Double Integral

曲顶柱体的体积,平面薄片的质量,基本思路:二重积分转为二次积分

  • 为被积函数
  • 为积分表达式
  • 为面积元素
  • 为积分区域

基本性质

\iint \limits_{D} f(x,y)\mathrm{d}\sigma=\iint \limits_{D_{1}} f(x,y)\mathrm{d}\sigma+\iint \limits_{D_{2}} f(x,y)\mathrm{d}\sigma \end{align}$$ $$\begin{align} \iint \limits_{D} f(x,y)\mathrm{d}\sigma=f(\xi,\eta )\sigma \end{align}$$ $$\begin{align} \left\lvert \iint \limits_{D} f(x,y)\mathrm{d}\sigma \right\rvert\leq \iint \limits_{D}\left\lvert f(x,y) \right\rvert \mathrm{d}\sigma \end{align}$$ #### 二重积分的对称性 积分区域 $D$ 关于 $y$ 轴对称,$D_{1}$ 为积分的右半部分 $$\begin{align} f(-x,y)=-f(x,y)\quad \Rightarrow\quad \iint \limits_{D}f(x,y)=0 \end{align}$$ $$\begin{align} f(-x,y)=f(x,y)\quad \Rightarrow \quad \iint \limits_{D}f(x,y)=2 \iint \limits_{D_{1}}f(x,y) \end{align}$$ 积分区域 $D$ 关于 $x$ 轴对称,$D_{1}$ 为积分的上半部分 $$\begin{align} f(x,-y)=-f(x,y)\quad \Rightarrow\quad \iint \limits_{D}f(x,y)=0 \end{align}$$ $$\begin{align} f(x,-y)=f(x,y)\quad \Rightarrow \quad \iint \limits_{D}f(x,y)=2 \iint \limits_{D_{1}}f(x,y) \end{align}$$ 积分区域 $D$ 关于原点对称 $$\begin{align} f(-x,-y)=-f(x,y)\quad \Rightarrow\quad \iint \limits_{D}f(x,y)=0 \end{align}$$ $$\begin{align} f(-x,-y)=f(x,y)\quad \Rightarrow \quad \iint \limits_{D}f(x,y)=2 \iint \limits_{D_{1}}f(x,y) \end{align}$$ ### 二重积分的计算 >确定好积分的区域,想明白积分的次序。以单次积分的理解将二重积分转为二次积分,最后转为[[定积分\|定积分]]的计算 #### 直角坐标 本质上是利用[[定积分的应用#2. 平行截面面积已知的立体的体积\|平行截面面积已知的立体的体积]]的计算方法 ![Pasted image 20241005133602.png](../img/user/Functional%20files/Photo%20Resources/Pasted%20image%2020241005133602.png) $$\begin{align} \iint \limits _{D}f(x,y)d\sigma= \int _{a}^{b} A(x)\, dx =\int _{a}^{b} \left[\int _{\varphi_{1}(x)}^{\varphi_{2}(x)} f(x,y)\, dy \right]\, dx \end{align}$$ $$\begin{align} \iint \limits _{D}f(x,y)d\sigma=\int _{a}^{b} A(y)\, dy =\int _{c}^{d} \left[\int _{\psi_{1}(y)}^{\psi_{2}(y)} f(x,y)\, dx \right]\, dy \end{align}$$ #### 极坐标 $$\begin{align} \begin{cases} x =\rho \cos \theta \\ \\ y =\rho \sin \theta \end{cases} \quad \Rightarrow dxdy \to \rho d\rho d \theta \\ \end{align}$$ $$\begin{align} \iint \limits_{D}f(\rho \cos \theta,\rho \sin \theta)\rho d\rho d\theta &=\int _{\alpha}^{\beta } \, d\theta \int _{\varphi_{1}(\theta)}^{\varphi_{2}(\theta)} f(\rho \cos \theta,\rho \sin \theta) \rho\, d\rho \end{align}$$