【微分積分】4-7-3 面積・曲線の長さの計算｜問題集

曲線\(C\)は\(x\)軸、\(y\)軸に関して対称であり、
\(\left\{\begin{array}{l}x=a\cos^3t \\ y=a\sin^3t\end{array}\right.\)
\((0\leqq t\leqq 2\pi)\)
とパラメータ表示できる。
第\(1\)象限の面積を\(4\)倍すればいよいので、求める面積\(S\)は
\(\displaystyle S=4\int_0^{a}ydx\)
\(\displaystyle =4\int_{\frac{\pi}{2}}^{0}a\sin^3t・-3a\cos^2t\sin tdt\)
\(\displaystyle =12a^2\int_0^{\frac{\pi}{2}}\sin^4t+\cos^2tdt\)
\(\displaystyle =12a^2\int_0^{\frac{\pi}{2}}(\sin^4t-\sin^6t)dt\)
\(\displaystyle =12a^2\left(\frac{3!!}{4!!}・\frac{\pi}{2}-\frac{5!!}{6!!}・\frac{\pi}{2}\right)\)
\(\displaystyle =\frac{3}{8}\pi a^2\)

求める弧長\(L\)は
\(\displaystyle L=4\int_0^{\frac{\pi}{2}}\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt\)
\(\displaystyle =4\int_0^{\frac{\pi}{2}}\sqrt{9a^2\cos^4t\sin^2t+9a^2\sin^4t\cos^2t}dt\)
\(\displaystyle =4\int_0^{\frac{\pi}{2}}\sqrt{9a^2\sin^2t\cos^2t}dt\)
\(\displaystyle =\int_0^{\frac{\pi}{2}}12a\sin t\cos tdt\)
\(\displaystyle =\left[-3a\cos2t\right]_0^{\frac{\pi}{2}}\)
\(\displaystyle =6a\)

曲線\(C\)は閉曲線で\(0\leqq \theta\leqq 2\pi\)のとき、\(r\geqq0\)なので、求める面積\(S\)は
\(\displaystyle S=\frac{1}{2}\int_0^{2\pi}r^2d\theta\)
\(\displaystyle =\frac{1}{2}\int_0^{2\pi}a^2(1+2\cos\theta+\cos^2\theta)d\theta\)
\(\displaystyle =a^2\int_0^{2\pi}\left(\frac{3}{4}+\cos\theta+\frac{1}{4}\cos2\theta\right)d\theta\)
\(\displaystyle =a^2\left[\frac{3}{4}\theta+\sin\theta+\frac{1}{8}\sin2\theta\right]_0^{2\pi}\)
\(\displaystyle =\frac{3}{2}\pi a^2\)

求める弧長\(L\)は
\(\displaystyle L=2\int_0^{\pi}\sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}d\theta\)
\(\displaystyle =2\int_0^{\pi}\sqrt{a^2(1+\cos\theta)^2+(-a\sin\theta)^2}d\theta\)
\(\displaystyle =2\int_0^{\pi}\sqrt{4a^2\cos^2\frac{\theta}{2}}d\theta\)
\(\displaystyle =\int_0^{\pi}4a\cos\frac{\theta}{2}d\theta\)
\(\displaystyle =\left[8a\sin\frac{\theta}{2}\right]_0^{\pi}\)
\(\displaystyle =8a\)