2.3. Find the center of gravity of the arc $AB$:

a) $AB: x^{2 / 3} + y^{2 / 3} = a^{2 / 3}$ (astroid), $A(0, a), B(a, 0)$, the linear density at each point of the arc is equal to its abscissa;

b) $AB: x = \cos t, y = \sin t, z = \frac{H}{2 \pi} t, t_{A} = 0, t_{B} = 2 \pi$ (helical line with pitch $H$), $\rho(x, y) = 1$.
S o l u t i o n.

a) First, we find the mass of the line $A B$ using a line integral of the first kind, considering that $P(x, y)=x$. According to the problem statement,

$$
\begin{gathered}
M_{A B}=\int_{A B} p(x, y) d l \\
M_{A B}=\int_{A B} x d l=\int_{0}^{a} x \cdot \sqrt{1+\left(y^{\prime}\right)^{2}} d x= \\
=\int_{0}^{a} x \cdot \sqrt{1+\frac{a^{2 / 3}-x^{2 / 3}}{x^{2 / 3}}} d x=a^{1 / 3} \int_{0}^{a} x^{2 / 3} d x=\frac{3}{5} a^{2} .
\end{gathered}
$$

Now we find the coordinates of the center of mass:

$$
\begin{aligned}
& x_{c}=\frac{1}{M_{A B}} \int_{A B} x \cdot p(x, y) d l \\
& y_{c}=\frac{1}{M_{A B}} \int_{A B} y \cdot p(x, y) d l
\end{aligned}
$$

Substituting the given density $p(x, y)=x$ and performing the integration:

$$
\begin{aligned}
& x_{c}=\frac{1}{M_{A B}} \int_{0}^{a} x^{2} \cdot \sqrt{1+\left(y^{\prime}\right)^{2}} d x=\frac{5}{a^{2}} \int_{0}^{a} a^{1 / 3} x^{5 / 3} d x=\frac{5}{8} a \\
& y_{c}=\frac{1}{M_{A B}} \int_{A B} x y \cdot \sqrt{d x^{2}+d y^{2}}:
\end{aligned}
$$

To find the ordinate of the center of mass $y_{c}$, it is convenient to represent the astroid using parametric equations:

$$
\begin{aligned}
& x=a \cos ^{3} \varphi, y=a \sin ^{3} \varphi, \varphi \in\left[0, \frac{\pi}{2}\right] . \text { We get } \\
& y_{c}=\frac{1}{M_{A B}} \int_{0}^{\frac{\pi}{2}} a^{2} \cos ^{3} \varphi \sin ^{3} \varphi \sqrt{\left(3 a \cos ^{2} \varphi \sin \varphi\right)^{2}+\left(3 a \sin ^{2} \varphi \cos \varphi\right)^{2}} d \varphi= \\
& =5 a \int_{0}^{\frac{\pi}{2}} \cos ^{4} \varphi \sin ^{4} \varphi d \varphi=\frac{5 a}{32} \int_{0}^{\pi} \sin ^{4} t d t
\end{aligned}
$$

To compute the last integral, we use known properties of trigonometric functions. First, it is not difficult to see that

$$
\int_{0}^{\pi} \sin ^{4} t d t=\int_{0}^{\pi} \cos ^{4} t d t
$$

Secondly, from the fundamental trigonometric identity and the power reduction formula, it follows that:

$$
\begin{gathered}
\pi=\int_{0}^{\pi} d t=\int_{0}^{\pi}\left(\sin ^{2} t+\cos ^{2} t\right)^{2} d t= \\
=2 \int_{0}^{\pi} \sin ^{4} t d t+\int_{0}^{\pi} \frac{1-\cos 4 t}{4} d t=2 \int_{0}^{\pi} \sin ^{4} t d t+\frac{\pi}{4} .
\end{gathered}
$$

Therefore, the desired integral is

$$
\int_{0}^{\pi} \sin ^{4} t d t=\frac{3 \pi}{8} .
$$

The ordinate of the center of mass of the astroid is

$$
y_{c}=\frac{5 \pi a}{128} .
$$

b) Since the helical line is homogeneous and $p(x, y)=1$, the coordinates of its center of mass are calculated using the formulas:

$$
\begin{aligned}
& x_{c}=\int_{A B} x d l, \quad y_{c}=\int_{A B} y d l, \quad z_{c}=\int_{A B} z d l, \\
& d l=\sqrt{d x^{2}+d y^{2}+d z^{2}}=\sqrt{\dot{x}^{2}+\dot{y}^{2}+\dot{z}^{2}} d t=\sqrt{1+\frac{H^{2}}{4 \pi^{2}}} d t .
\end{aligned}
$$

Considering the given parametrization of the helical line, we reduce the line integrals of the first kind to definite integrals:

$$
\begin{aligned}
& x_{c}=\int_{0}^{2 \pi} \cos t \sqrt{1+\frac{H^{2}}{4 \pi^{2}}} d t=0 \\
& y_{c}=\int_{0}^{2 \pi} \sin t \sqrt{1+\frac{H^{2}}{4 \pi^{2}}} d t=0 \\
& z_{c}=\int_{0}^{2 \pi} \frac{H}{2 \pi} t \sqrt{1+\frac{H^{2}}{4 \pi^{2}}} d t=H \pi \sqrt{1+\frac{H^{2}}{4 \pi^{2}}} .
\end{aligned}
$$

## Calculation of line integrals of the second kind.