回転体の体積と表面積の計算方法

$$\begin{array}{rl}V& =\pi {\displaystyle {\int }_a^{b}f(x{)}^{2}dx}\\ \\ S& =2\pi {\displaystyle {\int }_a^{b}f(x)\sqrt{1+f^{\prime }(x{)}^2}dx}\end{array}$$

$$\begin{array}{rl}V& =\pi {\displaystyle {\int }_p^q{f}^{\prime }(t)g(t{)}^{2}dt}\\ \\ S& =2\pi {\displaystyle {\int }_p^{q}g(t)\sqrt{f^{\prime }(t{)}^2+g^{\prime }(t{)}^2}dt}\end{array}$$

$$\begin{array}{rl}V& ={\displaystyle \frac{2}{3}\pi {\displaystyle {\int }_{\alpha }^{\beta }f(\theta {)}^3\sin \theta d\theta }}\\ \\ S& =2\pi {\displaystyle {\int }_{\alpha }^{\beta }f(\theta )\sin \theta \sqrt{f(\theta {)}^2+f^{\prime }(\theta {)}^2}d\theta }\end{array}$$

$$\begin{array}{cc}y=f(x)=\cosh x& (0\le x\le \log 2)\end{array}$$

$$\begin{array}{rl}V& =\pi {\displaystyle {\int }_0^{\log 2}f(x{)}^{2}dx}\\ \\ & =\pi {\displaystyle {\int }_0^{\log 2}{\cosh }^{2}xdx}\\ \\ & ={\displaystyle \frac{1}{2}\pi {\displaystyle {\int }_0^{\log 2}\cosh 2x+1dx}}\\ \\ & ={\displaystyle \frac{1}{2}\pi {\left[{\displaystyle \frac{1}{2}\sinh 2x+x}\right]}_0^{\log 2}}\\ \\ & ={\displaystyle \frac{1}{2}\pi \left\{{\displaystyle \frac{1}{2}\sinh (2\log 2)+\log 2}\right\}}\\ \\ & ={\displaystyle \frac{1}{2}\pi \left({\displaystyle \frac{15}{16}+\log 2}\right)}\end{array}$$

$$f^{\prime }(x)=\sinh x$$

$$\begin{array}{rl}S& =2\pi {\displaystyle {\int }_0^{\log 2}f(x)\sqrt{1+f^{\prime }(x{)}^2}dx}\\ \\ & =2\pi {\displaystyle {\int }_0^{\log 2}\cosh x\sqrt{1+{\sinh }^{2}x}dx}\\ \\ & =2\pi {\displaystyle {\int }_0^{\log 2}\cosh x\sqrt{{\cosh }^{2}x}dx}\\ \\ & =2\pi {\displaystyle {\int }_0^{\log 2}\cosh x|\cosh x|dx}\\ \\ & =2\pi {\displaystyle {\int }_0^{\log 2}{\cosh }^{2}xdx}\\ \\ & =\pi \left({\displaystyle \frac{15}{16}+\log 2}\right)\end{array}$$

$$\begin{array}{cc}\begin{array}{l}x=f(t)=t-\sin t\\ y=g(t)=1-\cos t\end{array}& (0\le t\le 2\pi )\end{array}$$

$$f^{\prime }(t)=1-\cos t$$

$$\begin{array}{rl}V& =\pi {\displaystyle {\int }_0^{2\pi }{f}^{\prime }(t)g(t{)}^{2}dt}\\ \\ & =\pi {\displaystyle {\int }_0^{2\pi }(1-\cos t)(1-\cos t{)}^{2}dt}\\ \\ & =\pi {\displaystyle {\int }_0^{2\pi }1-3\cos t+3{\cos }^{2}t-{\cos }^{3}tdt}\\ \\ & =\pi {\displaystyle {\int }_0^{2\pi }{\displaystyle \frac{5}{2}-{\displaystyle \frac{15}{4}\cos t+{\displaystyle \frac{3}{2}\cos 2t-{\displaystyle \frac{1}{4}\cos 3tdt}}}}}\\ \\ & =\pi {\left[{\displaystyle \frac{5}{2}t-{\displaystyle \frac{15}{4}\sin t+{\displaystyle \frac{3}{4}\sin 2t-{\displaystyle \frac{1}{12}\sin 3t}}}}\right]}_0^{2\pi }\\ \\ & =5{\pi }^2\end{array}$$

