式を回転させてから積分する

問

$\mathrm{区}\mathrm{間}0\le x\le 1\mathrm{に}\mathrm{お}\mathrm{い}\mathrm{て}y=x\mathrm{と}y=x^2\mathrm{で}\mathrm{囲}\mathrm{ま}\mathrm{れ}\mathrm{た}\mathrm{領}\mathrm{域}\mathrm{を}y=x\mathrm{を}\mathrm{軸}\mathrm{と}\mathrm{し}\mathrm{て}\mathrm{回}\mathrm{転}\mathrm{さ}\mathrm{せ}\mathrm{た}\mathrm{時}\mathrm{の}\mathrm{体}\mathrm{積}$

$y=x,y=x^2$を$-\frac{\pi }{4}$回転させる

求めたい$-\frac{\pi }{4}$回転させた側を$(x,y)$とすると，それを$\frac{\pi }{4}$回転させた結果として$(X,Y)$として考える． $$\begin{array}{rcl}\left[\begin{array}{c}X\\ Y\end{array}\right]& =& \left[\begin{array}{cc}\cos \left(\frac{\pi }{4}\right)& -\sin \left(\frac{\pi }{4}\right)\\ \sin \left(\frac{\pi }{4}\right)& \cos \left(\frac{\pi }{4}\right)\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]\\ & =& \left[\begin{array}{cc}\frac{\sqrt{2}}{2}& -\frac{\sqrt{2}}{2}\\ \frac{\sqrt{2}}{2}& \frac{\sqrt{2}}{2}\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]\\ & =& \frac{\sqrt{2}}{2}\left[\begin{array}{cc}1& -1\\ 1& 1\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]\end{array}$$ $$\{\begin{array}{rcl}X& =& \frac{\sqrt{2}}{2}(x-y)\\ Y& =& \frac{\sqrt{2}}{2}(x+y)\end{array}$$ $$\{\begin{array}{rcl}X& =& \alpha (x-y)\\ Y& =& \alpha (x+y)\end{array}\cdots \alpha =\frac{\sqrt{2}}{2}$$ $$\begin{array}{rcl}f& =& Y-X\\ & =& \alpha (x+y)-\left\{\alpha (x-y)\right\}\\ & =& \alpha (x+y-x+y)\\ & =& \alpha 2y\\ & =& \sqrt{2}y\\ f(y)& =& 0\\ \sqrt{2}y& =& 0\\ y& =& \frac{0}{\sqrt{2}}=0\cdots X\mathrm{軸}\\ \\ g& =& Y-X^2\\ & =& \alpha (x+y)-{\left\{\alpha (x-y)\right\}}^2\\ & =& \alpha (x+y)-{\alpha }^2(x^2-2xy+y^2)\\ & =& \alpha x+\alpha y-{\alpha }^2{x}^2+2{\alpha }^{2}xy-{\alpha }^2{y}^2\\ & =& -{\alpha }^2{y}^2+(\alpha +2{\alpha }^{2}x)y+(\alpha x-{\alpha }^2{x}^2)\\ g(y)& =& 0\\ y& =& \frac{-(\alpha +2{\alpha }^{2}x)\pm \sqrt{{(\alpha +2{\alpha }^{2}x)}^2-4\cdot (-{\alpha }^2)\cdot (\alpha x-{\alpha }^2{x}^2)}}{2(-{\alpha }^2)}\\ & =& \frac{-\alpha (1+2\alpha x)\pm \sqrt{{\alpha }^2+4{\alpha }^{3}x\cancel{+4{\alpha }^4{x}^2}+4{\alpha }^{3}x\cancel{-4{\alpha }^{3}x}}}{-2{\alpha }^2}\\ & =& \frac{-\alpha (1+2\alpha x)\pm \sqrt{{\alpha }^2+8{\alpha }^{3}x}}{-2{\alpha }^2}\\ & =& \frac{-\alpha (1+2\alpha x)\pm \sqrt{{\alpha }^2(1+8\alpha x)}}{-2{\alpha }^2}\\ & =& \frac{-\alpha (1+2\alpha x)\pm \alpha \sqrt{1+8\alpha x}}{-2{\alpha }^2}\\ & =& \frac{(1+2\alpha x)\pm \sqrt{1+8\alpha x}}{2\alpha }\\ & =& \frac{1+2\alpha x\pm \sqrt{1+8\alpha x}}{2\alpha }\\ & =& \frac{1}{2\alpha }+\frac{2\alpha x}{2\alpha }\pm \frac{\sqrt{1+8\alpha x}}{2\alpha }\\ & =& \frac{1}{2\alpha }+x\pm \frac{\sqrt{1+8\alpha x}}{2\alpha }\\ & =& \frac{1}{2\frac{\sqrt{2}}{2}}+x\pm \frac{\sqrt{1+8\frac{\sqrt{2}}{2}x}}{2\frac{\sqrt{2}}{2}}\\ & =& \frac{1}{\sqrt{2}}\frac{\sqrt{2}}{\sqrt{2}}+x\pm \frac{\sqrt{1+4\sqrt{2}x}}{\sqrt{2}}\frac{\sqrt{2}}{\sqrt{2}}\\ & =& \frac{\sqrt{2}}{2}+x\pm \frac{\sqrt{2}\sqrt{1+4\sqrt{2}x}}{2}\\ & =& \frac{\sqrt{2}}{2}+x\pm \frac{\sqrt{2(1+4\sqrt{2}x)}}{2}\\ & =& \frac{\sqrt{2}}{2}+x\pm \frac{\sqrt{2+8\sqrt{2}x}}{2}\end{array}$$ $\mathrm{区}\mathrm{間}0\le x\le 1\mathrm{に}\mathrm{お}\mathrm{け}\mathrm{る}\mathrm{回}\mathrm{転}\mathrm{軸}y=x\mathrm{の}\mathrm{長}\mathrm{さ}\mathrm{は}\sqrt{2}．$ $\mathrm{こ}\mathrm{れ}\mathrm{よ}\mathrm{り}，-\frac{\pi }{4}\mathrm{回}\mathrm{転}\mathrm{さ}\mathrm{せ}\mathrm{た}\mathrm{後}\mathrm{は}，\mathrm{区}\mathrm{間}0\mathrm{か}\mathrm{ら}\sqrt{2}\mathrm{を}\mathrm{扱}\mathrm{う}\mathrm{こ}\mathrm{と}\mathrm{に}\mathrm{な}\mathrm{る}．$
$x=0$のとき$+$側の式は $$\begin{array}{rcl}{\frac{\sqrt{2}}{2}+x+\frac{\sqrt{2+8\sqrt{2}x}}{2}|}_{x=0}& =& \frac{\sqrt{2}}{2}+0+\frac{\sqrt{2+8\sqrt{2}\cdot 0}}{2}\\ & =& \frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\\ & =& \sqrt{2}>0\end{array}$$ $-$側の式は $$\begin{array}{rcl}{\frac{\sqrt{2}}{2}+x-\frac{\sqrt{2+8\sqrt{2}x}}{2}|}_{x=0}& =& \frac{\sqrt{2}}{2}+0-\frac{\sqrt{2+8\sqrt{2}\cdot 0}}{2}\\ & =& \frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}\\ & =& 0\end{array}$$ であり,$x=0$で$y=0$となる，原点を通る$-$側の式をここでは用いる．
ちなみに，第三項の分子の平方根内の値が非負となる区間は $$\begin{array}{rcl}2+8\sqrt{2}x& \ge & 0\\ 8\sqrt{2}x& \ge & -2\\ x& \ge & -\frac{2}{8\sqrt{2}}\\ x& \ge & -\frac{1}{4\sqrt{2}}\end{array}$$ $x\ge -\frac{1}{4\sqrt{2}}$となり，特に問題ない．

