\pi\left(\frac{\sqrt{e}}{2}-2\sqrt{e}+2\right)=\pi\left(2-\frac{6}{5}\sqrt{e}\right)$