版块 9 定积分应用

A.实用解题工具

第 1 组：特殊曲线

心形线：

表达式：

极坐标：$r=a(1+\cos \theta ),\theta \in [0,2\pi ],a>0$

直角坐标：$x^2+y^2-ax=a\sqrt{x^2+y^2},a>0$

参数方程：

$$\{\begin{array}{c}x=a(1+cos\theta )cos\theta \\ y=a(1+cos\theta )sin\theta \end{array}(a>0,\theta \in [0,2\pi ])$$

图形：

弧长$L$：

$$\begin{array}{rl}& L=2{\int }_0^{\pi }\sqrt{{\rho }^2(\theta )+[{\rho }^{\prime }(\theta ){]}^2}d\theta \\ & =2{\int }_0^{\pi }\sqrt{a^2(1+cos\theta {)}^2+a^{2}si{n}^2\theta }\cdot d\theta \\ & =2a{\int }_0^{\pi }\sqrt{(1+cos\theta {)}^2+si{n}^2\theta }\cdot d\theta (a>0)\\ & =2\sqrt{2}a{\int }_0^{\pi }\sqrt{1+cos\theta }\cdot d\theta \\ & =4a{\int }_0^{\pi }\left|cos\frac{\theta }{2}\right|d\theta =8a\end{array}$$

面积$A$:

$$\begin{array}{rl}& A=2{\int }_0^{\pi }\frac{1}{2}[\rho (\theta ){]}^{2}d\theta ={\int }_0^{\pi }{a}^2(1+cos\theta {)}^{2}d\theta \\ & =a^2{\int }_0^{\pi }(1+co{s}^2\theta +2cos\theta )d\theta \\ & =a^2(\pi +2\cdot \frac{1}{2}\cdot \frac{\pi }{2})=\frac{3\pi }{2}{a}^2\end{array}$$

体积$V$:

$$\begin{array}{rl}& D=\{(r,\theta )|r=a(1-cos\theta ),\theta \in [0,\pi ]\}\\ & r(x,y)=f(x)=y\\ & V=2\pi {\iint }_{D}r(x,y)\mathrm{d}\sigma \\ & =2\pi {\iint }_{D}y\mathrm{d}xdy\\ & =2\pi {\int }_0^{\pi }d\theta {\int }_0^{r(\theta )}rsin\theta \cdot rdr\\ & =2\pi {\int }_0^{\pi }sin\theta d\theta {\int }_0^{r(\theta )}{r}^{2}dr\\ & =2\pi {\int }_0^{\pi }sin\theta \cdot \frac{1}{3}{r}^3{|}_0^{r(\theta )}d\theta \\ & =\frac{2\pi }{3}{\int }_0^{\pi }{r}^3\left(\theta \right)sin\theta d\theta \\ & =\frac{8\pi }{3}{a}^3\end{array}$$

表面积$S$:

$$\begin{array}{rl}& S=2\pi {\int }_{\alpha }^{\beta }|r(\theta )sin\theta |\sqrt{[r(\theta ){]}^2+[r^{\prime }(\theta ){]}^2}d\theta \\ & =2\pi {\int }_0^{\pi }|a(1-cos\theta )sin\theta |\sqrt{[a(1-cos\theta ){]}^2+[asin\theta {]}^2}d\theta \\ & =2\pi a^2{\int }_0^{\pi }(1-cos\theta )sin\theta \sqrt{[(1-cos\theta ){]}^2+[sin\theta {]}^2}d\theta \\ & =2\sqrt{2}\pi a^2{\int }_0^{\pi }(1-cos\theta )sin\theta \sqrt{1-cos\theta }d\theta \\ & =2\sqrt{2}\pi a^2{\int }_{\pi }^0(1-cos\theta )\sqrt{1-cos\theta }d(cos\theta )\\ & =2\sqrt{2}\pi a^2{\int }_{-1}^1(1-u{)}^{\frac{3}{2}}du\\ & =\frac{32\pi }{5}{a}^2\end{array}$$

阿基米德螺旋线：

表达式：

极坐标：$r=a\theta$，$\theta \in [0,+\infty ),a>0$

参数方程：

$$\{\begin{array}{c}x=a\theta \cdot cos\theta \\ y=a\theta \cdot sin\theta \end{array}(\theta \in [0,+\infty ),a>0)$$

