Section04_定积分的应用

几何应用

面积

1. $\text{L: }y=f(x)\ge 0\quad x\in[a,b]$ $$A = \int_{a}^{b}f(x)\cdot dx$$
2. $\text{L: } r= r(\theta)\quad \theta\in[\alpha,\beta]$
   1. 取 $\displaystyle[\theta,\theta+d\theta]\in[\alpha,\beta]$
   2. $\displaystyle dA = \frac{d\theta}{2}r^{2}(\theta)$
   3. $\displaystyle A =\int_{\alpha}^{\beta} \frac{r^{2}(\theta)}{2}\cdot d\theta$ $$\displaystyle A =\frac{1}{2}\int_{\alpha}^{\beta} r^{2}(\theta)\cdot d\theta$$
3. $\text{L}_{1}\text{: } r_{1}=r_{1}(\theta); \text{L}_{2}\text{: }r_{2} = r_{2}(\theta)\quad \theta\in[\alpha,\beta]$ 求两条曲线之间的面积
   1. 取 $\displaystyle [\theta,\theta+d\theta]\in[\alpha,\beta]$
   2. $\displaystyle dA = \frac{r_{2}^{2}(\theta)-r_{1}^{2}(\theta)}{2}d\theta$
   3. $\displaystyle A = \int_{\alpha}^{\beta}\frac{r^{2}_{1}(\theta)-r^{2}_{2}(\theta)}{2}\cdot d\theta$ $$A = \frac{1}{2} \int_{\alpha}^{\beta}[r_{2}^{2}(\theta)-r_{1}^{2}(\theta)]\cdot d\theta$$
4. 求$y=f(x)$绕$x$轴旋转一周后形成立体图形的侧面积
   1. 取$\displaystyle [x,x+dx]\in[a,b]$
   2. $\displaystyle dA = 2\pi f(x)\cdot ds = 2\pi f(x)\sqrt{1+[f'(x)]^{2}}\cdot dx$
   3. $\displaystyle A = 2\pi\int_{a}^{b} f(x)\sqrt{1+[f'(x)]^{2}}\cdot dx$ $$A = 2\pi \int_{a}^{b}f(x)\sqrt{1+[f'(x)]^{2}}\cdot dx$$

Notes

此处是 $ds$ 而非 $dx$

体积

1. 求$y=f(x)$绕$x$轴旋转一周后形成立体图形的体积
   1. 取 $\displaystyle[x,x+dx]\in[a,b]$
   2. $\displaystyle dV = \pi f^{2}(x)\cdot dx$
   3. $\displaystyle V = \pi\int_{a}^{b}f^{2}(x)\cdot dx$ $$V = \pi\int_{a}^{b}f^{2}(x)\cdot dx$$
2. 求$y=f(x)$绕$y$轴旋转一周后形成立体图形的体积
   1. 取 $\displaystyle[x,x+dx]\in[a,b]$
   2. $\displaystyle dV = 2\pi xf(x)dx$
   3. $\displaystyle V = 2\pi\int_{a}^{b}xf^{}(x)\cdot dx$ $$V = 2\pi \int_{a}^{b}xf(x)\cdot dx$$
3. 已知该立体图形在$x$处的截面积为$A(x)$求其体积
   1. 取 $\displaystyle[x,x+dx]\in[a,b]$
   2. $\displaystyle dV = A(x)dx$
   3. $\displaystyle V = \int_{a}^{b}A(x)\cdot dx$ $$V = \int_{a}^{b}A(x)\cdot dx$$

弧长

1. $\text{L: } y=f(x) \quad x\in[a,b]$
   1. 取 $\displaystyle[x,x+dx]\in[a,b]$
   2. $\displaystyle ds = \sqrt{(dx)^{2}+(dy)^{2}} = \sqrt{1+[f'(x)]^{2}}\cdot dx$
   3. $\displaystyle s = \int_{a}^{b}\sqrt{1+[f'(x)]^{2}}\cdot dx$
2. $\text{L: } \begin{cases} y=\phi(t) \\ x = \psi(t)\end{cases} t\in[\alpha,\beta]$
   1. 取 $\displaystyle[t,t+dt]\in[\alpha,\beta]$
   2. $\displaystyle ds = \sqrt{(dx)^{2}+(dy)^{2}} = \sqrt{[\psi'(t)]^{2}+ [\phi'(t)]^{2}}\cdot dt$
   3. $\displaystyle s = \int_{\alpha}^{\beta}\sqrt{[\psi'(x)]^{2}+[\phi'(x)]^{2}}\cdot dt$

