Step 1: Describe the same region the other way round.
From $y = x^2/4$, as $x$ goes from $0$ to $2$, $y$ climbs from $0$ up to a peak of $1$ at $x = 2$.
For a fixed $y$ between $0$ and $1$, $x$ must satisfy $x^2/4 \geq y$, so $x \geq 2\sqrt{y}$, while staying under the bound $x \leq 2$.
Step 2: Integrate over $x$ first.
\[ \int_{2\sqrt{y}}^{2} y \, dx = y\left(2 - 2\sqrt{y}\right) = 2y - 2y^{3/2} \]
This collapses the inner integral into a function of $y$ alone.
Step 3: Integrate over $y$.
\[ \int_0^1 \left(2y - 2y^{3/2}\right) dy = \left[y^2 - \frac{4}{5}y^{5/2}\right]_0^1 = 1 - \frac{4}{5} = \frac{1}{5} \]
Final Answer:
Swapping the order of integration gives the same clean value, confirming the region and setup were correct.
\[ \boxed{\dfrac{1}{5}} \]