Answer
$\int_Cf(x,y)ds = -\frac{h}{4}[2\sqrt 5 +\ln (2+\sqrt 5)]$
Work Step by Step
The path is given by ,
$C: y = 1-x^2$ from (1,0) to (0,1).
This can be written as $\textbf r = x\textbf i + y \textbf j = x\textbf i + (1-x^2)\textbf j \\
\Rightarrow d\textbf r = (\textbf i - 2x\textbf j ) dx\\
\Rightarrow \vert d\textbf r \vert = (1+4x^2)^{1/2}dx = ds$
$f(x,y) = h$
Lateral surface area $= \int_C f(x,y) ds\\
= \int_{x=1}^0 h(1+4x^2)^{1/2}dx\\
= \frac{h}{4}[2x\sqrt{4x^2+1}+\ln (2x+\sqrt{4x^2+1})]_1^0\\
= -\frac{h}{4}[2\sqrt 5 +\ln (2+\sqrt 5)]$