图形：

注：这里只展示θ在[0,2π]上的图像。

弧长$L$：

$$\begin{array}{rl}& L={\int }_0^{2\pi }\sqrt{[r(\theta ){]}^2+[r^{\prime }(\theta ){]}^2}d\theta \\ & ={\int }_0^{2\pi }\sqrt{a^2{\theta }^2+a^2}d\theta \\ & =a^2{\int }_0^{2\pi }\sqrt{{\theta }^2+1}d\theta \\ & =\frac{a}{2}[2\pi \sqrt{1+4{\pi }^2}+ln(2\pi +\sqrt{1+4{\pi }^2})]\end{array}$$

不定积分公式：

$$\begin{array}{rl}& \int \sqrt{x^2+a^2}dx\\ & =\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}ln(x+\sqrt{x^2+a^2})+C\end{array}$$

面积$A$：

$$\begin{array}{rl}& A={\int }_0^{2\pi }\frac{1}{2}[r(\theta ){]}^{2}d\theta =\frac{a^2}{2}{\int }_0^{2\pi }{\theta }^{2}d\theta \\ & =\frac{4{\pi }^3}{3}{a}^2\end{array}$$

笛卡尔叶形线：

表达式：

直角坐标：$x^3+y^3=3axy,a>0$

极坐标：

$$r=\frac{3asin\theta cos\theta }{si{n}^3\theta +co{s}^3\theta },\theta \in [0,\frac{3\pi }{4})\cup (\frac{3\pi }{4},\pi ],a>0$$

参数方程：

$$\{\begin{array}{c}x=\frac{3at}{1+t^3}\\ y=\frac{3a{t}^2}{1+t^3}\end{array}(a>0,t\ne -1)$$

图形：

渐近线：

当 t→-1时，x→∞，y→∞，故叶形线没有水平与铅直渐近线

$$\begin{array}{rl}& \underset{t\to -1}{\lim }\frac{y}{x}=\underset{t\to -1}{\lim }t=-1=k\\ & \underset{t\to -1}{\lim }\left(y-kx\right)\\ & =\underset{t\to -1}{\lim }(\frac{3a{t}^2}{1+t^3}+\frac{3at}{1+t^3})\\ & =\underset{t\to -1}{\lim }\frac{3at(t+1)}{1+t^3}=-a\end{array}$$

故叶形线有唯一的斜渐近线$y=-x-a$

笔记→参数方程求渐近线：

对于曲线的参数方程

$$\{\begin{array}{l}x=x(t)\\ y=y(t)\end{array}$$

①若$\underset{t\to t_0}{\lim }x(t)=\infty ,\underset{t\to t_0}{\lim }y(t)=b,$

则$y=b$为水平渐近线。

②若$\underset{t\to t_0}{\lim }x(t)=c,\underset{t\to t_0}{\lim }y(t)=\infty ,$

③若$\underset{t\to t_0}{\lim }x(t)=\infty ,\underset{t\to t_0}{\lim }y(t)=\infty ,$

$$\begin{array}{rl}& \underset{t\to t_0}{\lim }\frac{y(t)}{x(t)}=k\ne 0,\\ & \\ & \underset{t\to t_0}{\lim }\left[y(t)-kx(t)\right]=b,\end{array}$$

则$y=kx+b$为斜渐近线。

面积$A$：

记第一象限围成的面积为$A$

$$\begin{array}{rl}& A={\int }_0^{\frac{\pi }{2}}\frac{1}{2}[r(\theta ){]}^{2}d\theta =\frac{1}{2}{\int }_0^{\frac{\pi }{2}}(\frac{3asin\theta cos\theta }{si{n}^3\theta +co{s}^3\theta }{)}^{2}d\theta \\ & =\frac{9a^2}{2}{\int }_0^{\frac{\pi }{2}}\frac{si{n}^2\theta co{s}^2\theta }{(si{n}^3\theta +co{s}^3\theta {)}^2}d\theta \\ & =\frac{9a^2}{2}{\int }_0^{\frac{\pi }{2}}\frac{si{n}^2\theta co{s}^2\theta \cdot se{c}^6\theta }{(si{n}^3\theta +co{s}^3\theta {)}^2\cdot se{c}^6\theta }d\theta \\ & =\frac{9a^2}{2}{\int }_0^{\frac{\pi }{2}}\frac{ta{n}^2\theta \cdot se{c}^2\theta }{(ta{n}^3\theta +1{)}^2}d\theta \\ & =\frac{9a^2}{2}{\int }_0^{\frac{\pi }{2}}\frac{ta{n}^2\theta }{(ta{n}^3\theta +1{)}^2}d(tan\theta )\\ & =\frac{9a^2}{2}{\int }_0^{+\infty }\frac{u^2}{(u^3+1{)}^2}du\\ & =\frac{3a^2}{2}{\int }_0^{+\infty }\frac{d(u^3+1)}{(u^3+1{)}^2}\\ & ={\frac{3a^2}{2}\cdot \frac{1}{u^3+1}|}_{u=+\infty }^{u=0}\end{array}$$ $$=\frac{3a^2}{2}$$

