笛卡尔叶形曲线所围图形的面积

求笛卡尔叶形曲线 $x^3+y^3=3axy\ (a>0)$ 所围图形的面积.    

解答:  在极坐标  $$\bex  x=r\cos t,\quad y=r\sin t  \eex$$  下, 笛卡尔叶形线为  $$\beex  \bea  r^3\cos^3 t+r^3\sin^3t&=3ar^2\sin t\cos t,\\  r\cos^3t+r\sin^t&=3a\sin t\cos t,\\  r&=\frac{3a\sin t\cos t}{\cos^3t+\sin^3t}.  \eea  \eeex$$  由于 $r\geq 0$, 而 $0\leq t\leq \pi/2$, 而所求为  $$\beex  \bea  S&=\frac{1}{2}\int_0^{\pi/2} r^2(t)\rd t\\  &=\frac{9a^2}{2}\int_0^{\pi/2} \frac{\sin^2t\cos^2t}{(\cos^3t+\sin^3t)^2}\rd t\\  &=\frac{9a^2}{2}\int_0^{\pi/2}  \frac{\tan^2t\sec^2t}{(1+\tan^3t)^2}\rd t\\  &=\frac{9a^2}{2}\int_0^{\pi/2} \frac{\tan^2t}{(1+\tan^3t)^2}\rd \tan t\\  &=\frac{3a^2}{2}\int_0^{\pi/2} \frac{\rd \tan^3t}{(1+\tan^3t)^2}\\  &=\frac{3a^2}{2}\cdot\left.\frac{-1}{1+\tan^3t}\right|_0^{\pi/2}\\  &=\frac{3a^2}{2}.  \eea  \eeex$$    

转载于:https://www.cnblogs.com/zhangzujin/p/3540777.html