The given expression is a double integral:

$\int_1^4 \int_0^{y^2} \frac{\sin(\sqrt{yz})}{\sqrt{z}} \, dz \, dy$

Let's break this down:

1. The outer integral runs from $y = 1$ to $y = 4$.
2. The inner integral runs from $z = 0$ to $z = y^2$.
3. The integrand is $\frac{\sin(\sqrt{yz})}{\sqrt{z}}$.

To solve this double integral:

1. Inner integral with respect to $z$: $\int_0^{y^2} \frac{\sin(\sqrt{yz})}{\sqrt{z}} \, dz$ We'll need to apply substitution and integration techniques to solve this part.

2. Outer integral with respect to $y$: After integrating with respect to $z$, we will evaluate the outer integral from $y = 1$ to $y = 4$.

I'll now work through the inner integral first.