玫瑰线：

表达式：

$$r=a\sin 3\theta (\text{三叶玫瑰线});r=a\sin 2\theta (\text{四叶玫瑰线})$$ $$\theta \in [0,\pi ],a>0$$

图形：

三叶玫瑰线

四叶玫瑰线

面积$A$：

三叶玫瑰线所围成的面积$A_1$：

$$\begin{array}{rl}& A_1=3\cdot {\int }_0^{\frac{\pi }{3}}\frac{1}{2}[r(\theta ){]}^{2}d\theta =\frac{3a^2}{2}{\int }_0^{\frac{\pi }{3}}si{n}^{2}3\theta d\theta \\ & =\frac{a^2}{2}{\int }_0^{\frac{\pi }{3}}si{n}^{2}3\theta d(3\theta )=\frac{a^2}{2}{\int }_0^{\pi }si{n}^{2}udu\\ & =\frac{a^2}{2}\cdot 2\cdot \frac{1}{2}\cdot \frac{\pi }{2}=\frac{\pi a^2}{4}\end{array}$$

四叶玫瑰线所围成的面积$A_2$：

$$\begin{array}{rl}& A_2=4\cdot {\int }_0^{\frac{\pi }{2}}\frac{1}{2}[r(\theta ){]}^{2}d\theta =2a^2{\int }_0^{\frac{\pi }{2}}si{n}^{2}2\theta d\theta \\ & =a^2{\int }_0^{\frac{\pi }{2}}si{n}^{2}2\theta d(2\theta )=a^2{\int }_0^{\pi }si{n}^{2}udu\\ & =a^2\cdot 2\cdot \frac{1}{2}\cdot \frac{\pi }{2}=\frac{\pi a^2}{2}\end{array}$$

星形线：

表达式：

直角坐标：$x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}},a>0$

极坐标：$\rho =\frac{a}{(co{s}^{\frac{2}{3}}\theta +si{n}^{\frac{2}{3}}\theta {)}^{\frac{3}{2}}},\theta \in [0,2\pi ],a>0$

参数方程：

$$\{\begin{array}{l}x=a{\cos }^3\theta \\ y=a{\sin }^3\theta \end{array}$$ $$(a>0,\theta \in [0,2\pi ])$$

图形：

弧长$L$:

$$\begin{array}{rl}& \{\begin{array}{l}{x}^{\prime }=-3a\cdot co{s}^{2}t\cdot sint\\ y^{\prime }=3a\cdot si{n}^{2}t\cdot cost\end{array}(a>0,t\in [0,2\pi ])\\ & L=4{\int }_0^{\frac{\pi }{2}}\sqrt{[x^{\prime }(t){]}^2+[y^{\prime }(t){]}^2}dt\\ & =4{\int }_0^{\frac{\pi }{2}}\sqrt{[-3a\cdot co{s}^{2}t\cdot sint{]}^2+[3a\cdot si{n}^{2}t\cdot cost{]}^2}dt\\ & =4{\int }_0^{\frac{\pi }{2}}\sqrt{9a^2(co{s}^{4}t\cdot si{n}^{2}t+si{n}^{4}t\cdot co{s}^{2}t)}dt\\ & =4\cdot 3a{\int }_0^{\frac{\pi }{2}}\sqrt{co{s}^{2}t\cdot si{n}^{2}t(si{n}^{2}t+co{s}^{2}t)}dt\\ & =12a{\int }_0^{\frac{\pi }{2}}\left|sint\cdot cost\right|dt\\ & =12a{\int }_0^{\frac{\pi }{2}}\frac{1}{2}|sin2t|dt\\ & =6a{\int }_0^{\frac{\pi }{2}}sin2t\cdot dt\\ & =6a.\end{array}$$

