早稲田大学 創造理工学研究科 経営システム工学専攻 2021年8月実施 微积分

标签：

• Waseda-University
• Double-Integral
• Integration-by-Substitution
• Integration

Author

思齐塾, 祭音Myyura

Description

領域(domain)Dを以下のように設定する。

$$D = \{(x,y)| -1 \leq x + y \leq 1, -1 \leq x - y \leq 1\}$$

次の二重積分(double integral)を求めよ。

$$\iint_{D} \sqrt{1-(x+y)^2} dxdy$$

题目描述

定义区域

$$D=\{(x,y)\mid-1\leq x+y\leq1,\ -1\leq x-y\leq1\}.$$

求二重积分

$$\iint_D\sqrt{1-(x+y)^2}\,dx\,dy.$$

Kai

Let $u = x + y$ and $v = x - y$ . Then $x = \frac{u+v}{2}$ and $y = \frac{u-v}{2}$ . The Jacobian is given by

$$\frac{\partial(x, y)}{\partial(u, v)} = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} = \begin{vmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2} \end{vmatrix} = -\frac{1}{4} - \frac{1}{4} = -\frac{1}{2}$$

Therefore, $\left| \frac{\partial(x, y)}{\partial(u, v)} \right| = \frac{1}{2}$ . The region D is transformed into the region R such that $-1 \leq u \leq 1$ and $-1 \leq v \leq 1$ . Then, the double integral becomes

$$\iint_{D} \sqrt{1-(x+y)^2} dxdy = \iint_{R} \sqrt{1-u^2} \left| \frac{\partial(x, y)}{\partial(u, v)} \right| dudv = \int_{-1}^{1} \int_{-1}^{1} \sqrt{1-u^2} \cdot \frac{1}{2} dudv$$

$$= \frac{1}{2} \int_{-1}^{1} \sqrt{1-u^2} du \int_{-1}^{1} dv = \frac{1}{2} \cdot (2) \int_{-1}^{1} \sqrt{1-u^2} du = \int_{-1}^{1} \sqrt{1-u^2} du$$

Let $u = \sin\theta$ . Then $du = \cos\theta d\theta$ . When $u = -1$ , $\theta = -\frac{\pi}{2}$ . When $u = 1$ , $\theta = \frac{\pi}{2}$ .

$$\int_{-1}^{1} \sqrt{1-u^2} du = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{1-\sin^2\theta} \cos\theta d\theta = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2\theta d\theta = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1+\cos(2\theta)}{2} d\theta$$

$$= \frac{1}{2} \left[ \theta + \frac{\sin(2\theta)}{2} \right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \frac{1}{2} \left[ (\frac{\pi}{2} + 0) - (-\frac{\pi}{2} + 0) \right] = \frac{1}{2} (\pi) = \frac{\pi}{2}$$

Therefore, the double integral is $\frac{\pi}{2}$ .