第8章 8.26

(1)

$z = \sqrt{a^2 - x^2 - y^2}$

$z_x = \frac{-x}{\sqrt{a^2-x^2-y^2}} \qquad z_y = \frac{-y}{\sqrt{a^2-x^2-y^2}}$

$z=b$ として

$x^2+y^2+b^2=a^2$

$x=r\cos\theta$ $y=r\sin\theta$ とおくと $0 \le r \le \sqrt{a^2-b^2}$ $0 \le \theta \le 2\pi$

$\int_0^{2\pi}\int_0^{\sqrt{a^2-b^2}} \sqrt{\frac{r^2}{a^2-r^2}+1}\cdot r\,dr\,d\theta = \int_0^{2\pi}\int_0^{\sqrt{a^2-b^2}} r\sqrt{\frac{a^2}{a^2-r^2}}\,d\theta = \int_0^{2\pi}\left[a\left(-\sqrt{a^2-r^2}\right)\right]_0^{\sqrt{a^2-b^2}}d\theta$

$= \int_0^{2\pi} -ab+a^2\,d\theta = 2\pi a(a-b)$

(2)

$z = \tan^{-1}\dfrac{y}{x}$

$z_x = \frac{y}{x^2+y^2} \qquad z_y = \frac{x}{x^2+y^2}$

$x=r\cos\theta$ $y=r\sin\theta$ とおくと $0 \le r \le 1$ $0 \le \theta \le \dfrac{\pi}{2}$

$\int_0^{\frac{\pi}{2}}\int_0^{1} \sqrt{\frac{\sin^2\theta}{r^2}+\frac{\cos^2\theta}{r^2}+1}\cdot r\,dr\,d\theta = \int_0^{\frac{\pi}{2}}\int_0^{1} \sqrt{r^2+1}\,dr\,d\theta$

$= \int_0^{\frac{\pi}{2}}\left[\frac{1}{2}\left\{r\sqrt{r^2+1}+\log\left|r+\sqrt{r^2+1}\right|\right\}\right]_0^1 d\theta = \int_0^{\frac{\pi}{2}} \frac{1}{2}\{\sqrt{2}+\log(1+\sqrt{2})\}\,d\theta$

$= \frac{\pi}{4}\left(\sqrt{2}+\log(1+\sqrt{2})\right)$