Question

Find the area of the surface generated by revolving the curve xequals=StartFraction e Superscript y Baseline plus e Superscript negative y Over 2 EndFraction ey+e−y 2 in the interval 0 less than or equals y less than or equals ln 20≤y≤ln2 about the​ y-axis.

Answer 1

Solution :

$x=f(y) = \frac{e^y + e^{-y}}{2} , \ \ \ \ \ 0 \leq y \leq \ln 2$

$\frac{dx}{dy} = \frac{e^y + e^{-y}}{2}$

$\left(\frac{dx}{dy}\right)^2 = \frac{e^{2y} - 2 + e^{-2y}}{4}$

$1+\left(\frac{dx}{dy}\right)^2 = 1+\frac{e^{2y} - 2 + e^{-2y}}{4} = \frac{e^{2y} + 2 + e^{-2y}}{4}$

                  $= \left(\frac{e^y + e^{-y}}{2}\right)^2$

$\sqrt{1+\left(\frac{dx}{dy}\right)^2} = \sqrt{\left(\frac{e^y + e^{-y}}{2}\right)^2}=\frac{e^y + e^{-y}}{2}$

$S = \int_{y=a}^b 2 \pix \sqrt{1+\left(\frac{dx}{dy}\right)^2 } \ dy$

  $=\int_{0}^{\ln2} 2 \pi \left(\frac{e^y+e^{-y}}{2}\right) \left(\frac{e^y+e^{-y}}{2}\right) \ dy$

  $=\frac{\pi}{2}\int_{0}^{\ln 2}(e^y+e^{-y})^2 \ dy = \frac{\pi}{2}\int_{0}^{\ln 2}(e^{2y}+e^{-2y}+2) \ dy$

  $=\frac{\pi}{2} \left[ \frac{e^{2y}}{2} + \frac{e^{-2y}}{-2} + 2y \right]_2^{\ln 2}$

  $=\frac{\pi}{2} \left[ \left(\frac{e^{2 \ln 2}}{2} + \frac{e^{-2\ln2}}{-2} + 2 \ln2 \right) - \left( \frac{e^0}{2} + \frac{e^0}{-2}+0\right) \right]$

  $=\frac{\pi}{2}\left[ \frac{e^{\ln4}}{2} - \frac{e^{\ln(1/4)}}{2} + \ln 4 - \left( \frac{1}{2} - \frac{1}{2} + 0 \right) \right]$

  $=\frac{\pi}{2} \left[\frac{4}{2} -\frac{1/4}{2} + \ln 4 \right]$

  $=\frac{\pi}{2} \left[ 2-\frac{1}{8} + \ln 4 \right]$

  $=\left( \frac{15}{8} + \ln 4 \right) \frac{\pi}{2}$

Therefore, $S = \frac{15}{16} \pi + \pi \ln 2$