$$g^{\prime }(t)=\sin t$$

$$\begin{array}{rl}S& =2\pi {\displaystyle {\int }_0^{2\pi }g(t)\sqrt{f^{\prime }(t{)}^2+g^{\prime }(t{)}^2}dt}\\ \\ & =2\pi {\displaystyle {\int }_0^{2\pi }(1-\cos t)\sqrt{(1-\cos t{)}^2+{\sin }^{2}t}dt}\\ \\ & =2\pi {\displaystyle {\int }_0^{2\pi }(1-\cos t)\sqrt{2(1-\cos t)}dt}\\ \\ & =2\sqrt{2}\pi {\displaystyle {\int }_0^{2\pi }(1-\cos t{)}^{\frac{3}{2}}dt}\\ \\ & =2\sqrt{2}\pi {\displaystyle {\int }_0^{2\pi }{\left(2{\sin }^2{\displaystyle \frac{t}{2}}\right)}^{\frac{3}{2}}dt}\\ \\ & =8\pi {\displaystyle {\int }_0^{2\pi }{\sin }^3{\displaystyle \frac{t}{2}dt}}\\ \\ & =8\pi {\displaystyle {\int }_0^{2\pi }(1-{\cos }^2{\displaystyle \frac{t}{2}})\sin {\displaystyle \frac{t}{2}dt}}\end{array}$$

$$\begin{array}{rl}u& =\cos {\displaystyle \frac{t}{2}}\\ \\ du& =-{\displaystyle \frac{1}{2}\sin {\displaystyle \frac{t}{2}dt}}\end{array}$$

$$\begin{array}{ccccc}t& :& 0& \to & 2\pi \\ \\ u& :& 1& \to & -1\end{array}$$

$$\begin{array}{rl}8\pi {\displaystyle {\int }_0^{2\pi }(1-{\cos }^2{\displaystyle \frac{t}{2}})\sin {\displaystyle \frac{t}{2}dt}}& =8\pi {\displaystyle {\int }_1^{-1}(1-u^2)(-2)du}\\ \\ & =16\pi {\displaystyle {\int }_{-1}^{1}1-u^{2}du}\\ \\ & =16\pi {[u-{\displaystyle \frac{1}{3}{u}^3}]}_{-1}^1\\ \\ & ={\displaystyle \frac{64}{3}\pi }\end{array}$$

$$\begin{array}{cc}r=f(\theta )=1+\cos \theta & (0\le \theta \le \pi )\end{array}$$

$$\begin{array}{rl}V& ={\displaystyle \frac{2}{3}\pi {\displaystyle {\int }_0^{\pi }f(\theta {)}^3\sin \theta d\theta }}\\ \\ & ={\displaystyle \frac{2}{3}\pi {\displaystyle {\int }_0^{\pi }(1+\cos \theta {)}^3\sin \theta d\theta }}\end{array}$$

$$\begin{array}{rl}t& =\cos \theta \\ \\ dt& =-\sin \theta d\theta \end{array}$$

$$\begin{array}{ccccc}\theta & :& 0& \to & \pi \\ \\ t& :& 1& \to & -1\end{array}$$

$$\begin{array}{rl}{\displaystyle \frac{2}{3}\pi {\displaystyle {\int }_0^{\pi }(1+\cos \theta {)}^3\sin \theta d\theta }}& ={\displaystyle \frac{2}{3}\pi {\displaystyle {\int }_1^{-1}(1+t^3)(-1)dt}}\\ \\ & ={\displaystyle \frac{2}{3}\pi {\displaystyle {\int }_{-1}^{1}1+t^{3}dt}}\\ \\ & ={\displaystyle \frac{2}{3}\pi {\left[{\displaystyle \frac{1}{4}(1+t{)}^4}\right]}_{-1}^1}\\ \\ & ={\displaystyle \frac{8}{3}\pi }\end{array}$$

