Answer
Surface area of the wall = $6670$ ft$^2$
Work Step by Step
The path of the wall on (x-y) plane can be written as $\textbf r = x\textbf i + y \textbf j$ where $y = x^{3/2}$
$\therefore d\textbf r = dx \textbf i + \textbf j = dx \textbf i + \frac{3}{2}x^{1/2} dx \textbf j\\
\Rightarrow \vert d\textbf r \vert = ds = (1+\frac{9}{4}x)^{1/2}dx$
height of the wall $z = 20+\frac{x}{4}$
surface of the wall $= \int z .ds = \int_{x=0}^40 (20+\frac{x}{4}). (1+\frac{9}{4}x)^{1/2}dx\\
= [\frac{1}{1620}((4+9x)^{5/2}+1193.3(4+9x)^{3/2})]_0^{40}\\
=6675.9-5.9 = 6670 ft^2$
