高専の数学解説

第10章 10.11

(1)

y=0y=0 のとき x2+1=4x^2+1=4 x2=3x^2=3 x=±3x=\pm\sqrt{3}

V=2π∫03(4−x2−1)2 dx=2π∫03(4−x2−24−x2+1)dx=2π∫03(5−x2)dx−4π∫034−x2 dxV=2\pi\int_0^{\sqrt{3}}(\sqrt{4-x^2}-1)^2\,dx=2\pi\int_0^{\sqrt{3}}(4-x^2-2\sqrt{4-x^2}+1)dx=2\pi\int_0^{\sqrt{3}}(5-x^2)dx-4\pi\int_0^{\sqrt{3}}\sqrt{4-x^2}\,dx

第２項で x=2sin⁡tx=2\sin t とおくと dx=2cos⁡t dtdx=2\cos t\,dt x:0→3x:0\to\sqrt{3} t:0→π3t:0\to\dfrac{\pi}{3} これから

=2π[5x−13x3]03−4π∫0π34−4sin⁡2t⋅2cos⁡t dt=2π(53−3)−16π∫0π3cos⁡2t dt=2\pi\left[5x-\frac{1}{3}x^3\right]_0^{\sqrt{3}}-4\pi\int_0^{\frac{\pi}{3}}\sqrt{4-4\sin^2 t}\cdot 2\cos t\,dt=2\pi(5\sqrt{3}-\sqrt{3})-16\pi\int_0^{\frac{\pi}{3}}\cos^2 t\,dt

=8π3−8π∫0π3(1+cos⁡2t)dt=8π3−8π[t+12sin⁡2t]0π3=8π3−8π(π3+34)=63π−83π2=8\pi\sqrt{3}-8\pi\int_0^{\frac{\pi}{3}}(1+\cos 2t)dt=8\pi\sqrt{3}-8\pi\left[t+\frac{1}{2}\sin 2t\right]_0^{\frac{\pi}{3}}=8\pi\sqrt{3}-8\pi\left(\frac{\pi}{3}+\frac{\sqrt{3}}{4}\right)=6\sqrt{3}\pi-\frac{8}{3}\pi^2

(2)

2π∫02{(1−x24+1)2−(−1−x24+1)2}dx2\pi\int_0^2\left\{\left(\sqrt{1-\frac{x^2}{4}}+1\right)^2-\left(-\sqrt{1-\frac{x^2}{4}}+1\right)^2\right\}dx

x=2sin⁡tx=2\sin t とおくと dx=2cos⁡t dtdx=2\cos t\,dt x:0→2x:0\to 2 t:0→π2t:0\to\dfrac{\pi}{2} これから

=2π∫0π2{(cos⁡t+1)2−(−cos⁡t+1)2}⋅2cos⁡t dt=4π∫0π2(cos⁡2t+2cos⁡t+1−cos⁡2t+2cos⁡t−1)cos⁡t dt=2\pi\int_0^{\frac{\pi}{2}}\{(\cos t+1)^2-(-\cos t+1)^2\}\cdot 2\cos t\,dt=4\pi\int_0^{\frac{\pi}{2}}(\cos^2 t+2\cos t+1-\cos^2 t+2\cos t-1)\cos t\,dt

=16π∫0π2cos⁡2t dt=16π⋅12⋅π2=4π2=16\pi\int_0^{\frac{\pi}{2}}\cos^2 t\,dt=16\pi\cdot\frac{1}{2}\cdot\frac{\pi}{2}=4\pi^2

(3)

V=π∫0e(1)2dx−π∫1e(log⁡x)2dx=π[x]0e−π{[log⁡x⋅(xlog⁡x−x)]1e−∫1e1x(xlog⁡x−x)dx}V=\pi\int_0^e(1)^2dx-\pi\int_1^e(\log x)^2dx=\pi[x]_0^e-\pi\left\{[\log x\cdot(x\log x-x)]_1^e-\int_1^e\frac{1}{x}(x\log x-x)dx\right\}

=eπ−π{(e−e)−∫1e(log⁡x−1)dx}=eπ+π[xlog⁡x−x−x]1e=eπ+π(e−e−e+1+1)=2π=e\pi-\pi\{(e-e)-\int_1^e(\log x-1)dx\}=e\pi+\pi[x\log x-x-x]_1^e=e\pi+\pi(e-e-e+1+1)=2\pi

(4)

V=π∫a−ra+r(r2−(x−a)2+b)2−(−r2−(x−a)2+b)2 dxV=\pi\int_{a-r}^{a+r}(\sqrt{r^2-(x-a)^2}+b)^2-(-\sqrt{r^2-(x-a)^2}+b)^2\,dx

x−a=rsin⁡tx-a=r\sin t とおくと dx=rcos⁡tdx=r\cos t x:a−r→a+rx:a-r\to a+r t:−π2→π2t:-\dfrac{\pi}{2}\to\dfrac{\pi}{2} これから

=π∫−π2π2{(rcos⁡t+b)2−(−rcos⁡t+b)2}rcos⁡t dt= \pi\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \{(r\cos t+b)^2-(-r\cos t+b)^2\}r\cos t\,dt

=π∫−π2π2(r2cos⁡2t+2rbcos⁡t+b2−r2cos⁡2t+2rbcos⁡t−b2)rcos⁡t dt=π∫−π2π24rbcos⁡t⋅rcos⁡t dt= \pi\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (r^2\cos^2 t+2rb\cos t+b^2-r^2\cos^2 t+2rb\cos t-b^2)r\cos t\,dt = \pi\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 4rb\cos t\cdot r\cos t\,dt

=4πr2b∫−π2π2cos⁡2t dt=2πr2b∫−π2π2(1+cos⁡2t)dt=2πr2b[t+12cos⁡2t]−π2π2=2πr2b(π2+π2)=2π2r2b= 4\pi r^2 b\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^2 t\,dt = 2\pi r^2 b\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(1+\cos 2t)dt = 2\pi r^2 b\left[t+\frac{1}{2}\cos 2t\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = 2\pi r^2 b\left(\frac{\pi}{2}+\frac{\pi}{2}\right) = 2\pi^2 r^2 b

解説: r31bn1z