第5章 5.9

(1)

$x = \frac{1}{6}t^6 \qquad y = 2 - \frac{t^4}{4} \qquad [0, \sqrt[4]{8}]$

$x' = t^5 \qquad y' = -t^3$

$S = \int_0^{\sqrt[4]{8}} \sqrt{(t^5)^2 + (-t^3)^2}\, dt = \int_0^{\sqrt[4]{8}} \sqrt{t^{10} + t^6}\, dt = \int_0^{\sqrt[4]{8}} t^3 \sqrt{t^4 + 1}\, dt$

$t^4 + 1 = v$ とおくと $4t^3 dt = dv$ $t : 0 \to \sqrt[4]{8}$ $v : 1 \to 9$ これから

$= \int_1^9 \frac{1}{4}\sqrt{v}\, dv = \frac{1}{4}\left[\frac{2}{3}\sqrt{v^3}\right]_1^9 = \frac{1}{6}(27 - 1) = \frac{13}{3}$

(2)

$x = 2 + 3t^2 \qquad y = 3t - t^3$

$x' = 6t \qquad y' = 3 - 3t^2$

$S = \int_0^t \sqrt{(6t)^2 + (3-3t^2)^2}\, dt = \int_0^t \sqrt{36t^2 + 9 - 18t^2 + 9t^4}\, dt = \int_0^t 3\sqrt{1 + 2t^2 + t^4}\, dt = \int_0^t 3(1+t^2)\, dt$

$= 3\left[t + \frac{1}{3}t^3\right]_0^t = 3t + t^3$