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Question Number 29116 by tawa tawa last updated on 04/Feb/18

$$\mathrm{Find}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{region}\mathbf{R}\mathrm{bounded}\mathrm{by}\mathrm{the}\mathrm{curve}\mathrm{y}=\cosh \left(\mathrm{x}\right),\mathrm{the}\mathrm{line}\mathrm{x}={\log }_{\mathrm{e}}\left(2\right)$$$$\mathrm{and}\mathrm{the}\mathrm{coordinate}\mathrm{axis}.\mathrm{Find}\mathrm{also}\mathrm{the}\mathrm{volume}\mathrm{obtained}\mathrm{when}\mathbf{R}\mathrm{is}\mathrm{rotated}$$$$\mathrm{completely}\mathrm{about}\mathrm{the}\mathrm{x}-\mathrm{axis}.$$

Answered by ajfour last updated on 04/Feb/18

$$Area={\int }_0^{\ln 2}\left(\frac{e^x+e^{-x}}{2}\right)dx$$$$=\frac{1}{2}(e^x-e^{-x}){\mid }_0^{\ln 2}$$$$=\frac{1}{2}(2-\frac{1}{2})=\frac{3}{4}$$$$Volume=\pi {\int }_0^{\ln 2}{\left(\frac{e^x+e^{-x}}{2}\right)}^{2}dx$$$$=\frac{\pi }{4}{\int }_0^{\ln 2}(e^{2x}+e^{-2x}+2)dx$$$$=\frac{\pi }{4}{[\frac{e^{2x}}{2}-\frac{e^{-2x}}{2}+2x]}_0^{\ln 2}$$$$=\frac{\pi }{8}(4-\frac{1}{4}+4\ln 2)$$$$Volume=\frac{\pi }{32}(15+16\ln 2).$$

Commented by tawa tawa last updated on 04/Feb/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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