第10章 10.6

(1)

２つの放物線の交点は

$x^2 = 4y \qquad y = \frac{x^2}{4}$

これをもう１つの式に代入して

$\left(\frac{x^2}{4}\right)^2 = 4x \qquad x^4 = 64x \qquad x(x^3 - 64) = 0$

$x = 0, 4$ これから

$V = \pi\int_0^4 4x\,dx - \pi\int_0^4 \left(\frac{x^2}{4}\right)^2 dx = \pi[2x^2]_0^4 - \pi\int_0^4 \frac{x^4}{16}\,dx = 32\pi - \pi\left[\frac{x^5}{80}\right]_0^4 = 32\pi - \frac{64}{5}\pi = \frac{96}{5}\pi$

(2)

$V = \pi\int_0^{\frac{\pi}{4}} \tan^2 x\,dx = \pi\int_0^{\frac{\pi}{4}} \frac{\sin^2 x}{\cos^2 x}\,dx = \pi\int_0^{\frac{\pi}{4}} \frac{1 - \cos^2 x}{\cos^2 x}\,dx = \pi\int_0^{\frac{\pi}{4}} \left(\frac{1}{\cos^2 x} - 1\right) dx = \pi[\tan x - x]_0^{\frac{\pi}{4}}$

$= \pi\left(1 - \frac{\pi}{4}\right)$

(3)

２曲線の交点は

$\frac{1}{2}x^2 = 3 - x^2 \qquad \frac{3}{2}x^2 = 3 \qquad x^2 = 2 \qquad x = \pm\sqrt{2}$

$V = \pi\int_{-\sqrt{2}}^{\sqrt{2}} (3 - x^2)^2\,dx - \pi\int_{-\sqrt{2}}^{\sqrt{2}} \left(\frac{1}{2}x^2\right)^2 dx = \pi\int_{-\sqrt{2}}^{\sqrt{2}} (9 - 6x^2 + x^4)\,dx - \pi\int_{-\sqrt{2}}^{\sqrt{2}} \frac{1}{4}x^4\,dx$

$= \pi\left[9x - 2x^3 + \frac{1}{5}x^5\right]_{-\sqrt{2}}^{\sqrt{2}} - \pi\left[\frac{1}{20}x^5\right]_{-\sqrt{2}}^{\sqrt{2}} = \pi\left(\frac{58}{5}\sqrt{2} - \frac{2}{5}\sqrt{2}\right) = \frac{56\sqrt{2}}{5}\pi$

(4)

$\sqrt{\frac{x}{2}} + \sqrt{\frac{y}{3}} = 1 \qquad \text{定義域} : x, y \geq 0$

与式と x 軸との交点は

$\sqrt{\frac{x}{2}} + \sqrt{\frac{0}{3}} = 1 \qquad \sqrt{\frac{x}{2}} = 1 \qquad x = 2$

$\sqrt{\dfrac{y}{3}} = 1 - \sqrt{\dfrac{x}{2}} \qquad \dfrac{y}{3} = 1 - 2\sqrt{\dfrac{x}{2}} + \dfrac{x}{2} \qquad y = 3\left(1 - 2\sqrt{\dfrac{x}{2}} + \dfrac{x}{2}\right)$

$y^2 = 9\left(1 - 2\sqrt{\dfrac{x}{2}} + \dfrac{x}{2}\right)\left(1 - 2\sqrt{\dfrac{x}{2}} + \dfrac{x}{2}\right) = 9\left(1 - 2\sqrt{\dfrac{x}{2}} + \dfrac{x}{2} - 2\sqrt{\dfrac{x}{2}} + 4\cdot\dfrac{x}{2} - 2\sqrt{\dfrac{x}{2}}\cdot\dfrac{x}{2} + \dfrac{x}{2} - 2\sqrt{\dfrac{x}{2}}\cdot\dfrac{x}{2} + \dfrac{x^2}{4}\right)$

$= 9\left(1 - 4\sqrt{\dfrac{x}{2}} + 3x - 2\sqrt{\dfrac{x}{2}}\cdot x + \dfrac{x^2}{4}\right) = 9\left(1 - 2\sqrt{2}\cdot\sqrt{x} + 3x - \sqrt{2}\cdot\sqrt{x^3} + \dfrac{1}{4}x^2\right)$

これから

$V = \pi\int_0^2 y^2\,dx = 9\pi\int_0^2\left(1 - 2\sqrt{2}\cdot\sqrt{x} + 3x - \sqrt{2}\cdot\sqrt{x^3} + \dfrac{1}{4}x^2\right)dx$

$= 9\pi\left[x - \dfrac{4\sqrt{2}}{3}\cdot\sqrt{x^3} + \dfrac{3}{2}x^2 - \dfrac{2\sqrt{2}}{5}\sqrt{x^5} + \dfrac{1}{12}x^3\right]_0^2 = 9\pi\left\{2 - \dfrac{16}{3} + 6 - \dfrac{16}{5} + \dfrac{2}{3}\right\}$

$= 9\pi\left(\dfrac{30 - 80 + 90 - 48 + 10}{15}\right) = 9\pi\cdot\dfrac{2}{15} = \dfrac{6}{5}\pi$