Example 2.67. Calculate the mass of the plane curve $y=\frac{x^{3}}{3}$ over the segment from $x=0$ to $x=0.1$ units of length, if the linear density $\rho$ depends on the coordinate $x$ as follows: $\rho(x)=1+x^{2}$.
Solution. The elementary mass of a curve segment from $x=$ $=0$ to $x=0.1$ is determined by the formula $d m=\rho(x) d l=$ $=\rho(x) \sqrt{1+\left(\frac{d y}{d x}\right)^{2}} d x$, hence the total mass

$$
\begin{gathered}
m=\int_{0}^{0.1} \rho(x) \sqrt{1+\left(\frac{d y}{d x}\right)^{2}} d x=\int_{0}^{0.1}\left(1+x^{2}\right) \sqrt{1+x^{2}} d x= \\
=\left|\begin{array}{l}
x=\operatorname{sh} t, 1+x^{2}=1+\operatorname{sh}^{2} t=\operatorname{ch}^{2} t, \\
d x=\operatorname{ch} t d t, x_{1}=0 \rightarrow t_{1}=0, x_{2}=0.1 \rightarrow t_{2}=\operatorname{Arsh} 0.1
\end{array}\right|= \\
=\int_{0}^{\operatorname{Arsh} 0.1} \operatorname{ch}^{4} t d t=\left|\operatorname{ch}^{2} t=\frac{\operatorname{ch} 2 t+1}{2}\right|=
\end{gathered}
$$

$$
\begin{gathered}
=\int_{0}^{\operatorname{Arsh} 0.1}\left(\frac{1+\operatorname{ch} 2 t}{2}\right)^{2} d t=\int_{0}^{\operatorname{Arsh} 0.1}\left(1+2 \operatorname{ch} 2 t+\mathrm{ch}^{2} 2 t\right) d t= \\
=0.25 \int_{0}^{\text {Arsh } 0.1}(1+2 \operatorname{ch} 2 t+0.5 \operatorname{ch} 4 t) d t= \\
=\left.(0.375 t+0.25 \operatorname{sh} 2 t+0.03125 \operatorname{sh} 4 t)\right|_{0} ^{\text {Arsh } 0.1}= \\
=\left|\operatorname{Arsh} t=\ln \left(t+\sqrt{1+t^{2}}\right)\right|=0.099985655
\end{gathered}
$$