Example 2. Find the mass of the plate $D$ with surface density $\mu=x^{2} /\left(x^{2}+y^{2}\right)$, bounded by the curves

$$
y^{2}-4 y+x^{2}=0, \quad y^{2}-8 y+x^{2}=0, \quad y=\frac{x}{\sqrt{3}}, \quad x=0
$$
## Solution.

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

$$
m=\iint_{D} \frac{x^{2}}{x^{2}+y^{2}} d x d y
$$

2. We compute the given double integral:
a) since the region $D$ is bounded by circles and lines passing through the origin, it is easier to solve the problem in polar coordinates

$$
\left\{\begin{array}{l}
x=\varrho \cos \varphi \\
y=\varrho \sin \varphi
\end{array}\right.
$$

In this case, the region $D$ will transform into the region $D^{\prime}$, bounded by the lines

$$
\varrho=4 \sin \varphi, \quad \varrho=8 \sin \varphi, \quad \varphi=\frac{\pi}{6}, \quad \varphi=\frac{\pi}{2}
$$

and the desired mass is determined by the formula

$$
m=\iint_{D} \frac{x^{2}}{x^{2}+y^{2}} d x d y=\iint_{D^{\prime}} \cos ^{2} \varphi \varrho d \varrho d \varphi
$$

We define the region $D^{\prime}$ in polar coordinates using inequalities:

![](https://cdn.mathpix.com/cropped/2024_05_22_5878ce417e2ca061000bg-313.jpg?height=131&width=637&top_left_y=915&top_left_x=306)

b) we convert the double integral to an iterated integral

$$
m=\iint_{D^{\prime}} \cos ^{2} \varphi \varrho d \varrho d \varphi=\int_{\pi / 6}^{\pi / 2} \cos ^{2} \varphi d \varphi \int_{4 \sin \varphi}^{8 \sin \varphi} \varrho d \varrho
$$

c) integrating sequentially, we get

$$
\begin{aligned}
& m=\int_{\pi / 6}^{\pi / 2} \cos ^{2} \varphi d \varphi \int_{4 \sin \varphi}^{8 \sin \varphi} \varrho d \varrho=\left.\int_{\pi / 6}^{\pi / 2} \cos ^{2} \varphi \frac{\varrho^{2}}{2}\right|_{4 \sin \varphi} ^{8 \sin \varphi} d \varphi= \\
& \quad=24 \int_{\pi / 6}^{\pi / 2} \sin ^{2} \varphi \cos ^{2} \varphi d \varphi=\left.\left(3 \varphi-\frac{3}{4} \sin 4 \varphi\right)\right|_{\pi / 6} ^{\pi / 2}=\pi+\frac{3 \sqrt{3}}{8}
\end{aligned}
$$

Answer. $m=\pi+3 \sqrt{3} / 8$ units of mass.