887. Find the coordinates of the center of gravity of the arc $A B$ of the helical line $x=a \cos t, y=a \sin t, z=b t$, if at each point of it the linear density is proportional to the ordinate of that point; $t_{A}=0, t_{B}=\pi$.
Solution. We apply formulas (4). We will compute the curvilinear integrals contained in these formulas, transforming them into ordinary integrals with the variable $t$:

$$
\begin{aligned}
& \dot{x}=-a \sin t, \dot{y}=a \cos t, \quad \dot{z}=b ; d l=\sqrt{\dot{x}^{2}+\dot{y}^{2}+\dot{z}^{2}} d t= \\
& =\sqrt{a^{2}+b^{2}} d t \\
& I_{1}=\int_{A B} x \delta d l=\int_{0}^{\pi} a \cos t \cdot k b t \sqrt{a^{2}+b^{2}} d t= \\
& =\left.a k b \sqrt{a^{2}+b^{2}}(t \sin t+\cos t)\right|_{0} ^{\pi}=-2 a b k \sqrt{a^{2}+b^{2}} \\
& I_{2}==\int_{A B} \delta d l=\int_{0}^{\pi} k b t \sqrt{a^{2}+b^{2}} d t=\left.k b \sqrt{a^{2}+b^{2}} \frac{t^{2}}{2}\right|_{u} ^{\pi}= \\
& =\frac{b / 2 \pi^{2}}{2} \sqrt{a^{2}+b^{2}} \\
& I_{3}=\int_{A B} y \delta d l=\int_{0}^{\pi} a \sin t \cdot k b t \sqrt{a^{2}+b^{2}} d t= \\
& =\left.a b k \sqrt{a^{2}+b^{2}}(\sin t-t \cos t)\right|_{0} ^{\pi}=a b k \pi \sqrt{a^{2}+b^{2}} \\
& l_{4}=\int_{A B} z \delta d l=\int_{0}^{\pi} b t k b t \sqrt{a^{2}+1-b^{2}} d t=\frac{k t^{2} \pi^{3}}{3} \sqrt{a^{2}+b^{2}}
\end{aligned}
$$

Therefore, $x_{C}=\frac{I_{1}}{I_{2}}=-\frac{4 a}{\pi^{2}} ; \quad y_{C}=\frac{I_{3}}{I_{2}}=\frac{2 a}{\pi} ; z_{C}=\frac{I_{4}}{I_{2}}=\frac{2}{3} b \pi$.