Example 1. Find the mass of the plate $D$ with surface density $\mu=16 x+9 y^{2} / 2$, bounded by the curves

$$
x=\frac{1}{4}, \quad y=0, \quad y^{2}=16 x \quad(y \geq 0)
$$
Solution.

1. The mass of the plate $D$ with surface density $\mu=16 x+9 y^{2} / 2$ is determined by the formula

$$
m=\iint_{D}\left(16 x+\frac{9 y^{2}}{2}\right) d x d y
$$

2. We compute the obtained double integral in Cartesian coordinates:

a) define the region $D$ by a system of inequalities:

$$
\left\{\begin{array}{l}
0 \leq x \leq 1 / 4 \\
0 \leq y \leq 4 \sqrt{x}
\end{array}\right.
$$

The inequality $0 \leq x$ follows from $y^{2}=16 x$, i.e., $x$ is non-negative

b) convert the double integral to an iterated integral:

$$
m=\iint_{D}\left(16 x+\frac{9 y^{2}}{2}\right) d x d y=\int_{0}^{1 / 4} d x \int_{0}^{4 \sqrt{x}}\left(16 x+\frac{9 y^{2}}{2}\right) d y
$$

c) integrate sequentially, using the properties of the definite integral:

$$
\begin{aligned}
m=\int_{0}^{1 / 4} d x \int_{0}^{4 \sqrt{x}}(16 x & \left.+\frac{9 y^{2}}{2}\right) d y= \\
& =\left.\int_{0}^{1 / 4}\left(16 x y+\frac{3 y^{3}}{2}\right)\right|_{0} ^{4 \sqrt{x}} d x=160 \int_{0}^{1 / 4} x^{3 / 2} d x=2
\end{aligned}
$$

Answer. $m=2$ units of mass.