星形线全长为$6a$，那么每一象限的星形线弧长就为$1.5a$

面积$A$：

$$\begin{array}{rl}& A=4{\int }_0^{a}ydx\\ & =4{\int }_{\frac{\pi }{2}}^{0}y(t)x^{\prime }(t)dt\\ & =4{\int }_{\frac{\pi }{2}}^{0}asi{n}^{3}t\cdot (-3a)\cdot co{s}^{2}t\cdot sintdt\\ & =12a^2{\int }_0^{\frac{\pi }{2}}si{n}^{4}t\cdot co{s}^{2}tdt\\ & =12a^2{\int }_0^{\frac{\pi }{2}}si{n}^{4}t\cdot (1-si{n}^{2}t)dt\\ & =12a^2{\int }_0^{\frac{\pi }{2}}(si{n}^{4}t-si{n}^{6}t)dt\\ & =12a^2(\frac{3}{4}\cdot \frac{1}{2}\cdot \frac{\pi }{2}-\frac{5}{6}\cdot \frac{3}{4}\cdot \frac{1}{2}\cdot \frac{\pi }{2})\\ & =\frac{3\pi }{8}{a}^2\end{array}$$ $$\begin{array}{rl}& A=4{\int }_0^{a}ydx\\ & =4{\int }_0^a(a^{\frac{2}{3}}-x^{\frac{2}{3}}{)}^{\frac{3}{2}}dx\\ & =4{\int }_0^{\frac{\pi }{2}}(a^{\frac{2}{3}}-a^{\frac{2}{3}}si{n}^2\theta {)}^{\frac{3}{2}}\cdot 3asi{n}^2\theta cos\theta \cdot d\theta (x=asi{n}^3\theta )\\ & =12a^2{\int }_0^{\frac{\pi }{2}}si{n}^2\theta \cdot co{s}^4\theta d\theta \\ & =12a^2{\int }_0^{\frac{\pi }{2}}(1-co{s}^2\theta )\cdot co{s}^4\theta d\theta \\ & =12a^2{\int }_0^{\frac{\pi }{2}}(co{s}^4\theta -co{s}^6\theta )d\theta \\ & =12a^2(\frac{3}{4}\cdot \frac{1}{2}\cdot \frac{\pi }{2}-\frac{5}{6}\cdot \frac{3}{4}\cdot \frac{1}{2}\cdot \frac{\pi }{2})\\ & =\frac{3\pi }{8}{a}^2\end{array}$$

体积$V$：

$$\begin{array}{rl}& V=2\cdot \pi {\int }_0^a{y}^{2}dx\\ & =2\pi {\int }_0^a(a^{\frac{2}{3}}-x^{\frac{2}{3}}{)}^{3}dx\\ & =2\pi {\int }_0^a(a^2-3a^{\frac{4}{3}}{x}^{\frac{2}{3}}+3a^{\frac{2}{3}}{x}^{\frac{4}{3}}-x^2)dx\\ & =2\pi [a^{2}x-3a^{\frac{4}{3}}\cdot \frac{3}{5}{x}^{\frac{5}{3}}+3a^{\frac{2}{3}}\cdot {\frac{3}{7}{x}^{\frac{7}{3}}-\frac{1}{3}{x}^3]|}_{x=0}^{x=a}\\ & =2\pi (a^3-\frac{9a^3}{5}+\frac{9a^3}{7}-\frac{a^3}{3})\\ & =\frac{32\pi }{105}{a}^3\end{array}$$ $$\begin{array}{rl}& V=2\cdot \pi {\int }_{\frac{\pi }{2}}^0{y}^2(t)x^{\prime }(t)dt\\ & =2\pi {\int }_{\frac{\pi }{2}}^0{a}^{2}si{n}^{6}t\cdot (-3a)\cdot co{s}^{2}t\cdot sintdt\\ & =6a^3\pi {\int }_0^{\frac{\pi }{2}}si{n}^{7}t\cdot co{s}^{2}tdt\\ & =6a^3\pi {\int }_0^{\frac{\pi }{2}}(si{n}^{7}t-si{n}^{9}t)dt\\ & =6a^3\pi (\frac{6}{7}\cdot \frac{4}{5}\cdot \frac{2}{3}-\frac{8}{9}\cdot \frac{6}{7}\cdot \frac{4}{5}\cdot \frac{2}{3})\\ & =6a^3\pi \cdot \frac{1}{9}\cdot \frac{6}{7}\cdot \frac{4}{5}\cdot \frac{2}{3}\\ & =\frac{32\pi }{105}{a}^3\end{array}$$