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$f-g$となる領域をX軸まわりに回転させた体積を求める

$$\begin{array}{rcl}V& =& 2\pi {\int }_0^{\sqrt{2}}{[0-(x+\frac{\sqrt{2}}{2}-\frac{\sqrt{2+8\sqrt{2}x}}{2})]}^2\mathrm{d}x\\ & =& 2\pi {\int }_0^{\sqrt{2}}{(-x-\frac{\sqrt{2}}{2}+\frac{\sqrt{2+8\sqrt{2}x}}{2})}^2\mathrm{d}x\\ & =& 2\pi {\int }_0^{\sqrt{2}}\{-x(-x-\frac{\sqrt{2}}{2}+\frac{\sqrt{2+8\sqrt{2}x}}{2})+\frac{-\sqrt{2}}{2}(-x-\frac{\sqrt{2}}{2}+\frac{\sqrt{2+8\sqrt{2}x}}{2})+\frac{\sqrt{2+8\sqrt{2}x}}{2}(-x-\frac{\sqrt{2}}{2}+\frac{\sqrt{2+8\sqrt{2}x}}{2})\}\mathrm{d}x\\ & =& 2\pi {\int }_0^{\sqrt{2}}\{x^2+x\frac{\sqrt{2}}{2}-x\frac{\sqrt{2+8\sqrt{2}x}}{2}+\frac{\sqrt{2}}{2}x+\frac{\sqrt{2}}{2}\frac{\sqrt{2}}{2}+\frac{-\sqrt{2}}{2}\frac{\sqrt{2+8\sqrt{2}x}}{2}-\frac{\sqrt{2+8\sqrt{2}x}}{2}x-\frac{\sqrt{2+8\sqrt{2}x}}{2}\frac{\sqrt{2}}{2}+\frac{\sqrt{2+8\sqrt{2}x}}{2}\frac{\sqrt{2+8\sqrt{2}x}}{2}\}\mathrm{d}x\\ & =& 2\pi {\int }_0^{\sqrt{2}}\{x^2-2x\frac{\sqrt{2+8\sqrt{2}x}}{2}-2\frac{\sqrt{2}}{2}\frac{\sqrt{2+8\sqrt{2}x}}{2}+{\left(\frac{\sqrt{2+8\sqrt{2}x}}{2}\right)}^2+2\frac{\sqrt{2}}{2}x+{\left(\frac{\sqrt{2}}{2}\right)}^2\}\mathrm{d}x\\ & =& 2\pi {\int }_0^{\sqrt{2}}\{x^2-x\sqrt{2+8\sqrt{2}x}-\frac{\sqrt{2}}{2}\sqrt{2+8\sqrt{2}}x+\frac{2+8\sqrt{2}x}{4}+\sqrt{2}x+\frac{2}{4}\}\mathrm{d}x\\ & =& 2\pi {\int }_0^{\sqrt{2}}\{x^2-x\sqrt{2+8\sqrt{2}x}-\frac{\sqrt{2}}{2}\sqrt{2+8\sqrt{2}}x+\frac{2}{4}+\frac{8\sqrt{2}x}{4}+\sqrt{2}x+\frac{1}{2}\}\mathrm{d}x\\ & =& 2\pi {\int }_0^{\sqrt{2}}\{x^2-x\sqrt{2+8\sqrt{2}x}-\frac{\sqrt{2}}{2}\sqrt{2+8\sqrt{2}}x+2\sqrt{2}x+\sqrt{2}x+\frac{1}{2}+\frac{1}{2}\}\mathrm{d}x\\ & =& 2\pi {\int }_0^{\sqrt{2}}\{x^2-x\sqrt{2+8\sqrt{2}x}-\frac{\sqrt{2}}{2}\sqrt{2+8\sqrt{2}}x+3\sqrt{2}x+1\}\mathrm{d}x\\ & =& 2\pi [{\int }_0^{\sqrt{2}}{x}^2\mathrm{d}x-{\int }_0^{\sqrt{2}}x\sqrt{2+8\sqrt{2}x}\mathrm{d}x-\frac{\sqrt{2}}{2}{\int }_0^{\sqrt{2}}\sqrt{2+8\sqrt{2}x}\mathrm{d}x+3\sqrt{2}{\int }_0^{\sqrt{2}}x\mathrm{d}x+{\int }_0^{\sqrt{2}}\mathrm{d}x]\\ & =& 2\pi (\frac{2}{3}\sqrt{2}-\frac{149}{60}\sqrt{2}-\frac{\sqrt{2}}{2}\cdot \frac{13}{3}+3\sqrt{2}\cdot 1+\sqrt{2})\\ & =& 2\pi (\frac{2}{3}-\frac{149}{60}-\frac{13}{6}+4)\sqrt{2}\\ & =& 2\pi \frac{40-149-130+240}{60}\sqrt{2}\\ & =& 2\pi \frac{\sqrt{2}}{60}\\ & =& \frac{\pi \sqrt{2}}{30}\end{array}$$

