方企勤 第三章 一元函数积分学 第3题
📝 题目
解法 1 如图 3.8 所示. 所围图形面积为
$$ S = - {\int }_{0}^{2\pi }y\left( t\right) {x}^{\prime }\left( t\right) \mathrm{d}t = \frac{4}{3}{\pi }^{3}{a}^{2} + \pi {a}^{2}. $$
解法 $\displaystyle{2S} = \frac{1}{2}{\int }_{0}^{2\pi }x\mathrm{\;d}y - y\mathrm{\;d}x +}$ $\bigtriangleup {OAB}$ 的面积,
$$ \bigtriangleup {OAB}\text{ 的面积 } = \frac{1}{2} \cdot a \cdot {2\pi a} = \pi {a}^{2}\text{ , } $$
$$ \frac{1}{2}{\int }_{0}^{2\pi }x\mathrm{\;d}y - y\mathrm{\;d}x = \frac{1}{2}{\int }_{0}^{2\pi }{a}^{2}\left\lbrack {\left( {\cos t + t\sin t}\right) t\sin t}\right. $$
$$ - \left( {\sin t - t\cos t}\right) t\cos t\rbrack \mathrm{d}t $$
$$ = \frac{1}{2}{\int }_{0}^{2\pi }{a}^{2}{t}^{2}\mathrm{\;d}t = \frac{4}{3}{\pi }^{3}{a}^{2}, $$
即得 $S = \frac{4}{3}{\pi }^{3}{a}^{2} + \pi {a}^{2}$ .
解法 3 用极坐标. $S = \frac{1}{2}{\int }_{0}^{{\theta }_{0}}{r}^{2}\left( \theta \right) \mathrm{d}\theta + \bigtriangleup {OAB}$ 的面积,其中 $r =$ $r\left( \theta \right)$ 为曲线的极坐标方程, ${\theta }_{0}$ 为向径 ${OB}$ 的极角 $\left( {0 < {\theta }_{0} < {2\pi }}\right)$ . 当 $0 \leq$ $\theta \leq {\theta }_{0}$ 时, $0 \leq t \leq {2\pi }$ ,
$$ {r}^{2} = {x}^{2} + {y}^{2} = {a}^{2}{\left( \cos t + t\sin t\right) }^{2} + {a}^{2}{\left( \sin t - t\cos t\right) }^{2} $$
$$ = {a}^{2}\left( {1 + {t}^{2}}\right) \text{ . } $$
又
$$ \tan \theta = \frac{y}{x} = \frac{\sin t - t\cos t}{\cos t + t\sin t} \Rightarrow \mathrm{d}\theta = \frac{{t}^{2}\mathrm{\;d}t}{1 + {t}^{2}}, $$
于是
$$ S = \frac{1}{2}{\int }_{0}^{2\pi }{a}^{2}\left( {1 + {t}^{2}}\right) \cdot \frac{{t}^{2}}{1 + {t}^{2}}\mathrm{\;d}t + \pi {a}^{2} $$
$$ = \frac{4}{3}{\pi }^{3}{a}^{2} + \pi {a}^{2}. $$
求曲线的弧长. 因为 ${x}^{\prime }\left( t\right) = {at}\cos t,{y}^{\prime }\left( t\right) = {bt}\sin t$ ,所以弧长为
$$ s = {\int }_{0}^{2\pi }\sqrt{{x}^{\prime 2}\left( t\right) + {y}^{\prime 2}\left( t\right) }\mathrm{d}t = {\int }_{0}^{2\pi }{at}\mathrm{\;d}t = 2{\pi }^{2}a. $$
💡 答案解析
解法 1 如图 3.8 所示. 所围图形面积为
$$ S = - {\int }_{0}^{2\pi }y\left( t\right) {x}^{\prime }\left( t\right) \mathrm{d}t = \frac{4}{3}{\pi }^{3}{a}^{2} + \pi {a}^{2}. $$
解法 $\displaystyle{2S} = \frac{1}{2}{\int }_{0}^{2\pi }x\mathrm{\;d}y - y\mathrm{\;d}x +}$ $\bigtriangleup {OAB}$ 的面积,
$$ \bigtriangleup {OAB}\text{ 的面积 } = \frac{1}{2} \cdot a \cdot {2\pi a} = \pi {a}^{2}\text{ , } $$
$$ \frac{1}{2}{\int }_{0}^{2\pi }x\mathrm{\;d}y - y\mathrm{\;d}x = \frac{1}{2}{\int }_{0}^{2\pi }{a}^{2}\left\lbrack {\left( {\cos t + t\sin t}\right) t\sin t}\right. $$
$$ - \left( {\sin t - t\cos t}\right) t\cos t\rbrack \mathrm{d}t $$
$$ = \frac{1}{2}{\int }_{0}^{2\pi }{a}^{2}{t}^{2}\mathrm{\;d}t = \frac{4}{3}{\pi }^{3}{a}^{2}, $$
即得 $S = \frac{4}{3}{\pi }^{3}{a}^{2} + \pi {a}^{2}$ .
解法 3 用极坐标. $S = \frac{1}{2}{\int }_{0}^{{\theta }_{0}}{r}^{2}\left( \theta \right) \mathrm{d}\theta + \bigtriangleup {OAB}$ 的面积,其中 $r =$ $r\left( \theta \right)$ 为曲线的极坐标方程, ${\theta }_{0}$ 为向径 ${OB}$ 的极角 $\left( {0 < {\theta }_{0} < {2\pi }}\right)$ . 当 $0 \leq$ $\theta \leq {\theta }_{0}$ 时, $0 \leq t \leq {2\pi }$ ,
$$ {r}^{2} = {x}^{2} + {y}^{2} = {a}^{2}{\left( \cos t + t\sin t\right) }^{2} + {a}^{2}{\left( \sin t - t\cos t\right) }^{2} $$
$$ = {a}^{2}\left( {1 + {t}^{2}}\right) \text{ . } $$
又
$$ \tan \theta = \frac{y}{x} = \frac{\sin t - t\cos t}{\cos t + t\sin t} \Rightarrow \mathrm{d}\theta = \frac{{t}^{2}\mathrm{\;d}t}{1 + {t}^{2}}, $$
于是
$$ S = \frac{1}{2}{\int }_{0}^{2\pi }{a}^{2}\left( {1 + {t}^{2}}\right) \cdot \frac{{t}^{2}}{1 + {t}^{2}}\mathrm{\;d}t + \pi {a}^{2} $$
$$ = \frac{4}{3}{\pi }^{3}{a}^{2} + \pi {a}^{2}. $$
求曲线的弧长. 因为 ${x}^{\prime }\left( t\right) = {at}\cos t,{y}^{\prime }\left( t\right) = {bt}\sin t$ ,所以弧长为
$$ s = {\int }_{0}^{2\pi }\sqrt{{x}^{\prime 2}\left( t\right) + {y}^{\prime 2}\left( t\right) }\mathrm{d}t = {\int }_{0}^{2\pi }{at}\mathrm{\;d}t = 2{\pi }^{2}a. $$