第8章 8.27

(1)

$(x-y)^2+(x+2y)^2$ $D:|x-y|\leq 2$ $|x+2y|\leq 1$

$|x-y|=|t|$ $|x+2y|=|k|$ とおくと $-2\leq t\leq 2$ $-1\leq k\leq 1$

$x=\frac{1}{3}k+\frac{2}{3}t \qquad y=\frac{1}{3}k-\frac{1}{3}t$

$\left|\frac{\partial(x,y)}{\partial(t,k)}\right| = \begin{vmatrix} \dfrac{2}{3} & \dfrac{1}{3} \\ -\dfrac{1}{3} & \dfrac{1}{3} \end{vmatrix} = \frac{1}{3}$

$\int_{-2}^{2}\int_{-1}^{1}(t^2+k^2)\frac{1}{3}\,dk\,dt = \int_{-2}^{2}\left[t^2k+\frac{1}{3}k^3\right]_{-1}^{1}\frac{1}{3}\,dt = \int_{-2}^{2}\frac{1}{3}\left(2t^2+\frac{2}{3}\right)dt$

$=\frac{1}{3}\left[\frac{2}{3}t^3+\frac{2}{3}t\right]_{-2}^{2} = \frac{40}{9}$

(2)

$x^2+y^2$ $D:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}\leq 1$

$x=ar\cos\theta$ $y=br\sin\theta$ とおくと $0\leq r\leq 1$ $0\leq\theta\leq 2\pi$

$\left|\frac{\partial(x,y)}{\partial(r,\theta)}\right| = \begin{vmatrix} a\cos\theta & -ar\sin\theta \\ b\sin\theta & br\cos\theta \end{vmatrix} = abr$

$\int_0^{2\pi}\int_0^1 (a^2r^2\cos^2\theta+b^2r^2\sin^2\theta)abr\,dr\,d\theta = \int_0^{2\pi} ab(a^2\cos^2\theta+b^2\sin^2\theta)\left[\frac{1}{4}r^4\right]_0^1 d\theta$

$= \int_0^{2\pi}\frac{1}{4}ab(a^2\cos^2\theta+b^2\sin^2\theta)\,d\theta = \int_0^{2\pi}\frac{1}{4}ab\left(a^2\frac{1+\cos2\theta}{2}+b^2\frac{1-\cos2\theta}{2}\right)d\theta$

$=\frac{1}{4}ab(\pi a^2+\pi b^2) = \frac{\pi}{4}ab(a^2+b^2)$

(3)

$(x+y)e^{-x+y} \qquad D:0\le x+y\le 1 \quad 0\le x-y\le 1$

$x+y=t$ $x-y=k$ とおくと $0\le t\le 1$ $0\le k\le 1$

$x=\frac{1}{2}(t+k) \qquad y=\frac{1}{2}(t-k)$

$\left|\frac{\partial(x,y)}{\partial(t,k)}\right|=\frac{1}{2}$

$\int_0^1\int_0^1 te^{-k}\cdot\frac{1}{2}\,dk\,dt=\int_0^1\frac{1}{2}\Big[-te^{-k}\Big]_0^1dt=\int_0^1\frac{1}{2}(-te^{-1}+t)\,dt=\frac{1}{2}\left[-\frac{1}{2}t^2e^{-1}+\frac{1}{2}t^2\right]_0^1=\frac{1}{4}\left(1-\frac{1}{e}\right)$

(4)

$(x+y)\sin(x+y) \qquad 0\le x-y\le \pi \quad 0\le x+y\le \pi$

$x-y=t$ $x+y=k$ とおくと $0\le t\le \pi$ $0\le k\le \pi$

$x=\frac{1}{2}(k+t) \qquad y=\frac{1}{2}(k-t) \qquad \left|\frac{\partial(x,y)}{\partial(t,k)}\right|=\frac{1}{2}$

$\int_0^\pi\int_0^\pi \frac{1}{2}t\sin k\,dk\,dt=\int_0^\pi\frac{1}{2}\Big[-t\cos k\Big]_0^\pi dt=\int_0^\pi t\,dt=\frac{\pi^2}{2}$