3. Evaluate each line integral:
(i) $\int_{C}\left(x^{2}+y^{2}+z^{2}\right) d s$
C: $x=t, \quad y=\cos 2 t, \quad z=\sin 2 t, \quad 0 \leq t \leq 2 \pi$
(ii) $\quad \int_{C} \vec{F} \cdot d \vec{r}, \quad \vec{F}=x z \vec{\imath}+z^{3} \vec{\jmath}+y \vec{k}$
$C: \quad \vec{r}(t)=e^{t} \vec{\imath}+e^{2 t} \vec{\jmath}+e^{-t} \vec{k}, \quad-1 \leq t \leq 1$

Step 1 ：Given the curve $C$ parametrized by $x = t$, $y = \cos(2t)$, and $z = \sin(2t)$ for $0 \leq t \leq 2\pi$, we are asked to evaluate the line integral $\int_{C}(x^{2}+y^{2}+z^{2}) ds$.

Step 2 ：The line integral of a scalar function along a curve $C$ is given by $\int_C f(x, y, z) ds$, where $ds$ is the differential arc length along the curve.

Step 3 ：We can express $ds$ in terms of $dt$ using the parametrization of the curve. The differential arc length $ds$ is given by $ds = \sqrt{(dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2} dt$.

Step 4 ：Differentiating the parametric equations for $x$, $y$, and $z$ with respect to $t$, we find $dx/dt = 1$, $dy/dt = -2\sin(2t)$, and $dz/dt = 2\cos(2t)$.

Step 5 ：Substituting $x = t$, $y = \cos(2t)$, $z = \sin(2t)$, $dx/dt = 1$, $dy/dt = -2\sin(2t)$, and $dz/dt = 2\cos(2t)$ into the integral, we get $\int_{0}^{2\pi} (t^2 + \cos^2(2t) + \sin^2(2t)) \sqrt{1 + 4\sin^2(2t) + 4\cos^2(2t)} dt$.

Step 6 ：Evaluating this integral from $t = 0$ to $t = 2\pi$, we find the value of the line integral to be $2\sqrt{5}\pi + \frac{8\sqrt{5}\pi^3}{3}$.

Step 7 ：Final Answer: The value of the line integral $\int_{C}(x^{2}+y^{2}+z^{2}) ds$ along the curve $C$ parametrized by $x = t$, $y = \cos(2t)$, and $z = \sin(2t)$ for $0 \leq t \leq 2\pi$ is $\boxed{2\sqrt{5}\pi + \frac{8\sqrt{5}\pi^3}{3}}$.