第10章 10.5

(1)

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

(2)

$y=0$ のとき $x=\pm\sqrt{3}$ これから

$V=\pi\int_{-\sqrt{3}}^{\sqrt{3}}(3-x^2)^2\,dx=2\pi\int_0^{\sqrt{3}}(9-6x^2+x^4)\,dx=2\pi\left[9x-2x^3+\frac{1}{5}x^5\right]_0^{\sqrt{3}}=2\pi\left(9\sqrt{3}-6\sqrt{3}+\frac{9}{5}\sqrt{3}\right)=\frac{48\sqrt{3}}{5}\pi$

(3)

$V=\pi\int_0^3 x(3-x)\,dx=\pi\left[\frac{3}{2}x^2-\frac{1}{3}x^3\right]_0^3=\pi\left(\frac{27}{2}-9\right)=\frac{9}{2}\pi$

(4)

$y=0$ のとき $x=1,-\dfrac{2}{3}$ これから

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

$=\pi\left[\frac{3}{4}x^4-\frac{4}{3}x^3-\frac{1}{2}x^2+2x\right]_{-\frac{2}{3}}^{1}=\pi\left(\frac{3}{4}-\frac{4}{3}-\frac{1}{2}+2-\frac{4}{27}-\frac{32}{81}+\frac{2}{9}+\frac{4}{3}\right)=\frac{625}{324}\pi$