$$f^{\prime }(\theta )=-\sin \theta$$

$$\begin{array}{rl}S& =2\pi {\displaystyle {\int }_0^{\pi }f(\theta )\sin \theta \sqrt{f(\theta {)}^2+f^{\prime }(\theta {)}^2}d\theta }\\ \\ & =2\pi {\displaystyle {\int }_0^{\pi }(1+\cos \theta )\sin \theta \sqrt{(1+\cos \theta {)}^2+(-\sin \theta {)}^2}d\theta }\\ \\ & =2\pi {\displaystyle {\int }_0^{\pi }(1+\cos \theta )\sin \theta \sqrt{2(1+\cos \theta )}d\theta }\\ \\ & =2\sqrt{2}\pi {\displaystyle {\int }_0^{\pi }(1+\cos \theta {)}^{\frac{3}{2}}\sin \theta d\theta }\end{array}$$

$$\begin{array}{rl}t& =\cos \theta \\ \\ dt& =-\sin \theta d\theta \end{array}$$

$$\begin{array}{ccccc}\theta & :& 0& \to & \pi \\ \\ t& :& 1& \to & -1\end{array}$$

$$\begin{array}{rl}2\sqrt{2}\pi {\displaystyle {\int }_0^{\pi }(1+\cos \theta {)}^{\frac{3}{2}}\sin \theta d\theta }& =2\sqrt{2}\pi {\displaystyle {\int }_1^{-1}(1+t{)}^{\frac{3}{2}}(-1)dt}\\ \\ & =2\sqrt{2}\pi {\displaystyle {\int }_{-1}^1(1+t{)}^{\frac{3}{2}}dt}\\ \\ & =2\sqrt{2}\pi {\left[{\displaystyle \frac{2}{5}(1+t{)}^{\frac{5}{2}}}\right]}_{-1}^1\\ \\ & ={\displaystyle \frac{32}{5}\pi }\end{array}$$

$$\begin{array}{lll}f(x)=\sqrt{1-x^2}& & (0\le x\le {\displaystyle \frac{1}{\sqrt{2}}})\\ \\ f(\theta )=1& & \left({\displaystyle \frac{1}{4}\pi \le x\le {\displaystyle \frac{1}{2}\pi }}\right)\end{array}$$

$$\begin{array}{rl}V& =\pi {\displaystyle {\int }_a^{b}f(x{)}^{2}dx}\\ \\ & =\pi {\displaystyle {\int }_0^{\frac{1}{\sqrt{2}}}1-x^{2}dx}\\ \\ & =\pi {[x-{\displaystyle \frac{1}{3}{x}^3}]}_0^{\frac{1}{\sqrt{2}}}\\ \\ & ={\displaystyle \frac{5\sqrt{2}}{12}\pi }\end{array}$$

$$\begin{array}{rl}S& =2\pi {\displaystyle {\int }_a^{b}f(x)\sqrt{1+f^{\prime }(x{)}^2}dx}\\ \\ & =2\pi {\displaystyle {\int }_0^{\frac{1}{\sqrt{2}}}\sqrt{1-x^2}{\displaystyle \frac{1}{\sqrt{1-x^2}}dx}}\\ \\ & =2\pi {\displaystyle {\int }_0^{\frac{1}{\sqrt{2}}}dx}\\ \\ & =2\pi {\left[x\right]}_0^{\frac{1}{\sqrt{2}}}\\ \\ & =\sqrt{2}\pi \end{array}$$

