Section02_积分方法

直角坐标法

方法

1. $x$ 型区域 $\mathbb{D} = \{(x,y)\vert a\le x\le b, \phi_{1}(x)\le y\le \phi_{2}(x)\}$ $$\iint\limits_{\mathbb{D}} f(x,y)\cdot d\sigma = \int_{a}^{b}dx\int_{\phi_{1}(x)}^{\phi_{2}(x)} f(x,y) \cdot dy$$
2. $y$ 型区域 $\mathbb{D} = \{(x,y)\vert \psi_{1}(y)\le x\le \psi_{2}(y), c\le y\le d\}$ $$\iint\limits_{\mathbb{D}} f(x,y)\cdot d\sigma = \int_{c}^{d}dy\int_{\psi_{1}(y)}^{\psi_{2}(y)} f(x,y) \cdot dx$$

例题

1. $\mathbb{D}$ 如下图所示，求 $\displaystyle I = \iint\limits_{\mathbb{D}}x^{2}y\cdot d\sigma$ $$\begin{array}{ll} & \mathbb{D} = \{(x,y)\vert 0\le x\le 1, 0\le y\le x\} \\\\ \therefore & \displaystyle I = \int_{0}^{1}dx \int_{0}^{y} x^{2}y\cdot dy \\ & = \displaystyle \frac{1}{2}\int_{0}^{1} x^{2} \times y^{2}\vert^{x}_{0}\cdot dx \\ & = \displaystyle \frac{1}{2}\int_{0}^{1}x^{4}\cdot dx = \frac{1}{10} \end{array}$$
2. $\displaystyle I = \iint\limits_{\mathbb{D}}(x+y)\cdot d\sigma$， $\mathbb{D}$ 由 $x=y^{2}$ 与 $y=x-2$ 围成 $$\begin{array}{ll} & \mathbb{D} = \{(x,y)\vert y^{2} \le x \le y+2, -1 \le y\le 2\} \\\\ \therefore & \displaystyle I = \int_{-1}^{2}dy \int_{y^{2}}^{y+2} (x+y)\cdot dx \\ & \displaystyle = \int_{-1}^{2} (\frac{1}{2}x^{2} + yx)\vert_{y^{2}}^{y+2} \cdot dy \\ & \displaystyle = \bigg(-\frac{1}{10}y^{5}-\frac{1}{4}y^{4}+\frac{1}{2}y^{3} + 2y^{2} + 2y\bigg)\bigg\vert_{-1}^{2} = 9 \frac{9}{20} \end{array}$$
3. $\displaystyle I =\int_{0}^{1}dy\int_{y}^{1}e^{x^{2}}\cdot dx$ $$\begin{array}{ll} & \text{由题可知} \mathbb{D} \text{的图像} \\ \therefore & \displaystyle I = \int_{0}^{1}dx \int_{0}^{x}e^{x^{2}}dy\\ & \displaystyle = \int_{0}^{1} xe^{x^{2}}\cdot dx = \left. \frac{1}{2}e^{x^{2}}\right|_{0}^{1} = \frac{1}{2}(e-1) \\ \end{array}$$

Notes 无法直接计算需要重新考虑积分顺序的情形

1. $x^{2n} e^{\pm x^{2}}\cdot dx$
2. $e^{\frac{k}{x}}\cdot dx$
3. $\cos \frac{k}{x}\cdot dx;\ \sin \frac{k}{x}\cdot dx$

极坐标法

方法

1. 特征
   • $\mathbb{D}$ 的边界曲线方程包含 $x^{2} + y^{2}$
   • $f(x,y)$ 包含 $x^{2}+y^{2}$
2. 变换
   • $\begin{cases} x = r\cos \theta \\ y = r\sin \theta\end{cases}$
   • $\mathbb{D} = \{(r,\theta)\vert r_{1}(\theta)\le r\le r_{2}(\theta), \alpha\le\theta \le \beta\}$
   • $\mathbb{D} = \{(r,\theta)\vert 0 \le r \le r_{}(\theta), \alpha\le\theta \le \beta\}$
3. 取 $(\theta,\theta + d\theta)\subset [\alpha,\beta]$，取 $(r,r+dr)\subset [0, r(\theta)]$ $$d\sigma = dr\cdot r\cdot d\theta$$ $$\iint\limits_{\mathbb{D}} f(x,y)\cdot d\sigma = \int_{\alpha}^{\beta} d\theta \int_{r_{1}(\theta)}^{r_{2}(\theta)} rf(r\cos \theta,r\sin \theta)\cdot dr$$

