Step 1: Alternative using x-limits:
The line $y = 2$ meets the curve where $x = \frac{1}{\sqrt2}$, and $y = 4$ where $x = 1$.
Step 2: Subtract rectangles and the area under the curve:
Area $= $(rectangle to $y = 4$ over $0 \le x \le 1$) minus (rectangle to $y = 2$ over $0 \le x \le \frac1{\sqrt2}$) minus the area under the curve... which is easier done by the y-integral above.
Step 3: Use the y-integral:
$\int_2^4 \frac{\sqrt y}{2}dy = \frac{1}{3}(8 - 2\sqrt2)\approx \frac{1}{3}(5.17) = 1.72$. This lies between the rectangle areas and matches option (B).
Final Answer:
The area is (1/3)(8 - 2 root 2), option (B).
\[ \boxed{\frac{1}{3}\left[8-2\sqrt{2}\right]\text{ sq. units}} \]