Q. 72 - Calculus

Question

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

Step-by-Step Solution

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 1. Given Information

The curve, $f\left(x\right)=\cos x$

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

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