第8章 8.18

(1)

$x$ $D: 0 \le x+y \le 1 \quad 0 \le x-y \le 1$

$\int_0^{\frac{1}{2}}\int_0^x x\,dy\,dx + \int_{\frac{1}{2}}^1\int_0^{1-x} x\,dy\,dx + \int_0^{\frac{1}{2}}\int_{-x}^0 x\,dy\,dx + \int_{\frac{1}{2}}^1\int_{x-1}^0 x\,dy\,dx$

$= \int_0^{\frac{1}{2}} x^2\,dx + \int_{\frac{1}{2}}^1 x(1-x)\,dx + \int_0^{\frac{1}{2}} x^2\,dx + \int_{\frac{1}{2}}^1 x(1-x)\,dx$

$= \left[\frac{1}{3}x^3\right]_0^{\frac{1}{2}} + \left[\frac{1}{2}x^2 - \frac{1}{3}x^3\right]_{\frac{1}{2}}^1 + \left[\frac{1}{3}x^3\right]_0^{\frac{1}{2}} + \left[\frac{1}{2}x^2 - \frac{1}{3}x^3\right]_{\frac{1}{2}}^1 = \frac{1}{4}$

(2)

$y$ $D: x\ge0 \quad y\ge0 \quad \sqrt{\dfrac{x}{2}}+\sqrt{\dfrac{y}{3}} \le 1$

$\sqrt{\frac{x}{2}}+\sqrt{\frac{y}{3}}=1 \qquad \sqrt{\frac{y}{3}}=1-\sqrt{\frac{x}{2}} \qquad \frac{y}{3}=\left(1-\sqrt{\frac{x}{2}}\right)^2 \qquad y=3\left(1-\sqrt{\frac{x}{2}}\right)^2$

$\int_0^2\int_0^{3\left(1-\sqrt{\frac{x}{2}}\right)^2} y\,dy\,dx = \int_0^2 \frac{9}{2}\left(1-\sqrt{\frac{x}{2}}\right)^4 dx$

$\sqrt{\dfrac{x}{2}}=t$ とおくと $dx=4t\,dt$ $x:0\to2$ $t:0\to1$ これから

$=\int_0^1 \frac{9}{2}(1-t)^4 \cdot 4t\,dt = 18\left\{\left[t\cdot\left(-\frac{1}{5}\right)(1-t)^5\right]_0^1 + \int_0^1 \frac{1}{5}(1-t)^5\,dt\right\} = 18\left[-\frac{1}{30}(1-t)^6\right]_0^1 = \frac{3}{5}$

(3)

$\cos(x-3y)$ $D: 0\le x\le \dfrac{\pi}{2} \quad 0\le y \le \dfrac{\pi}{2}$

$\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}} \cos(x-3y)\,dx\,dy = \int_0^{\frac{\pi}{2}} \left[\sin(x-3y)\right]_0^{\frac{\pi}{2}} dy = \int_0^{\frac{\pi}{2}} \sin\left(\frac{\pi}{2}-3y\right) - \sin(-3y)\,dy$

$= \left[\frac{1}{3}\cos\left(\frac{\pi}{2}-3y\right) - \frac{1}{3}\cos(-3y)\right]_0^{\frac{\pi}{2}} = \frac{1}{3}(-1-0-0+1)=0$

(4)

$\tan^{-1}\dfrac{y}{x} \qquad D: 0 < y \leq x \leq 1$

$\int_0^1\int_0^x \tan^{-1}\frac{y}{x}\,dy\,dx = \int_0^1 \left\{ \left[y\tan^{-1}\frac{y}{x}\right]_0^x - \int_0^x \frac{xy}{x^2+y^2}\,dy \right\} dx$

$= \int_0^1 \frac{\pi}{4}x - \frac{1}{2}x\log 2x^2 + \frac{1}{2}x\log x^2\,dx = \int_0^1 \frac{\pi}{4}x - \frac{1}{2}x\log 2\,dx = \left[\frac{\pi}{8}x^2 - \frac{1}{4}x^2\log 2\right]_0^1$

$= \frac{\pi}{8} - \frac{1}{4}\log 2$

(5)

$\frac{x+y}{x^2+y^2} \qquad D: 0 \leq x \leq 1 \quad 0 \leq y \leq x$

$\int_0^1\int_0^x \frac{x+y}{x^2+y^2}\,dy\,dx = \int_0^1 \left[\tan^{-1}\frac{y}{x}\right]_0^x + \left[\frac{1}{2}\log(x^2+y^2)\right]_0^x dx = \int_0^1 \frac{\pi}{4} + \frac{1}{2}\log 2\,dx = \frac{\pi}{4} + \frac{1}{2}\log 2$