第5章 5.18

(1)

$x = a(t - \sin t) \qquad y = a(1 - \cos t) \qquad 0 \le t \le 2\pi$

$S = \int_0^{2\pi} a(1-\cos t)\cdot a(1-\cos t)dt = \int_0^{2\pi} a^2(1-\cos t)^2 dt = \int_0^{2\pi} a^2(1-2\cos t+\cos^2 t)dt$

$= \int_0^{2\pi} a^2(1-2\cos t+\frac{1+\cos 2t}{2})dt = a^2\left[t-2\sin t+\frac{1}{2}t+\frac{1}{4}\sin 2t\right]_0^{2\pi} = 3\pi a^2$

$V = \pi\int_0^{2\pi} a^3(1-\cos t)^3 dt = \pi a^3\int_0^{2\pi}(1-3\cos t+3\cos^2 t-\cos^3 t)dt$

ここで$\cos t$、$\cos^3 t$を$0\sim2\pi$で積分すると$0$となるので

$= \pi a^3\int_0^{2\pi}(1+3\cos^2 t)dt = \pi a^3\int_0^{2\pi}(1+3\frac{1+\cos 2t}{2})dt = \pi a^3\left[t+\frac{3}{2}t+\frac{3}{4}\sin 2t\right]_0^{2\pi} = 5\pi^2 a^3$

$s = \int_0^{2\pi}\sqrt{a^2(1-\cos t)^2+a^2\sin^2 t}\,dt = \int_0^{2\pi}\sqrt{a^2-2a^2\cos t+a^2}\,dt = \int_0^{2\pi} a\sqrt{2-2\cos t}\,dt$

$= \int_0^{2\pi} 2a\sin\frac{t}{2}\,dt = 2a\left[-2\cos\frac{t}{2}\right]_0^{2\pi} = 4a(-\cos\pi+\cos 0) = 8a$

(2)

$x = a\cos^3 t \qquad y = a\sin^3 t \qquad x: 0 \to a \qquad t: \frac{\pi}{2}\to 0$

$S = 4\int_{\pi/2}^0 a\sin^3 t\cdot a\cdot 3\cos^2 t\cdot(-\sin t)dt = 4\int_0^{\pi/2} 3a^2\sin^4 t\cdot\cos^2 t\,dt = 12a^2\int_0^{\pi/2}\sin^4 t(1-\sin^2 t)dt$

$= 12a^2\int_0^{\pi/2}\sin^4 t-\sin^6 t\,dt = 12a^2\left(\frac{3}{4}\cdot\frac{1}{2}\cdot\frac{\pi}{2}-\frac{5}{6}\cdot\frac{3}{4}\cdot\frac{1}{2}\cdot\frac{\pi}{2}\right) = 12a^2\left(\frac{3}{16}\pi\cdot\frac{1}{6}\right) = \frac{3}{8}\pi a^2$

$V = \pi\int_{-a}^a y^2 dx = 2\pi\int_0^a y^2 dx = -2\pi\int_{\pi/2}^0 a^2\sin^6 t\cdot 3a\cos^2 t\cdot(-\sin t)dt = 6\pi a^3\int_0^{\pi/2}\sin^7 t\cdot\cos^2 t\,dt$

$= 6\pi a^3\int_0^{\pi/2}(\sin^7 t-\sin^9 t)dt = 6\pi a^3\left(\frac{6}{7}\cdot\frac{4}{5}\cdot\frac{2}{3}-\frac{8}{9}\cdot\frac{6}{7}\cdot\frac{4}{5}\cdot\frac{2}{3}\right) = 6\pi a^3\left(\frac{16}{35}\cdot\frac{1}{9}\right) = \frac{32}{105}\pi a^3$

$s = 4\int_0^{\pi/2}\sqrt{(3a\sin^2 t\cdot\cos t)^2+(-3a\sin t\cdot\cos^2 t)^2}\,dt = 4\int_0^{\pi/2}\sqrt{9a^2(\sin^2 t\cdot\cos^2 t)(\sin^2 t+\cos^2 t)}\,dt$

$= 4\int_0^{\pi/2} 3a\sin t\cdot\cos t\,dt \qquad \sin t = v \text{ とおくと} \qquad \cos t\,dt = dv \qquad t: 0\to\frac{\pi}{2} \qquad v: 0\to 1 \qquad \text{これから}$

$= 12a\int_0^1 t\,dt = 12a\left[\frac{1}{2}t^2\right]_0^1 = 6a$

(3)

$x = t^2 \qquad y = t\left(1-\frac{t^2}{3}\right) \qquad (|t| \le \sqrt{3})$

$S = \int_{-\sqrt{3}}^{\sqrt{3}} t\left(1-\frac{t^2}{3}\right)\cdot 2t\,dt = \int_{-\sqrt{3}}^{\sqrt{3}} \left(2t^2-\frac{2t^4}{3}\right)dt = \left[\frac{2}{3}t^3-\frac{2}{15}t^5\right]_{-\sqrt{3}}^{\sqrt{3}} = \frac{8}{5}\sqrt{3}$

体積は下図より

$V = \pi\int_0^{3} y^2\,dx = \pi\int_0^{\sqrt{3}} t^2\left(1-\frac{t^2}{3}\right)^2\cdot 2t\,dt = \pi\int_0^{\sqrt{3}} \left(2t^3-\frac{4}{3}t^5+\frac{2}{9}t^7\right)dt = \left[\frac{1}{2}t^4-\frac{2}{9}t^6+\frac{1}{36}t^8\right]_0^{\sqrt{3}} = \frac{3}{4}\pi$

$s = \int_{-\sqrt{3}}^{\sqrt{3}} \sqrt{\left\{1-\frac{t^2}{3}+t\left(-\frac{2}{3}t\right)\right\}^2+(2t)^2}\,dt = \int_{-\sqrt{3}}^{\sqrt{3}} \sqrt{(1-t^2)^2+4t^2}\,dt = \int_{-\sqrt{3}}^{\sqrt{3}} (1+t^2)\,dt$

$= \left[t+\frac{1}{3}t^3\right]_{-\sqrt{3}}^{\sqrt{3}} = 4\sqrt{3}$