Section01_定义与性质

背景

1. 已知密度函数 $\rho(x,y)$ 求钢板的质量
   1. 将 $\mathbb{D}$ 划分为 $\Delta \sigma_{1},\Delta \sigma_{2},\cdots,\Delta\sigma_{n}$
   2. $\forall\ (\zeta_{i},\eta_{i})\in \Delta\sigma_{i}\quad (i = 1,2,\cdots, n)$
      1. $\Delta m\approx \rho(\zeta_{i},\eta_{i})\Delta \sigma_{i}$
      2. $m \approx \sum_{i=1}^{n}\rho(\zeta_{i},\eta_{i})\Delta \sigma_{i}$
   3. $\lambda$ 为 $\Delta \sigma_{1},\Delta \sigma_{2},\cdots,\Delta\sigma_{n}$ 直径的最大者
      • $m = \lim\limits_{\lambda\to 0}\sum_{i=1}^{n}\rho(\zeta_{i},\eta_{i})\Delta \sigma_{i}$
2. $\Sigma:\ z=f(x,y)\ge 0, (x,y)\in\mathbb{D}$ 求体积 $V$
   1. 将 $\mathbb{D}$ 划分为 $\Delta \sigma_{1},\Delta \sigma_{2},\cdots,\Delta\sigma_{n}$
   2. $\forall\ (\zeta_{i},\eta_{i})\in \Delta\sigma_{i}\quad (i = 1,2,\cdots, n)$
      1. $\Delta V\approx f(\zeta_{i},\eta_{i})\Delta \sigma_{i}$
      2. $V \approx \sum_{i=1}^{n}f(\zeta_{i},\eta_{i})\Delta \sigma_{i}$
   3. $\lambda$ 为 $\Delta \sigma_{1},\Delta \sigma_{2},\cdots,\Delta\sigma_{n}$ 直径的最大者
      • $V = \lim\limits_{\lambda\to 0}\sum_{i=1}^{n}f(\zeta_{i},\eta_{i})\Delta \sigma_{i}$

定义

• $\mathbb{D}$ 为 $xoy$ 平面上的有界闭区域，$f(x,y)$ 在 $\mathbb{D}$ 上有界
  1. 将 $\mathbb{D}$ 划分为 $\Delta \sigma_{1},\Delta \sigma_{2},\cdots,\Delta\sigma_{n}$
  2. $\forall\ (\zeta_{i},\eta_{i})\in \Delta\sigma_{i}\quad (i = 1,2,\cdots, n)$ 作 $\sum_{i=1}^{n}f(\zeta_{i},\eta_{i})\Delta \sigma_{i}$
  3. $\lambda$ 为 $\Delta \sigma_{1},\Delta \sigma_{2},\cdots,\Delta\sigma_{n}$ 直径的最大者
• 若 $\lim\limits_{\lambda\to 0}\sum_{i=1}^{n} f(\zeta_{i},\eta_{i})\Delta \sigma_{i}\ \exists$，则称此极限为 $f(x,y)$ 在 $\mathbb{D}$ 上的二重积分，记为 $\displaystyle \iint\limits_{\mathbb{D}}f(x,y)\cdot d\sigma$，即 $$\iint\limits_{\mathbb{D}} f(x,y)\cdot d\sigma \triangleq \lim_{\lambda\to 0}\sum_{i=1}^{n}f(\zeta_{i},\eta_{i})\cdot \Delta \sigma_{i}$$

性质

1. $\displaystyle \iint\limits_{\mathbb{D}}[f(x,y)\pm g(x,y)]\cdot d\sigma = \iint\limits_{\mathbb{D}}f(x,y)\cdot d\sigma\pm\iint\limits_{\mathbb{D}}g(x,y)\cdot d\sigma$
2. $\displaystyle \iint\limits_{\mathbb{D}}kf(x,y)\cdot d\sigma = k\iint\limits_{\mathbb{D}}f(x,y)\cdot d\sigma$
3. $\displaystyle \iint\limits_{\mathbb{D}}f(x,y)\cdot d\sigma = \iint\limits_{\mathbb{D}_{1}}f(x,y)\cdot d\sigma+\iint\limits_{\mathbb{D}_{2}}f(x,y)\cdot d\sigma\quad (\mathbb{D} = \mathbb{D}_{1}+\mathbb{D}_{2})$
4. $\displaystyle \iint\limits_{D} 1\cdot d\sigma = A\quad A\text{为}\mathbb{D}\text{的面积}$
5. 重点 积分中值定理：$\mathbb{D}$ 为有界闭区间，$f(x,y)$ 在 $\mathbb{D}$ 上连续，则 $\exists\ (\zeta,\eta)\in\mathbb{D}$，使 $$\iint\limits_{\mathbb{D}} f(z,y)\cdot d\sigma = Af(\zeta,\eta)$$
   • 例 $\mathbb{D}:\ x^{2} + 4y^{2} \le t^{2}\quad (t^{2}> 0)$，求 $\displaystyle \lim_{t\to 0}\frac{\iint\limits_{\mathbb{D}}e^{x^{2}}\cos 2y\cdot d\sigma}{t^{2}}$ $$\begin{array}{ll} & \displaystyle x^{2}+4y^{2} = t^{2} \Rightarrow \frac{x^{2}}{t^{2}} + \frac{y^{2}}{(\frac{t}{2})^{2}} = 1 \\ \because & \displaystyle \exists\ (\zeta,\eta)\in \mathbb{D},\ \iint\limits_{\mathbb{D}}e^{x^{2}} \cos 2y\cdot d\sigma = \frac{t^{2}}{2}\pi \cdot e^{\zeta^{2}}\cos 2\eta \\ \therefore & \displaystyle \text{原式} = \lim_{t\to 0} \frac{\frac{t^{2}}{2}\cdot e^{\zeta^{2}}\cos 2\eta}{t^{2}} = \frac{1}{2}\lim_{t\to 0}e^{\zeta^{2}}\cos 2\eta \\ \because & \displaystyle \lim_{t\to 0} \mathbb{D} = (0,0) \\ \therefore & \displaystyle \frac{1}{2}\lim_{t\to 0}e^{\zeta^{2}}\cos 2\eta = \frac{1}{2} \\ \end{array}$$
6. 对称性
   1. $\mathbb{D}$ 关于左右对称（关于 $y$ 轴或关于 $x$ 对称），若右侧为 $\mathbb{D}_{1}$ $$\iint\limits_{\mathbb{D}}f(x,y)\cdot d\sigma = 2\iint\limits_{\mathbb{D}_{1}} f(x,y)\cdot d\sigma$$
   2. $\mathbb{D}$ 关于 $y=x$ 对称 $$\iint\limits_{\mathbb{D}}f(x,y)\cdot d\sigma = \iint\limits_{\mathbb{D}} f(y,x)\cdot d\sigma$$
      • 例 $\mathbb{D}$ 如下图所示，求 $\displaystyle I = \iint\limits_{\mathbb{D}}\frac{x^{3}}{x^{3}+y^{3}}\cdot d\sigma$ $$\begin{array}{ll} \because & \mathbb{D} \text{关于} y=x \text{对称} \\ \therefore & \displaystyle I = \iint\limits_{\mathbb{D}} \frac{x^{3}}{x^{3}+y^{3}} \cdot d\sigma = \iint\limits_{\mathbb{D}} \frac{y^{3}}{y^{3}+x^{3}} \cdot d\sigma \\ \therefore & \displaystyle 2I = \iint\limits_{\mathbb{D}}1\cdot d\sigma = A = 2 - \frac{1}{2} = \frac{3}{2} \\ \end{array}$$