Find the area of the surface obtained by revolving the given curve about the x-axis. $y=\frac{e^{x}+e^{-x}}{2}$ on $[0, \ln 5]$
A)
$\frac{156}{25} \pi$
B)$\frac{\pi}{5}\left(\frac{156}{25}+\ln 5\right)$
C) $\pi\left(\frac{156}{25}+\ln 5\right)$
D)$\pi \ln 5$

Question

Find the area of the surface obtained by revolving the given curve about the x-axis. $y=\frac{e^{x}+e^{-x}}{2}$ on $[0, \ln 5]$
A)
$\frac{156}{25} \pi$ 
B)$\frac{\pi}{5}\left(\frac{156}{25}+\ln 5\right)$ 
C) $\pi\left(\frac{156}{25}+\ln 5\right)$ 
D)$\pi \ln 5$

Quizplus · Accepted Answer

The surface area of revolution about the x-axis is given by $2\pi \int_{a}^{b} y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$. For $y = \frac{e^x + e^{-x}}{2}$, $\frac{dy}{dx} = \frac{e^x - e^{-x}}{2}$. The integral evaluates to $\pi\left(\frac{156}{25}+\ln 5\right)$.

Find the Area of the Surface Obtained by Revolving the Given