表面积$S$：

$$\begin{array}{rl}& S=2\cdot 2\pi {\int }_0^a\left|y\right|\sqrt{1+(y^{\prime }{)}^2}dx\\ & =4\pi {\int }_0^a\left|(a^{\frac{2}{3}}-x^{\frac{2}{3}}{)}^{\frac{3}{2}}\right|\sqrt{1+[-(a^{\frac{2}{3}}-x^{\frac{2}{3}}{)}^{\frac{1}{2}}\cdot x^{-\frac{1}{3}}{]}^2}dx\\ & =4\pi {\int }_0^a\left|(a^{\frac{2}{3}}-x^{\frac{2}{3}}{)}^{\frac{3}{2}}\right|\cdot \left|a^{\frac{1}{3}}\cdot x^{-\frac{1}{3}}\right|dx\\ & =4a^{\frac{1}{3}}\pi {\int }_0^a(a^{\frac{2}{3}}-x^{\frac{2}{3}}{)}^{\frac{3}{2}}\cdot x^{-\frac{1}{3}}dx\\ & =4a^{\frac{1}{3}}\pi \cdot \underset{t\to 0^{+}}{\lim }{\int }_t^a(a^{\frac{2}{3}}-x^{\frac{2}{3}}{)}^{\frac{3}{2}}\cdot x^{-\frac{1}{3}}dx\\ & =-4a^{\frac{1}{3}}\pi \cdot \frac{3}{2}\underset{t\to 0^{+}}{\lim }{\int }_t^a(a^{\frac{2}{3}}-x^{\frac{2}{3}}{)}^{\frac{3}{2}}d(a^{\frac{2}{3}}-x^{\frac{2}{3}})\\ & =-4a^{\frac{1}{3}}\pi \cdot \frac{3}{2}\cdot \frac{2}{5}\cdot \underset{t\to 0^{+}}{\lim }{(a^{\frac{2}{3}}-x^{\frac{2}{3}}{)}^{\frac{5}{2}}|}_{x=t}^{x=a}\\ & =-4a^{\frac{1}{3}}\pi \cdot \frac{3}{2}\cdot \frac{2}{5}\cdot [0-\underset{t\to 0^{+}}{\lim }(a^{\frac{2}{3}}-t^{\frac{2}{3}}{)}^{\frac{5}{2}}]\\ & =4a^{\frac{1}{3}}\pi \cdot \frac{3}{2}\cdot \frac{2}{5}\cdot a^{\frac{5}{3}}\end{array}$$ $$=\frac{12\pi }{5}{a}^2$$ $$\begin{array}{rl}& S=2\cdot 2\pi {\int }_0^{\frac{\pi }{2}}|y(t)|\sqrt{[x^{\prime }(t){]}^2+[y^{\prime }(t){]}^2}dt\\ & =4\pi {\int }_0^{\frac{\pi }{2}}\left|asi{n}^{3}t\right|\sqrt{[-3a\cdot co{s}^{2}t\cdot sint{]}^2+[3a\cdot si{n}^{2}t\cdot cost{]}^2}dt\\ & =12\pi a^2{\int }_0^{\frac{\pi }{2}}si{n}^{3}t\sqrt{co{s}^{4}t\cdot si{n}^{2}t+si{n}^{4}t\cdot co{s}^{2}t}dt\\ & =12\pi a^2{\int }_0^{\frac{\pi }{2}}si{n}^{3}t\sqrt{co{s}^{2}t\cdot si{n}^{2}t(si{n}^{2}t+co{s}^{2}t)}dt\\ & =12\pi a^2{\int }_0^{\frac{\pi }{2}}si{n}^{3}t\left|sint\cdot cost\right|dt\\ & =12\pi a^2{\int }_0^{\frac{\pi }{2}}si{n}^{4}t\cdot costdt\\ & ={12\pi a^2\cdot \frac{1}{5}si{n}^{5}t|}_{t=0}^{t=\frac{\pi }{2}}\\ & =\frac{12\pi }{5}{a}^2\end{array}$$

摆线：

表达式：

参数方程：

$$\{\begin{array}{l}x=a(\theta -\sin \theta )\\ y=a(1-\cos \theta )\end{array}$$ $$(a>0;\theta \in [0,2\pi ])$$

图形：

注：摆线图像以$T=2\pi a$为周期，这里只展示θ在$[0,2\pi ]$上的图像。

弧长$L$：

$$\begin{array}{rl}& \{\begin{array}{l}{x}^{\prime }=a(1-cos\theta )\\ y^{\prime }=asin\theta \end{array}(a>0,\theta \in [0,2\pi ])\\ & L=2\cdot {\int }_0^{\pi a}\sqrt{1+(y^{\prime }{)}^2}dx\\ & =2\cdot {\int }_0^{\pi }\sqrt{[x^{\prime }(\theta ){]}^2+[y^{\prime }(\theta ){]}^2}d\theta \\ & =2\cdot {\int }_0^{\pi }\sqrt{[a(1-cos\theta ){]}^2+[asin\theta {]}^2}d\theta \\ & =2\sqrt{2}a{\int }_0^{\pi }\sqrt{1-cos\theta }d\theta \\ & =4a{\int }_0^{\pi }\sqrt{si{n}^2\frac{\theta }{2}}d\theta \\ & =4a{\int }_0^{\pi }\left|sin\frac{\theta }{2}\right|d\theta \\ & =8a{\int }_0^{\pi }sin\frac{\theta }{2}d(\frac{\theta }{2})\\ & =8a\end{array}$$