$$\begin{array}{rl}V& ={\displaystyle \frac{2}{3}\pi {\displaystyle {\int }_{\alpha }^{\beta }f(\theta {)}^3\sin \theta d\theta }}\\ \\ & ={\displaystyle \frac{2}{3}\pi {\displaystyle {\int }_{\frac{1}{4}\pi }^{\frac{1}{2}\pi }\sin \theta d\theta }}\\ \\ & ={\displaystyle \frac{2}{3}\pi {[\cos \theta ]}_{\frac{1}{4}\pi }^{\frac{1}{2}\pi }}\\ \\ & ={\displaystyle \frac{\sqrt{2}}{3}\pi }\\ \\ & ={\displaystyle \frac{4\sqrt{2}}{12}\pi }\end{array}$$

$$\begin{array}{rl}S& =2\pi {\displaystyle {\int }_{\alpha }^{\beta }f(\theta )\sin \theta \sqrt{f(\theta {)}^2+f^{\prime }(\theta {)}^2}d\theta }\\ \\ & =2\pi {\displaystyle {\int }_{\frac{1}{4}\pi }^{\frac{1}{2}\pi }\sin \theta d\theta }\\ \\ & =2\pi {[\cos \theta ]}_{\frac{1}{4}\pi }^{\frac{1}{2}\pi }\\ \\ & =\sqrt{2}\pi \end{array}$$

$$V={\displaystyle \frac{\sqrt{2}}{12}\pi }$$

$$\mathrm{\Delta }V\simeq \pi y^2\mathrm{\Delta }x$$

$${\displaystyle \frac{\mathrm{\Delta }V}{\mathrm{\Delta }x}\simeq \pi y^2}$$

$${\displaystyle \frac{dV}{dx}=\pi y^2}$$

$$V=\pi {\displaystyle {\int }_a^b{y}^{2}dx}$$

$$l=2\pi y$$

$$\mathrm{\Delta }S\simeq 2\pi y\mathrm{\Delta }L$$

$$\begin{array}{rl}\mathrm{\Delta }L& \simeq \sqrt{(\mathrm{\Delta }x{)}^2+(\mathrm{\Delta }y{)}^2}\\ \\ & =\sqrt{1+{\left({\displaystyle \frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}}\right)}^2}\mathrm{\Delta }x\end{array}$$

$$\mathrm{\Delta }S\simeq 2\pi y\sqrt{1+{\left({\displaystyle \frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}}\right)}^2}\mathrm{\Delta }x$$

$${\displaystyle \frac{\mathrm{\Delta }S}{\mathrm{\Delta }x}\simeq 2\pi y\sqrt{1+{\left({\displaystyle \frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}}\right)}^2}}$$

$${\displaystyle \frac{dS}{dx}=2\pi y\sqrt{1+{\left({\displaystyle \frac{dy}{dx}}\right)}^2}}$$

$$S=2\pi {\displaystyle {\int }_a^{b}y\sqrt{1+{\left({\displaystyle \frac{dy}{dx}}\right)}^2}dx}$$

$$\begin{array}{rl}V& =\pi {\displaystyle {\int }_a^b{y}^{2}dx}\\ \\ & =\pi {\displaystyle {\int }_p^q{y}^2{\displaystyle \frac{dx}{dt}dt}}\end{array}$$

$$l=2\pi y$$

$$\mathrm{\Delta }S\simeq 2\pi y\mathrm{\Delta }L$$

$$\begin{array}{rl}\mathrm{\Delta }L& \simeq \sqrt{(\mathrm{\Delta }x{)}^2+(\mathrm{\Delta }y{)}^2}\\ \\ & =\sqrt{{\left({\displaystyle \frac{\mathrm{\Delta }x}{\mathrm{\Delta }t}}\right)}^2+{\left({\displaystyle \frac{\mathrm{\Delta }y}{\mathrm{\Delta }t}}\right)}^2}\mathrm{\Delta }t\end{array}$$