例题

1. 下图为$y=\sqrt{x-1}$的图像，$P$为$y=f(x)$上过原点的切线

   1. 求切线方程 $$\begin{array}{ll} & \text{设}P \text{点坐标为}(a,\sqrt{a-1})\\ \therefore & P \text{点切线斜率为} y'\vert_{x=a} = \frac{1}{2 \sqrt{a-1}} \\ & \text{方程为} y-\sqrt{a-1} = \frac{1}{2 \sqrt{a-1}}(x-a) \\ \because & \text{切线过原点} \\ \therefore & -\sqrt{a-1} = -\frac{a}{2 \sqrt{a-1}} \\ & \text{解得} a = 2 \\ \therefore & \text{切线方程为} y = \frac{1}{2}x \\ \end{array}$$
   2. 求阴影部分面积 $$\begin{array}{ll} & \displaystyle A = \int_{0}^{2} \frac{1}{2}x \cdot dx - \int_{1}^{2} \sqrt{x-1} \cdot dx \\ & \displaystyle =\left. \frac{1}{4}x^{2}\right\vert^{2}_{0} - \left.\frac{2}{3}(x-1)^{\frac{3}{2}}\right\vert_{1}^{2} = 1- \frac{2}{3} = \frac{1}{3} \end{array}$$
   3. 求阴影部分绕$x$轴一周图形的表面积 $$\begin{array}{ll} & A_{\text{总}} = A_{\text{内}} + A_{\text{外}} \\ \because &\displaystyle A_{\text{內}} = 2\pi\int_{1}^{2}\sqrt{x-1}\sqrt{1+[(\sqrt{x-1})']^{2}} \cdot dx \\ \therefore & \displaystyle A_{\text{内}} = 2\pi \int_{1}^{2} \frac{1}{2} \sqrt{4x-3 }\cdot dx = \pi\cdot \frac{1}{4} \int_{1}^{2} \sqrt{4x-3}\cdot d(4x-3)\\ & \displaystyle = \left.\frac{\pi}{4}\times \frac{2}{3}(4x-3)^{\frac{3}{2}}\right\vert_{1}^{2} = \frac{5 \sqrt{5} - 1}{6}\pi \\ \because & \displaystyle A_{\text{外}} = 2\pi \int_{0}^{2} \frac{1}{2}x \sqrt{1+\big[\big(\frac{1}{2}x\big)'\big]^{2}}\cdot dx \\ \therefore & \displaystyle A_{\text{外}} = \frac{\sqrt{5}}{2}\pi \int_{0}^{2}x\cdot dx = \left.\frac{\sqrt{5}\pi}{4}x^{2}\right\vert_{0}^{2} = \sqrt{5}\pi\\ \therefore & \displaystyle A_{\text{总}} = \frac{11 \sqrt{5} - 1}{6}\pi \\ \end{array}$$

2. 如下图所示，已知阴影为半个椭圆，求其绕$x=2$旋转一周后的图形的体积 $$\begin{array}{ll} & \text{取}[x,x+dx]\in[-2,2]\\ & dV = 2(2-x)\pi y dx \\ \because & \frac{x^{2}}{4} + y^{2} = 1 \\ \therefore & y = \sqrt{1-\frac{x^{2}}{4}} \\ & dV = 2\pi (2-x)\sqrt{1-\frac{x^{2}}{4}}\cdot dx\\ \therefore &\displaystyle V = 2\pi\int_{-2}^{2}(2-x)\sqrt{1-\frac{x^{2}}{4}}\cdot dx \\ & \displaystyle =2\pi \int_{0}^{2}\bigg[(2-x)\sqrt{1-\frac{x^{2}}{4}} + (2+x)\sqrt{1-\frac{x^{2}}{4}}\bigg]\cdot dx \\ & \displaystyle = 8\pi \int_{0}^{2} \sqrt{1-\frac{x^{2}}{4}} \cdot dx \xlongequal{x=2\sin t} 8\pi \int_{0}^{\frac{\pi}{2}}\cos t(2\cos t)\cdot dt \\ & \displaystyle = 16\pi I_{2} = 4\pi^{2} \end{array}$$

3. 如下图所示，求该图形绕$y$轴旋转一周后图形的体积 $$\begin{array}{ll} & \text{取}[y,y+dy] \in[-1,\frac{1}{2}] \\ & dV = \pi x^{2}dy \\ \because & x^{2} + y^{2} = 1 \\ \therefore & dV = \pi(1-y^{2})dy \\ & \displaystyle V = \pi\int_{-1}^{\frac{1}{2}}(1-y^{2})\cdot dy = \pi(y- \frac{1}{3}y^{3})\vert_{-1}^{\frac{1}{2}} = \frac{9}{8}\pi \end{array}$$