面积$A$：

$$\begin{array}{rl}& A=2{\int }_0^{\pi a}y(x)dx\\ & =2{\int }_0^{\pi }y(\theta )x^{\prime }(\theta )d\theta \\ & =2{\int }_0^{\pi }[a(1-cos\theta )][a(1-cos\theta )]d\theta \\ & =2a^2{\int }_0^{\pi }(1-cos\theta {)}^{2}d\theta \\ & =2a^2{\int }_0^{\pi }(1+co{s}^2\theta -2cos\theta )d\theta \\ & =2a^2(\pi +2\cdot \frac{1}{2}\cdot \frac{\pi }{2}+0)\\ & =3\pi a^2\end{array}$$

旋转体积$V_x;V_y$

注：红色为 x 轴，绿色为 y 轴，蓝色为 z 轴。

绕x轴旋转

绕 y 轴旋转 $$\begin{array}{rl}& V_x=2\cdot \pi {\int }_0^{\pi a}{y}^2(x)dx\\ & =2\pi {\int }_0^{\pi }{y}^2(\theta )\cdot x^{\prime }(\theta )d\theta \\ & =2\pi {\int }_0^{\pi }[a(1-cos\theta ){]}^2\cdot a(1-cos\theta )d\theta \\ & =2\pi a^3{\int }_0^{\pi }(1-cos\theta {)}^{3}d\theta \\ & =2\pi a^3{\int }_0^{\pi }(1-3cos\theta +3co{s}^2\theta -co{s}^3\theta )d\theta \\ & =2\pi a^3\left(\pi -0+3\cdot 2\cdot \frac{1}{2}\cdot \frac{\pi }{2}-0\right)\\ & =5{\pi }^2{a}^3\end{array}$$ $$\begin{array}{rl}& V_y=2\pi {\int }_0^{2\pi a}x\left|y(x)\right|dx\\ & =2\pi {\int }_0^{2\pi }x(\theta )|y(\theta )|x^{\prime }(\theta )d\theta \\ & =2\pi {\int }_0^{2\pi }a(\theta -sin\theta )\left|a(1-cos\theta )\right|a(1-cos\theta )d\theta \\ & =2\pi a^3{\int }_0^{2\pi }(\theta -sin\theta )(1-cos\theta {)}^{2}d\theta \\ & =2\pi a^3{\int }_0^{2\pi }(\theta +\theta co{s}^2\theta -2\theta cos\theta -sin\theta -sin\theta co{s}^2\theta +sin2\theta )d\theta \\ & =2\pi a^3{\int }_0^{2\pi }(\theta +\theta co{s}^2\theta -2\theta cos\theta )d\theta \\ & =2\pi a^3(2{\pi }^2+{\int }_0^{2\pi }\theta co{s}^2\theta d\theta -2{\int }_0^{2\pi }\theta cos\theta d\theta )\\ & =2\pi a^3\left(2{\pi }^2+I_2-2I_1\right)\end{array}$$ $$\begin{array}{rl}& I_1={\int }_0^{2\pi }\theta cos\theta d\theta \\ & ={\int }_{-\pi }^{\pi }(\theta +\pi )cos(\theta +\pi )d\theta \\ & =-{\int }_{-\pi }^{\pi }(\theta +\pi )cos\theta d\theta \\ & =-{\int }_{-\pi }^{\pi }\theta cos\theta d\theta -2\pi {\int }_0^{\pi }cos\theta d\theta \\ & =0\end{array}$$ $$\begin{array}{rl}& I_2={\int }_0^{2\pi }\theta co{s}^2\theta d\theta \\ & ={\int }_{-\pi }^{\pi }(\theta +\pi )co{s}^2(\theta +\pi )d\theta \\ & ={\int }_{-\pi }^{\pi }(\theta +\pi )co{s}^2\theta d\theta \\ & ={\int }_{-\pi }^{\pi }\theta co{s}^2\theta d\theta +2\pi {\int }_0^{\pi }co{s}^2\theta d\theta \\ & =2\pi \cdot 2\cdot \frac{1}{2}\cdot \frac{\pi }{2}={\pi }^2\end{array}$$ $$\therefore V_y=2\pi a^3(2{\pi }^2+{\pi }^2-0)=6{\pi }^3{a}^3$$