$$\mathrm{\Delta }S\simeq 2\pi y\sqrt{{\left({\displaystyle \frac{\mathrm{\Delta }x}{\mathrm{\Delta }t}}\right)}^2+{\left({\displaystyle \frac{\mathrm{\Delta }y}{\mathrm{\Delta }t}}\right)}^2}\mathrm{\Delta }t$$

$${\displaystyle \frac{\mathrm{\Delta }S}{\mathrm{\Delta }t}\simeq 2\pi y\sqrt{{\left({\displaystyle \frac{\mathrm{\Delta }x}{\mathrm{\Delta }t}}\right)}^2+{\left({\displaystyle \frac{\mathrm{\Delta }y}{\mathrm{\Delta }t}}\right)}^2}}$$

$${\displaystyle \frac{dS}{dx}=2\pi y\sqrt{{\left({\displaystyle \frac{dx}{dt}}\right)}^2+{\left({\displaystyle \frac{dy}{dt}}\right)}^2}}$$

$$S=2\pi {\displaystyle {\int }_p^{q}y\sqrt{{\left({\displaystyle \frac{dx}{dt}}\right)}^2+{\left({\displaystyle \frac{dy}{dt}}\right)}^2}dt}$$

$$\begin{array}{lll}(r,\theta )& \to & (r\cos \theta ,r\sin \theta )\\ \\ (r+\mathrm{\Delta }r,\theta +\mathrm{\Delta }\theta )& \to & ((r+\mathrm{\Delta }r)\cos (\theta +\mathrm{\Delta }\theta ),(r+\mathrm{\Delta }r)\sin (\theta +\mathrm{\Delta }\theta ))\end{array}$$

$$\mathrm{\Delta }A\simeq {\displaystyle \frac{1}{2}r(r+\mathrm{\Delta }r)\sin \mathrm{\Delta }\theta }$$

$$g\simeq {\displaystyle \frac{1}{3}\{r\sin \theta +(r+\mathrm{\Delta }r)\sin (\theta +\mathrm{\Delta }\theta )\}}$$

$$\begin{array}{rl}\mathrm{\Delta }V& \simeq 2\pi g\mathrm{\Delta }A\\ \\ & ={\displaystyle \frac{1}{3}\pi \{r\sin \theta +(r+\mathrm{\Delta }r)\sin (\theta +\mathrm{\Delta }\theta )\}r(r+\mathrm{\Delta }r)\sin \mathrm{\Delta }\theta }\end{array}$$

$${\displaystyle \frac{\mathrm{\Delta }V}{\mathrm{\Delta }\theta }\simeq {\displaystyle \frac{1}{3}\pi \{r\sin \theta +(r+\mathrm{\Delta }r)\sin (\theta +\mathrm{\Delta }\theta )\}r(r+\mathrm{\Delta }r){\displaystyle \frac{\sin \mathrm{\Delta }\theta }{\mathrm{\Delta }\theta }}}}$$

$$\begin{array}{rl}{\displaystyle \frac{dV}{d\theta }}& ={\displaystyle \frac{1}{3}\pi \{r\sin \theta +r\sin \theta \}r\cdot r\cdot 1}\\ \\ & ={\displaystyle \frac{2}{3}\pi r^3\sin \theta }\end{array}$$

$$V={\displaystyle \frac{2}{3}\pi {\displaystyle {\int }_{\alpha }^{\beta }{r}^3\sin \theta d\theta }}$$

$$l=2\pi r\sin \theta$$

$$\mathrm{\Delta }S\simeq 2\pi r\sin \theta \mathrm{\Delta }L$$

$${\displaystyle \frac{\mathrm{\Delta }S}{\mathrm{\Delta }\theta }\simeq 2\pi r\sin \theta {\displaystyle \frac{\mathrm{\Delta }L}{\mathrm{\Delta }\theta }}}$$

