Eufisky - The lost book

## 涉及到星形线的一个运动轨迹问题

（K神问题）长度为$L$的木棍从竖直位置开始沿墙壁一直滑到地面（墙壁与地面垂直），则木棍扫过的面积为？

$\left\{ \begin{array}{l}x = \alpha L\cos \theta \\y = \left( {1 - \alpha } \right)L\sin \theta\end{array} \right. \Rightarrow \frac{{{x^2}}}{{{\alpha ^2}{L^2}}} + \frac{{{y^2}}}{{{{\left( {1 - \alpha } \right)}^2}{L^2}}} = 1.$

$\Rightarrow y = \sqrt {{{\left( {1 - \alpha } \right)}^2}{L^2} - {x^2}{{\left( {\frac{{1 - \alpha }}{\alpha }} \right)}^2}} ,0 < \alpha < 1,x,L > 0.$

\begin{align*}y &= \sqrt {{{\left( {1 - \alpha } \right)}^2}{L^2} - {x^2}{{\left( {\frac{{1 - \alpha }}{\alpha }} \right)}^2}}  = \sqrt {{{\left( {1 - \alpha } \right)}^2}{L^2} - {x^2}{{\left( {\frac{1}{\alpha } - 1} \right)}^2}} \\&= \sqrt {{L^2} - {x^2} + {L^2}{\alpha ^2} - 2{L^2}\alpha  - \frac{{{x^2}}}{{{\alpha ^2}}} + 2\frac{{{x^2}}}{\alpha }}.\end{align*}

\begin{align*}g\left( \alpha  \right) &\le g\left( {{{\left( {\frac{x}{L}} \right)}^{\frac{2}{3}}}} \right) = {L^2} - {x^2} + {L^2}{\left( {\frac{x}{L}} \right)^{\frac{4}{3}}} - 2{L^2}{\left( {\frac{x}{L}} \right)^{\frac{2}{3}}} - {x^2}{\left( {\frac{L}{x}} \right)^{\frac{4}{3}}} + 2{x^2}{\left( {\frac{L}{x}} \right)^{\frac{2}{3}}}\\&= {L^2} - {x^2} + 3{L^{\frac{2}{3}}}{x^{\frac{4}{3}}} - 3{L^{\frac{4}{3}}}{x^{\frac{2}{3}}}.\end{align*}

$\int_0^L {\sqrt {{L^2} - {x^2} + 3{L^{\frac{2}{3}}}{x^{\frac{4}{3}}} - 3{L^{\frac{4}{3}}}{x^{\frac{2}{3}}}} dx} \underline{\underline {{\text{令}u = \frac{x}{L}}}} {L^2}\int_0^1 {\sqrt {1 - {u^2} + 3{u^{\frac{4}{3}}} - 3{u^{\frac{2}{3}}}} du} = \frac{{3\pi }}{{32}}{L^2}.$

\begin{align*}\int_0^1 {\sqrt {1 - {u^2} + 3{u^{\frac{4}{3}}} - 3{u^{\frac{2}{3}}}} du}  &= \int_0^1 {\sqrt {{{\left( {1 - {u^{\frac{2}{3}}}} \right)}^3}} du} \underline{\underline {{\text{令}u = {{\cos }^3}v}}} 3\int_0^{\frac{\pi }{2}} {{{\sin }^4}v{{\cos }^2}vdv} \\&= 3\int_0^{\frac{\pi }{2}} {{{\sin }^4}vdv}  - 3\int_0^{\frac{\pi }{2}} {{{\sin }^6}vdv}  = \frac{{3\pi }}{{32}}.\end{align*}