表面积$S_x;S_y$

$$\begin{array}{rl}& S_x=2\cdot 2\pi {\int }_0^{\pi a}\left|y(x)\right|\sqrt{1+[y^{\prime }(x){]}^2}dx\\ & =4\pi {\int }_0^{\pi }\left|y(\theta )\right|\sqrt{[x^{\prime }(\theta ){]}^2+[y^{\prime }(\theta ){]}^2}d\theta \\ & =4\pi {\int }_0^{\pi }\left|a(1-cos\theta )\right|\sqrt{[a(1-cos\theta ){]}^2+[asin\theta {]}^2}d\theta \\ & =4\sqrt{2}\pi a^2{\int }_0^{\pi }(1-cos\theta )\sqrt{1-cos\theta }d\theta \\ & =4\sqrt{2}\pi a^2{\int }_0^{\pi }2si{n}^2\frac{\theta }{2}\cdot \sqrt{2}\left|sin\frac{\theta }{2}\right|d\theta \\ & =16\pi a^2{\int }_0^{\pi }si{n}^3\frac{\theta }{2}d\theta \\ & =32\pi a^2{\int }_0^{\frac{\pi }{2}}si{n}^{3}udu\\ & =\frac{64\pi }{3}{a}^2\end{array}$$ $$\begin{array}{rl}& S_y={\int }_{L}2\pi x\mathrm{ds}\\ & =2\pi {\int }_0^{2\pi }\left|x(\theta )\right|\sqrt{[x^{\prime }(\theta ){]}^2+[y^{\prime }(\theta ){]}^2}d\theta \\ & =4\pi a^2{\int }_0^{2\pi }(\theta -sin\theta )sin\frac{\theta }{2}d\theta \\ & =16{\pi }^2{a}^2\end{array}$$

第 2 组: 形心坐标计算公式

1. 平面图形的形心坐标为$(\overline{x},\overline{y}),$其中$\bar{x}=\frac{\underset{D}{\iint }xdxdy}{\underset{D}{\iint }dxdy},\bar{y}=\frac{\underset{D}{\iint }ydxdy}{\underset{D}{\iint }dxdy}$

曲面、立体等的形心坐标公式可依此类推:

$\overline{x}=\frac{\underset{D}{\iint }xdS}{\underset{D}{\iint }dS}$( 曲面, 仅数一);$\overline{x}=\frac{\underset{D}{\iiint }xdV}{\underset{D}{\iiint }dV}$( 立体, 仅数一)

1. 已知形心, 可简化积分, 如:$\underset{D}{\iint }xdxdy=\overline{x}\cdot \underset{D}{\iint }dxdy$(其他同理)

第 3 组: 求解截面已知立体的体积

截面已知立体的体积$:V={\int }_a^{b}s(x)dx,$其中$s(x)$是垂直于 x 轴的平面截立体所得到的截面面积

B. 核心问题及解题方法总结

1. 利用古尔丁定理求体积

古尔丁定理：由平面形状绕和它的同一个平面上的轴旋转而产生的族转体的体积$V$,等于平面形状面积$S$乘以平面形状的几何中心经过的距离$d_1$的积:$V=S{d}_1$

例题 1. 设$a,b>1$,则圆域$D=\left\{(x,y)|(x-a{)}^2+(y-1{)}^2\le 1\right\}$绕直线$x=-b$旋转一周所形成的旋转体的体积是

【解】$V=S{d}_1=\pi \cdot 1^2\cdot 2\pi \cdot (a+b)=2{\pi }^2(a+b)$

【细节说明】圆心就是圆域的几何中心, 其坐标为$(a,1)$.圆域$D$绕直线旋转一周时, 几何中心走过的轨迹为圆$L$,其半径为$r=a+b$, 故几何中心经过的距离$d_1=2\pi r=2\pi (a+b)$。另一方面圆域的面积$S=\pi \cdot 1^2=\pi$。

所以$V=S{d}_{{}_1}=2{\pi }^2(a+b)$

2. 双纽线的上下限

计算双纽线${\left(x^2+y^2\right)}^2=a^2\left(x^2-y^2\right)$相关问题时。注意参数的上下限。具体参考下面例题。

例题 1. 曲线${\left(x^2+y^2\right)}^2=a^2\left(x^2-y^2\right)$所围成的区域的面积为$f(a)$,当$a\to 0^{+}$时, 与$f(a)$同阶的无穷小有几个 ( )

$$-2\ln (\sqrt{\cos a})\sqrt{1+a}+\sqrt{1-a}-2(1+a{)}^{\frac{1}{a}}-e{\int }_0^{a^2}\sin \sqrt{a^2-x}dx$$

