Answer
True.
Work Step by Step
We know that the maximum value that $\sin(x^2y^2)$ takes on is $1$.
We also know that the maximum value that $(x^2 + \sqrt{y})$ can obtain is $1^2 + \sqrt{4} = 3$.
Thus, the max value of the iterated integral is:
$\int_{1}^{4} \int_{0}^{1} (x^2 + \sqrt{y}) \sin (x^2y^2) \, dx \, dy \leq \int_{1}^{4} \int_{0}^{1} 3 \, dx \, dy = 3 * 3 = 9$.
