Answer
$\int_C f(x,y) ds = 2$
Work Step by Step
The path is given by ,
$C: x^2+y^2 =1$ from (1,0) to (0,1).
This can be written as $\textbf r = x\textbf i + y \textbf j = \cos t\textbf i + \sin t\textbf j $
where $x(t) = \cos t, y(t) = \sin t, 0\le\pi/2$
$\textbf r' = -\sin t \textbf i + \cos t \textbf j \Rightarrow \vert \textbf r' \vert = 1$
$f(x,y) = x+y = \cos t + \sin t$
Lateral surface area = $\int_C f(x,y) ds\\
=\int_C f(x,y).\vert \textbf r' \vert dt\\
= \int_{t=0}^{\pi/2}(\cos t + \sin t)dt\\
= [\sin t -\cos t]_0^{\pi/2}=1+1=2$