例题

1. $\displaystyle I = \iint\limits_{\mathbb{D}}x^{2}\cdot d\sigma$，$\mathbb{D}$ 如下图所示 $$\begin{array}{ll} & \displaystyle I = 2\iint\limits_{\mathbb{D}_{1}} x^{2}\cdot d\sigma \\ & \text{令}\begin{cases}x=r\cos \theta\\y=r\sin \theta\end{cases}\quad (0\le\theta\le\frac{\pi}{2}, 0\le r\le 2) \\ \because & \displaystyle \iint\limits_{\mathbb{D}_{1}}x^{2}\cdot d\sigma = \int_{0}^{\frac{\pi}{2}}d\theta \int_{0}^{2}r\cdot r^{2}\cos^{2}\theta\cdot dr \\ & = \displaystyle \frac{1}{4}\int_{0}^{\frac{\pi}{2}} 16\cos^{2}\theta \cdot d\theta = 4\int_{0}^{\frac{\pi}{2}}\cos^{2}\theta\cdot d\theta \\ & \displaystyle = 4\times\frac{1}{2}\times\frac{\pi}{2} = \pi \\ \therefore & I = 2\pi \\ \end{array}$$
2. $\displaystyle I = \iint\limits_{\mathbb{D}}x^{2}\cdot d\sigma$ $\mathbb{D}$ 如下图所示 $$\begin{array}{ll} & \text{令} \begin{cases} x=r\cos \theta \\ y=r\sin \theta \end{cases}\quad (0\le\theta\le \frac{\pi}{2}, 0\le r\le 2\cos \theta) \\ \therefore & \displaystyle I = \int_{0}^{\frac{\pi}{2}} d\theta \int_{0}^{2\cos \theta} r\cdot r^{2}\cos^{2}\theta \cdot dr \\ & \displaystyle = \frac{1}{4}\int_{0}^{\frac{\pi}{2}}16\cos^{6} \theta \cdot d\theta \\ & \displaystyle = 4\int_{0}^{\frac{\pi}{2}} \cos^{6} \theta \cdot d\theta \\ & \displaystyle = 4\times \frac{5}{6}\times \frac{3}{4}\times \frac{1}{2}\times \frac{\pi}{2} = \frac{5}{8}\pi \end{array}$$
3. $\displaystyle \iint\limits_{\mathbb{D}}\frac{d\sigma}{\sqrt{x^{2}+y^{2}}}$，$\mathbb{D}$ 如下图所示 $$\begin{array}{ll} & \text{令} \begin{cases} x=r\cos \theta \\ y=r\sin \theta \end{cases}\quad (0\le\theta\le \frac{\pi}{2}, \frac{1}{\cos\theta + \sin \theta} \le r\le 1) \\ \therefore & \displaystyle I = \int_{0}^{\frac{\pi}{2}} d\theta \int_{\frac{1}{\sin \theta + \cos \theta}}^{1} r\cdot \frac{1}{r} \cdot dr\\ & \displaystyle = \int_{0}^{\frac{\pi}{2}}\bigg(1-\frac{1}{\sin\theta +\cos\theta}\bigg)\cdot d\theta\\ & \displaystyle = \frac{\pi}{2} - \frac{1}{\sqrt{2}}\int_{0}^{\frac{\pi}{2}} \frac{d(\theta - \frac{\pi}{4})}{\cos(\theta - \frac{\pi}{4})} \\ & \displaystyle = \frac{\pi}{2} - \frac{1}{\sqrt{2}}\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{d\theta}{\cos\theta} = \frac{\pi}{2} - \sqrt{2}\int_{0}^{\frac{\pi}{4}}\frac{d\theta}{\cos\theta} \\ & \displaystyle = \frac{\pi}{2} - \left.\sqrt{2}\ln \vert \sec \theta + \tan \theta \vert\right\vert_{0}^{\frac{\pi}{4}} = \frac{\pi}{2} - \sqrt{2}\ln(1+\sqrt{2}) \end{array}$$
4. $\displaystyle I = \iint\limits_{\mathbb{D}}\sqrt{\vert x-y\vert}\cdot d\sigma$，$\mathbb{D}$ 如下图所示 $$\begin{array}{ll} & \displaystyle I = \iint\limits_{\mathbb{D}_{1}} \sqrt{x-y}\cdot d\sigma + \iint\limits_{\mathbb{D}_{2}}\sqrt{y-x}\cdot d\sigma \\ \because & \displaystyle \iint\limits_{\mathbb{D}_{1}} \sqrt{x-y}\cdot d\sigma = -\int_{0}^{1}dx \int_{0}^{x}\sqrt{x-y} \cdot d(x-y) \\ & \displaystyle = \frac{2}{3}\int_{0}^{1} x^{\frac{3}{2}}dx = \left. \frac{2}{3}\times \frac{2}{5} x^{\frac{5}{2}}\right\vert_{0}^{1} = \frac{4}{15} \\ & \displaystyle \iint\limits_{\mathbb{D}_{2}}\sqrt{y-x}\cdot d\sigma = -\int_{0}^{1}dy \int_{0}^{y} \sqrt{y-x}d(y-x) \\ & \displaystyle = \frac{2}{3} \int_{0}^{1} y^{\frac{3}{2}} \cdot dy = \left. \frac{2}{3}\times \frac{2}{5}y^{\frac{5}{2}}\right\vert_{0}^{1} = \frac{4}{15} \\ \therefore & I = \frac{8}{15} \\ \end{array}$$