874. Given points $A(3, -6, 0)$ and $B(-2, 4, 5)$. Compute the line integral $I=\int_{C} x y^{2} d x + y z^{2} d y - z x^{2} d z$:
1) along the straight line segment $O B$ and
2) along the arc $A B$ of the circle defined by the equations $x^{2} + y^{2} + z^{2} = 45, 2 x + y = 0$.
Solution. 1) First, we write the equation of the line passing through two points $\frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}}$, obtaining $\frac{x}{-2}=\frac{y}{4}=\frac{z}{5}$. By equating these equal ratios to the parameter $t$, we transform the canonical equations of the line $O B$ into parametric form: $x=-2 t, y=4 t, z=5 t$.

Next, using these equations, we transform the given line integral into an ordinary integral with the variable $t$, and then compute it

$$
\begin{gathered}
I_{1}=\int_{t_{0}=0}^{t_{B}=1}-2 t(4 t)^{2}(-2 d t)-4 t(5 t)^{2} 4 d t-5 t(-2 t)^{2} 5 d t= \\
=364 \int_{0}^{1} t^{3} d t=91
\end{gathered}
$$

2) Transform the given equations of the circle into parametric form. Setting $x=t$, we get $y=-2 t$ (from the second given equation), $z=\sqrt{45-5 t^{2}}$ (from the first equation). Hence, $d x=d t, d y=-2 d t, d z=-\frac{5 t d t}{\sqrt{4.5-5 t^{2}}}$

$$
\begin{gathered}
I_{2}=\int_{t_{A^{\prime}}=3}^{t_{B}=-2} t(-2 t)^{2} d t+(-2 t)\left(45-5 t^{2}\right)(-2 d t)- \\
-\sqrt{45-5 t^{2}} t^{2}\left(-\frac{5 t d t}{\sqrt{45-5 t^{2}}}\right)=\int_{3}^{-2}\left(180 t-17 t^{3}\right) d t=-173 \frac{3}{4}
\end{gathered}
$$