Question

$$\begin{gathered} \mathrm{In}\mathrm{Exercises}63–72,\mathrm{set}\mathrm{up}\mathrm{and}\mathrm{solve}\mathrm{a}\mathrm{definite}\mathrm{integral}\mathrm{to}\mathrm{find}\mathrm{the}\mathrm{exact} \\ \mathrm{area}\mathrm{of}\mathrm{each}\mathrm{surface}\mathrm{of}\mathrm{revolution}\mathrm{obtained}\mathrm{by}\mathrm{revolving}\mathrm{the}\mathrm{curve}y=f\left(x\right) \\ \mathrm{around}\mathrm{the}x-\mathrm{axis}\mathrm{on}\mathrm{the}\mathrm{interval}[a,b].f\left(x\right)=\sin x,[a,b]=[0,\pi ] \end{gathered}$$

Short answer

$$\begin{gathered} \mathrm{The}\mathrm{surface}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{solid}\mathrm{of}\mathrm{revolution}\mathrm{obtained}\mathrm{by}\mathrm{revolving}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{sinx} \\ \mathrm{around}\mathrm{the}\mathrm{x}-\mathrm{axis}\mathrm{from}\mathrm{x}=0\mathrm{to}\mathrm{x}=\mathrm{\pi }\mathrm{is}:\mathrm{S}=2\mathrm{\pi }\sqrt{2}+\mathrm{\pi }\ln (\sqrt{2}+1)-\mathrm{\pi }\ln (\sqrt{2}-1) \end{gathered}$$

Step-by-step solution

Step 1. Given Information

The curve,$f\left(x\right)=\sin x$

Step 2. Finding the surface area of the solid of revolution

$$\begin{gathered} \mathrm{The}\mathrm{surface}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{solid}\mathrm{of}\mathrm{revolution}\mathrm{obtained}\mathrm{by}\mathrm{revolving}f\left(x\right) \\ \mathrm{around}\mathrm{the}x-\mathrm{axis}\mathrm{from}x=a\mathrm{to}x=b\mathrm{is}:2\pi {\int }_a^{b}f\left(x\right)\sqrt{1+(f\text{'}{\left(x\right))}^2}dx \\ \mathrm{The}\mathrm{derivative}\mathrm{of}f\left(x\right)=sinx\mathrm{is}f\text{'}\left(x\right)=cosx \\ \mathrm{Therefore},\mathrm{the}\mathrm{surface}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{solid}\mathrm{of}\mathrm{revolution}\mathrm{obtained}\mathrm{by}\mathrm{revolving} \\ \mathrm{f}\left(\mathrm{x}\right)\mathrm{around}\mathrm{the}\mathrm{x}-\mathrm{axis}\mathrm{from}\mathrm{x}=0\mathrm{to}\mathrm{x}=\pi \mathrm{is}:2\pi {\int }_0^{\pi }sinx\sqrt{1+{(cosx)}^2}dx \\ \mathrm{Put},cosx=a,-sinxdx=da \\ \mathrm{Therefore},\mathrm{surface}\mathrm{area},\mathrm{S}=-2\pi {\int }_1^{-1}\sqrt{1+a^2}da \\ S=2\pi {\int }_{-1}^1\sqrt{1+a^2}da \\ S=2\pi \left[{\overline{)\frac{a}{2}\sqrt{1+a^2}+\frac{1}{2}\ln \left|a+\sqrt{1+a^2}\right|}}_{-1}^1\right] \\ S=2\pi \left[\left(\frac{1}{2}\sqrt{1+1^2}-\frac{-1}{2}\sqrt{1+{\left(-1\right)}^2}\right)+\frac{1}{2}\ln \left|1+\sqrt{1+1^2}\right|-\frac{1}{2}\ln \left|-1+\sqrt{1+{(-1)}^2}\right|\right] \\ S=2\mathrm{\pi }\left[\frac{2\sqrt{2}}{2}+\frac{1}{2}\ln (\sqrt{2}+1)-\frac{1}{2}\ln (\sqrt{2}-1)\right] \\ \mathrm{S}=2\mathrm{\pi }\sqrt{2}+\mathrm{\pi }\ln (\sqrt{2}+1)-\mathrm{\pi }\ln (\sqrt{2}-1) \end{gathered}$$