Answer
$\pi(e^2+2e-6)$
Work Step by Step
Recall: The area of the surface obtained by rotating the curve $x=f(t)$ and $y=g(t)$ for $a\leq t\leq b$ about the y-axis is formulated by $S=\int_a^b2\pi xds$ where $ds=\sqrt{(dx/dt)^2+(dy/dt)^2}dt$.
Find $ds$:
$ds=\sqrt{(dx/dt)^2+(dy/dt)^2}dt=\sqrt{(e^t-1)^2+(2e^{t/2})^2}dt=\sqrt{(e^t)^2-2e^t+1+4e^t}dt=\sqrt{(e^t)^2+2e^t+1}dt=\sqrt{(e^t+1)^2}dt=(e^t+1)dt$
Evaluate the area of the surface:
$S=\int_0^12\pi(e^t-t)\cdot (e^t+1)dt$
$S=\int_0^12\pi (e^{2t}+e^t-te^t-t)dt$
$S=\int_0^1\pi(2e^{2t}+2e^t-2te^t-2t)dt$
$S=[\pi(e^{2t}+2e^t-(2te^t-2e^t)-t^2)]_0^1$
$S=[\pi(e^{2t}+4e^t-2te^t-t^2)]_0^1$
$S=\pi(e^{2\cdot 1}+4e^1-2\cdot 1e^1-1^2)-\pi(e^{2\cdot 0}+4e^0-2\cdot0e^0-0^2)$
$S=\pi(e^2+4e-2e-1)-\pi(1+4-0-0)$
$S=\pi(e^2+2e-1)-5\pi$
$S=\pi(e^2+2e-6)$
Thus, the surface area is $\pi(e^2+2e-6)$.