(A) 1 个 (B) 2 个 (C) 3 个 (D) 4 个

【解】选 B

区域边界的极坐标方程为$r^2=a^2\cos 2\theta$.由$\cos 2\theta >0$得

$\theta \in \left(-\frac{\pi }{4},\frac{\pi }{4}\right)$或$\left(\frac{3}{4}\pi ,\frac{5}{4}\pi \right).$

$$A=4A_1=4\cdot {\int }_0^{\frac{\pi }{4}}\frac{1}{2}{a}^2\cos 2\theta d\theta =a^2$$

分析：

说法1：$-2\ln (\sqrt{\cos a})=-\ln (\cos a)=-\ln (1+\cos a-1)\sim 1-\cos a\sim \frac{a^2}{2}$

说法2：$\sqrt{1+a}+\sqrt{1-a}-2$

$$=1+\frac{1}{2}a-\frac{a^2}{8}+o(a^2)+1-\frac{1}{2}a-\frac{a^2}{8}+o(a^2)-2$$ $$=-\frac{a^2}{4}+o(a^2)$$

说法3：$(1+a{)}^{\frac{1}{a}}-e=e^{\frac{\ln (1+a)}{a}}-e=e(e^{\frac{\ln (1+a)}{a}-1}-1)\sim e\frac{\ln (1+a)-a}{a}\sim -\frac{ea}{2}$

说法4：${\int }_0^{a^2}\sin \sqrt{a^2-x}dx={\int }_0^{a^2}\sin \sqrt{u}du\sim {\int }_0^{a^2}\sqrt{u}du=\frac{2a^3}{3}$

所以 1.2 与$f(a)$同阶。故选 B

3. 求解定积分的证明题

定积分证明题的主要方法有：

1. 化为变限函数，从而利用函数单调性

2. 利用积分中值定理

3. 利用$F(x)={\int }_a^{x}f\left(t\right)dt,F^{\prime }(x)=f(x)$

例题 1. (1994,2)设$f(x)$在$[0,1]$上连续且递城, 证明: 当$0<\lambda <1$时,${\int }_0^{\lambda }f(x)dx\ge \lambda {\int }_0^{1}f(x)dx$

【解法 1】用函数单调性.

令$\phi (\lambda )={\int }_0^{\lambda }f(x)dx-\lambda {\int }_0^{1}f(x)dx,\lambda \in [0,1].\text{则}{\phi }^{\prime }(\lambda )=f(\lambda )-{\int }_0^{1}f(x)dx$

令${\phi }^{\prime }(\lambda )=0\Rightarrow f(\xi )={\int }_0^{1}f(x)dx,\xi \in (0,1)$

由$f(\lambda )$递减,$\lambda =\xi$为唯一驻点, 且${\phi }^{\prime }(\lambda )$在$\lambda =\xi$由正变负,$\lambda =\xi$是$\phi (\lambda )$的极大值点也是最大值点; 由此, 最小点必为端点$\lambda =0$或$1$

【笔记】连续函数在闭区间必能取到最大和最小值。若可导函数最小值点在区间内部, 则必为驻点, 此时与驻点唯一矛盾。从而有$\phi (\lambda )\ge \phi (0)=\phi (1)=0,0<\lambda <1.$

命题得证.

【笔记】定积分不等式证明问题, 将上限换成$x$, 就是函数单调性的问题。这是非常重要的命题方式

【解法 2】用积分中值定理。

为分清两中值的大小, 需要分别在$(0,\lambda ),(\lambda ,1)$两区间内用积分中值定理:

$${\int }_0^{1}f(x)dx={\int }_0^{\lambda }f(x)dx+{\int }_{\lambda }^{1}f(x)dx,$$ $$\text{由此},{\int }_0^{\lambda }f(x)dx-\lambda {\int }_0^{1}f(x)dx=(1-\lambda ){\int }_0^{\lambda }f(x)dx-\lambda {\int }_{\lambda }^{1}f(x)dx$$ $$=(1-\lambda )\cdot \lambda f({\xi }_1)-\lambda \cdot (1-\lambda )f({\xi }_2)=(1-\lambda )\cdot \lambda [f({\xi }_1)-f({\xi }_2)],$$

其中,$0<{\xi }_1<\lambda <{\xi }_2<1$: 又因$f(x)$递减,$f({\xi }_1)\ge f({\xi }_2)$.上式的右端大于零, 问题得证.

【笔记】积分中值定理是定积分或重积分中非常重要的命题方式