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個々の積分について

$$\begin{array}{rcl}{\int }_0^{\sqrt{2}}\mathrm{d}x& =& {\left[x\right]}_0^{\sqrt{2}}\\ & =& [\sqrt{2}-0]\\ & =& \sqrt{2}\end{array}$$ $$\begin{array}{rcl}{\int }_0^{\sqrt{2}}x\mathrm{d}x& =& {\left[\frac{1}{2}{x}^2\right]}_0^{\sqrt{2}}\\ & =& [\frac{1}{2}{\sqrt{2}}^2-\frac{1}{2}{0}^2]\\ & =& [1-0]\\ & =& 1\end{array}$$ $$\begin{array}{r}{\int }_0^{\sqrt{2}}\sqrt{2+8\sqrt{2}x}\mathrm{d}x\\ & =& \frac{1}{8\sqrt{2}}{\int }_2^{18}\sqrt{u}\mathrm{d}u\cdots u=2+8\sqrt{2}x,\frac{\mathrm{d}u}{\mathrm{d}x}=8\sqrt{2},\mathrm{d}x=\frac{1}{8\sqrt{2}}\mathrm{d}u,x=0\to t=2,x=\sqrt{2}\to t=18\\ & =& \frac{1}{8\sqrt{2}}{\left[\frac{1}{\frac{1}{2}+1}{u}^{\frac{1}{2}+1}\right]}_2^{18}\\ & =& \frac{1}{8\sqrt{2}}{\left[\frac{1}{\frac{3}{2}}{u}^{\frac{3}{2}}\right]}_2^{18}\\ & =& \frac{1}{8\sqrt{2}}{\left[\frac{2}{3}{u}^{\frac{3}{2}}\right]}_2^{18}\\ & =& \frac{1}{8\sqrt{2}}\frac{2}{3}{\left[u^{\frac{3}{2}}\right]}_2^{18}\\ & =& \frac{1}{8\sqrt{2}}\frac{2}{3}[{18}^{\frac{3}{2}}-2^{\frac{3}{2}}]\\ & =& \frac{1}{8\sqrt{2}}\frac{2}{3}(18\cdot {18}^{\frac{1}{2}}-2\cdot 2^{\frac{1}{2}})\\ & =& \frac{1}{8\sqrt{2}}\frac{2}{3}(18\cdot {(3^2\cdot 2)}^{\frac{1}{2}}-2\cdot 2^{\frac{1}{2}})\\ & =& \frac{1}{8\sqrt{2}}\frac{2}{3}(18\cdot 3\cdot 2^{\frac{1}{2}}-2\cdot 2^{\frac{1}{2}})\\ & =& \frac{1}{8\sqrt{2}}\frac{2}{3}(54\cdot 2^{\frac{1}{2}}-2\cdot 2^{\frac{1}{2}})\\ & =& \frac{1}{8\sqrt{2}}\frac{2}{3}(54-2)\cdot 2^{\frac{1}{2}}\\ & =& \frac{1}{8\sqrt{2}}\frac{2\cdot 52}{3}\sqrt{2}\\ & =& \frac{1}{8\sqrt{2}}\frac{104}{3}\sqrt{2}\cdots {\int }_2^{18}\sqrt{u}\mathrm{d}u=\frac{104}{3}\sqrt{2}\\ & =& \frac{104}{24}\\ & =& \frac{13}{3}\end{array}$$ $$\begin{array}{r}{\int }_0^{\sqrt{2}}x\sqrt{2+8\sqrt{2}x}\mathrm{d}x\\ & =& \frac{1}{8\sqrt{2}}{\int }_2^{18}\frac{u-2}{8\sqrt{2}}\sqrt{u}\mathrm{d}u\cdots u=2+8\sqrt{2}x,\frac{\mathrm{d}u}{\mathrm{d}x}=8\sqrt{2},\mathrm{d}x=\frac{1}{8\sqrt{2}}\mathrm{d}u,x=0\to t=2,x=\sqrt{2}\to t=18,x=\frac{u-2}{8\sqrt{2}}\\ & =& \frac{1}{8\sqrt{2}}\frac{1}{8\sqrt{2}}{\int }_2^{18}(u-2)\sqrt{u}\mathrm{d}u\\ & =& \frac{1}{128}{\int }_2^{18}(u\sqrt{u}-2\sqrt{u})\mathrm{d}u\\ & =& \frac{1}{128}{\int }_2^{18}(u^{\frac{3}{2}}-2u^{\frac{1}{2}})\mathrm{d}u\\ & =& \frac{1}{128}{\int }_2^{18}{u}^{\frac{3}{2}}\mathrm{d}u-\frac{2}{128}{\int }_2^{18}{u}^{\frac{1}{2}}\mathrm{d}u\\ & =& \frac{1}{128}{\left[\frac{1}{\frac{3}{2}+1}{u}^{\frac{3}{2}+1}\right]}_2^{18}-\frac{1}{64}\frac{104}{3}\sqrt{2}\cdots {\int }_2^{18}\sqrt{u}\mathrm{d}u=\frac{104}{3}\sqrt{2},\mathrm{前}\mathrm{述}\mathrm{の}\mathrm{積}\mathrm{分}\mathrm{結}\mathrm{果}\mathrm{よ}\mathrm{り}\\ & =& \frac{1}{128}{\left[\frac{1}{\frac{5}{2}}{u}^{\frac{5}{2}}\right]}_2^{18}-\frac{13}{24}\sqrt{2}\\ & =& \frac{1}{128}\frac{2}{5}{\left[u^{\frac{5}{2}}\right]}_2^{18}-\frac{13}{24}\sqrt{2}\\ & =& \frac{1}{128}\frac{2}{5}[{18}^{\frac{5}{2}}-2^{\frac{5}{2}}]-\frac{13}{24}\sqrt{2}\\ & =& \frac{1}{128}\frac{2}{5}[{18}^2\cdot {18}^{\frac{1}{2}}-2^2\cdot 2^{\frac{1}{2}}]-\frac{13}{24}\sqrt{2}\\ & =& \frac{1}{128}\frac{2}{5}[{18}^2\cdot (3^2\cdot 2{)}^{\frac{1}{2}}-2^2\cdot 2^{\frac{1}{2}}]-\frac{13}{24}\sqrt{2}\\ & =& \frac{1}{128}\frac{2}{5}[{18}^2\cdot 3\cdot 2^{\frac{1}{2}}-2^2\cdot 2^{\frac{1}{2}}]-\frac{13}{24}\sqrt{2}\\ & =& \frac{1}{128}\frac{2}{5}[({18}^2\cdot 3-2^2)\cdot 2^{\frac{1}{2}}]-\frac{13}{24}\sqrt{2}\\ & =& \frac{1}{128}\frac{2}{5}968\sqrt{2}-\frac{13}{24}\sqrt{2}\\ & =& \frac{121}{40}\sqrt{2}-\frac{13}{24}\sqrt{2}\\ & =& \frac{121\cdot 3-13\cdot 5}{120}\sqrt{2}\\ & =& \frac{363-65}{120}\sqrt{2}\\ & =& \frac{298}{120}\sqrt{2}\\ & =& \frac{149}{60}\sqrt{2}\end{array}$$ $$\begin{array}{rcl}{\int }_0^{\sqrt{2}}{x}^2\mathrm{d}x& =& {\left[\frac{1}{3}{x}^3\right]}_0^{\sqrt{2}}\\ & =& [\frac{1}{3}{\sqrt{2}}^3-\frac{1}{3}{0}^3]\\ & =& [\frac{1}{3}{\sqrt{2}}^3-0]\\ & =& \frac{1}{3}{\sqrt{2}}^3\\ & =& \frac{2}{3}\sqrt{2}\end{array}$$