Q. 71 - Calculus

Question

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

Step-by-Step Solution

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 1. Given Information

The curve, $f\left(x\right)=\sin x$

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

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