📝 题目
例 5 求曲线 ${\left( \frac{x}{a}\right) }^{\frac{2}{3}} + {\left( \frac{y}{b}\right) }^{\frac{2}{3}} = 1\left( {a > 0,b > 0}\right)$ 的全长.
💡 答案与解析
解 将曲线改写成参数方程, 并计算微弧:
$$ x = a{\cos }^{3}t,\;y = b{\sin }^{3}t, $$
$$ \frac{\mathrm{d}x}{\mathrm{\;d}t} = - {3a}{\cos }^{2}t\sin t,\;\frac{\mathrm{d}y}{\mathrm{\;d}t} = {3b}{\sin }^{2}t\cos t, $$
$$ \mathrm{d}s = \sqrt{{\left( \frac{\mathrm{d}x}{\mathrm{\;d}t}\right) }^{2} + {\left( \frac{\mathrm{d}y}{\mathrm{\;d}t}\right) }^{2}} = 3\cos t\sin t\sqrt{{a}^{2}{\cos }^{2}t + {b}^{2}{\sin }^{2}t}. $$
因此
$$ s = 4{\int }_{0}^{\frac{\pi }{2}}\sqrt{{\left( \frac{\mathrm{d}x}{\mathrm{\;d}t}\right) }^{2} + {\left( \frac{\mathrm{d}y}{\mathrm{\;d}t}\right) }^{2}}\mathrm{\;d}t $$
$$ = {12}{\int }_{0}^{\frac{\pi }{2}}\cos t\sin t\sqrt{{a}^{2}{\cos }^{2}t + {b}^{2}{\sin }^{2}t\mathrm{\;d}t} $$
$$ \overset{u = \sin t}{ = }{12}{\int }_{0}^{1}u\sqrt{{a}^{2}\left( {1 - {u}^{2}}\right) + {b}^{2}{u}^{2}}\mathrm{\;d}u $$
$$ \overset{v = {u}^{2}}{ = }6{\int }_{0}^{1}\sqrt{{a}^{2}\left( {1 - v}\right) + {b}^{2}v}\mathrm{\;d}v $$
$$ \overset{{c}^{2} = {b}^{2} - {a}^{2}}{ = }6{\int }_{0}^{1}\sqrt{{a}^{2} + {cv}}\mathrm{\;d}v = \frac{12}{c}{\int }_{a}^{b}{z}^{2}\mathrm{\;d}z $$
$$ = \frac{12}{c}\left( {\frac{1}{3}{b}^{3} - \frac{1}{3}{a}^{3}}\right) = \frac{4\left( {{b}^{3} - {a}^{3}}\right) }{{b}^{2} - {a}^{2}} = 4\frac{{ab} + {a}^{2} + {b}^{2}}{a + b}. $$