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

`$x=f(y) = (e^y + e^(-y))/(2) , \ \ \ \ \ 0 \leq y \leq \ln 2$`

`$(dx)/(dy) = (e^y + e^(-y))/(2)$`

`$\left((dx)/(dy)\right)^2 = (e^(2y) - 2 + e^(-2y))/(4)$`

`$1+\left((dx)/(dy)\right)^2 = 1+(e^(2y) - 2 + e^(-2y))/(4) = (e^(2y) + 2 + e^(-2y))/(4)$`

`$ = \left((e^y + e^(-y))/(2)\right)^2$`

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

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

`$=\int_(0)^(\ln2) 2 \pi \left((e^y+e^(-y))/(2)\right) \left((e^y+e^(-y))/(2)\right) \ dy$`

`$=(\pi)/(2)\int_(0)^(\ln 2)(e^y+e^(-y))^2 \ dy = (\pi)/(2)\int_(0)^(\ln 2)(e^(2y)+e^(-2y)+2) \ dy $`

`$=(\pi)/(2) \left[ (e^(2y))/(2) + (e^(-2y))/(-2) + 2y \right]_2^(\ln 2)$`

`$=(\pi)/(2) \left[ \left((e^(2 \ln 2))/(2) + (e^(-2\ln2))/(-2) + 2 \ln2 \right) - \left( (e^0)/(2) + (e^0)/(-2)+0\right) \right]$`

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

`$=(\pi)/(2) \left[(4)/(2) -(1/4)/(2) + \ln 4 \right]$`

`$=(\pi)/(2) \left[ 2-(1)/(8) + \ln 4 \right]$`

`$=\left( (15)/(8) + \ln 4 \right) (\pi)/(2)$`

Therefore,
`$S = (15)/(16) \pi + \pi \ln 2$`