Answer
$=\frac{1}{2} \pi [\sinh(6)+6]$
Work Step by Step
Given: $x=2+3t$ $\implies$ $\frac{dx}{dt}=3$
$y=\cosh (3t)$ $\implies$ $y=3 \sinh(3t)$
To calculate the area of the surface, we will use the equation:
$A=2 \pi \int_0^1 y(t) \sqrt{(dx/dt)^2+(dy/dt)^2}dt$
$=2 \pi \int_0^1 [\cosh (3t)] \sqrt{3^2+(3 \sinh(3t))^2}dt$
$=6 \pi \int_0^1 [\cosh (3t)] \sqrt{1+( \sinh^2(3t))}dt$
Since, $\cosh (3t) \gt 0$
Thus,
$=6 \pi \int_0^1 [\cosh^2 (3t)] dt$
$=6 \pi \int_0^1 (\dfrac{e^{3t}+e^{-3t}}{2})^2 dt$
$=\frac{3}{2} \pi [ \dfrac{e^{6t}}{6}+2t+\frac{e^{-6t}}{-6}]_0^1$
$=\frac{1}{2} \pi [\sinh(6)+6]$