$$\begin{array}{rl}(\mathrm{\Delta }L{)}^2& \simeq r^2+(r+\mathrm{\Delta }r{)}^2-2r(r+\mathrm{\Delta }r)\cos \mathrm{\Delta }\theta \\ \\ & =2r^2+2r\mathrm{\Delta }r+(\mathrm{\Delta }r{)}^2-(2r^2+2r\mathrm{\Delta }r)\cos \mathrm{\Delta }\theta \\ \\ & =(2r^2+2r\mathrm{\Delta }r)(1-\cos \mathrm{\Delta }\theta )+(\mathrm{\Delta }r{)}^2\\ \\ & =2r(r+\mathrm{\Delta }r)(1-\cos \mathrm{\Delta }\theta )+(\mathrm{\Delta }r{)}^2\end{array}$$

$${\left({\displaystyle \frac{\mathrm{\Delta }L}{\mathrm{\Delta }\theta }}\right)}^2\simeq 2r(r+\mathrm{\Delta }r){\displaystyle \frac{(1-\cos \mathrm{\Delta }\theta )}{(\mathrm{\Delta }\theta {)}^2}+{\left({\displaystyle \frac{\mathrm{\Delta }r}{\mathrm{\Delta }\theta }}\right)}^2}$$

$${\displaystyle \frac{\mathrm{\Delta }L}{\mathrm{\Delta }\theta }\simeq \sqrt{2r(r+\mathrm{\Delta }r){\displaystyle \frac{(1-\cos \mathrm{\Delta }\theta )}{(\mathrm{\Delta }\theta {)}^2}+{\left({\displaystyle \frac{\mathrm{\Delta }r}{\mathrm{\Delta }\theta }}\right)}^2}}}$$

$$\begin{array}{rl}{\displaystyle \frac{\mathrm{\Delta }S}{\mathrm{\Delta }\theta }}& \simeq 2\pi r\sin \theta {\displaystyle \frac{\mathrm{\Delta }L}{\mathrm{\Delta }\theta }}\\ \\ & =2\pi r\sin \theta \sqrt{2r(r+\mathrm{\Delta }r){\displaystyle \frac{(1-\cos \mathrm{\Delta }\theta )}{(\mathrm{\Delta }\theta {)}^2}+{\left({\displaystyle \frac{\mathrm{\Delta }r}{\mathrm{\Delta }\theta }}\right)}^2}}\end{array}$$

$$\begin{array}{rl}{\displaystyle \frac{dS}{d\theta }}& =2\pi r\sin \theta \sqrt{2r\cdot r\cdot {\displaystyle \frac{1}{2}+{\left({\displaystyle \frac{dr}{d\theta }}\right)}^2}}\\ \\ & =2\pi r\sin \theta \sqrt{r^2+{\left({\displaystyle \frac{dr}{d\theta }}\right)}^2}\end{array}$$

$$S=2\pi {\displaystyle {\int }_{\alpha }^{\beta }r\sin \theta \sqrt{r^2+{\left({\displaystyle \frac{dr}{d\theta }}\right)}^2}d\theta }$$

$$\begin{array}{rl}V& =\pi {\displaystyle {\int }_a^{b}f(x{)}^{2}dx}\\ \\ S& =2\pi {\displaystyle {\int }_a^{b}f(x)\sqrt{1+f^{\prime }(x{)}^2}dx}\end{array}$$

$$\begin{array}{rl}V& =\pi {\displaystyle {\int }_p^q{f}^{\prime }(t)g(t{)}^{2}dt}\\ \\ S& =2\pi {\displaystyle {\int }_p^{q}g(t)\sqrt{f^{\prime }(t{)}^2+g^{\prime }(t{)}^2}dt}\end{array}$$

$$\begin{array}{rl}V& ={\displaystyle \frac{2}{3}\pi {\displaystyle {\int }_{\alpha }^{\beta }f(\theta {)}^3\sin \theta d\theta }}\\ \\ S& =2\pi {\displaystyle {\int }_{\alpha }^{\beta }f(\theta )\sin \theta \sqrt{f(\theta {)}^2+f^{\prime }(\theta {)}^2}d\theta }\end{array}$$