자연계열 (오전) 1번

1. 넓이$\displaystyle =\pi(1-a^2\sin^2\theta)$. $\displaystyle \theta=\frac\pi6$이면, 원의 면적$\displaystyle =\pi\left(1-\frac{a^2}4\right)$.

2. $\displaystyle dt=\sec^2\theta\,d\theta,\ 1+\tan^2\theta=\sec^2\theta$이므로주어진 식은 $\displaystyle \pi\int_0^{\theta_0}(1-a^2\sin^2\theta)\sec^2\theta\,d\theta=\pi\int_0^{\theta_0}\sec^2\theta-a^2\tan^2\theta\,d\theta=\pi\int_0^{\theta_0}(1-a^2)\sec^2\theta+a^2\,d\theta=\pi((1-a^2)\tan\theta_0+a^2\theta_0)$

$\displaystyle \tan\theta_0=\frac1{\sqrt{a^2-1}}$이므로, 적분값은 $\displaystyle \pi\frac{\theta_0\tan^2\theta_0-\tan\theta_0+\theta_0}{\tan^2\theta_0}$

3.$\displaystyle \begin{cases}x(\theta)=-\frac{a\tan^2\theta}{\tan^2\theta+1}=-\frac{a\tan^2\theta}{\sec^2\theta}=-a\sin^2\theta\\y(\theta)=\frac{a\tan\theta}{\tan^2\theta+1}=a\sin\theta\cos\theta\end{cases}$$\displaystyle x^{\prime}(\theta)=-2a\sin\theta\cos\theta$$\displaystyle y^{\prime}(\theta)=a\cos^2\theta-a\sin^2\theta$$\displaystyle \{x^{\prime}(\theta)\}^2+\{y^{\prime}(\theta)\}^2=4a^2\sin^2\theta\cos^2\theta+a^2(\cos^2\theta-\sin^2\theta)^2=a^2(\cos^2\theta+\sin^2\theta)^2=a^2$

곡선의 길이$\displaystyle =\int_0^{\theta_0}a\,d\theta=a\theta_0$