Question

$$\begin{gathered} \mathrm{In}\mathrm{Exercises}63–72,\mathrm{set}\mathrm{up}\mathrm{and}\mathrm{solve}\mathrm{a}\mathrm{definite}\mathrm{integral}\mathrm{to}\mathrm{find}\mathrm{the}\mathrm{exact} \\ \mathrm{area}\mathrm{of}\mathrm{each}\mathrm{surface}\mathrm{of}\mathrm{revolution}\mathrm{obtained}\mathrm{by}\mathrm{revolving}\mathrm{the}\mathrm{curve}y=f\left(x\right) \\ \mathrm{around}\mathrm{the}x-\mathrm{axis}\mathrm{on}\mathrm{the}\mathrm{interval}[a,b].f\left(x\right)=\cos x,[a,b]=\left[-\frac{\pi }{2},\frac{\pi }{2}\right] \end{gathered}$$

Short answer

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

Step-by-step solution

Step 1. Given Information

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

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

$$\begin{gathered} \mathrm{The}\mathrm{surface}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{solid}\mathrm{of}\mathrm{revolution}\mathrm{obtained}\mathrm{by}\mathrm{revolving}f\left(x\right) \\ \mathrm{around}\mathrm{the}x-\mathrm{axis}\mathrm{from}x=a\mathrm{to}x=b\mathrm{is}:2\pi {\int }_a^{b}f\left(x\right)\sqrt{1+(f\text{'}{\left(x\right))}^2}dx \\ \mathrm{The}\mathrm{derivative}\mathrm{of}f\left(x\right)=\cos x\mathrm{is}f\text{'}\left(x\right)=-\sin x \\ \mathrm{Therefore},\mathrm{the}\mathrm{surface}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{solid}\mathrm{of}\mathrm{revolution}\mathrm{obtained}\mathrm{by}\mathrm{revolving} \\ \mathrm{f}\left(\mathrm{x}\right)\mathrm{around}\mathrm{the}\mathrm{x}-\mathrm{axis}\mathrm{from}\mathrm{x}=-\frac{\mathrm{\pi }}{2}\mathrm{to}\mathrm{x}=\frac{\pi }{2}\mathrm{is}:2\pi {\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\pi }{2}}\cos x\sqrt{1+{(-\sin x)}^2}dx \\ \mathrm{Put},\sin x=a,\cos xdx=da \\ \mathrm{Therefore},\mathrm{surface}\mathrm{area},\mathrm{S}=2\pi \times 2{\int }_0^1\sqrt{1+a^2}da \\ S=4\pi {\int }_0^1\sqrt{1+a^2}da \\ S=4\pi \left[{\overline{)\frac{a}{2}\sqrt{1+a^2}+\frac{1}{2}\ln \left|a+\sqrt{1+a^2}\right|}}_0^1\right] \\ S=4\pi \left[\left(\frac{1}{2}\sqrt{1+1^2}-\frac{0}{2}\sqrt{1+0^2}\right)+\frac{1}{2}\ln \left|1+\sqrt{1+1^2}\right|-\frac{1}{2}\ln \left|0+\sqrt{1+0^2}\right|\right] \\ S=4\mathrm{\pi }\left[\frac{\sqrt{2}}{2}+\frac{1}{2}\ln (\sqrt{2}+1)\right] \\ \mathrm{S}=2\mathrm{\pi }\sqrt{2}+2\mathrm{\pi }\ln (\sqrt{2}+1) \end{gathered}$$