Answer
$\int_C f(x,y) ds = 5h$
Work Step by Step
The path is given by ,
C: line from (0,0) to (3,4).
This can be written as $\textbf r = x\textbf i + y \textbf j = 3t\textbf i + 4 t\textbf j $
where $x(t) = 3t, y(t) = 4t, 0\le t \le 1$
$\textbf r' = 3 \textbf i + 4 \textbf j \Rightarrow \vert \textbf r' \vert = 5$
$f(x,y) = h$
Lateral surface area = $\int_C f(x,y) ds\\
=\int_C f(x,y).\vert \textbf r' \vert dt\\
= \int_{t=0}^{1}(5h)dt\\
= [5ht]_0^{1}=5h$
