To evaluate the line integral of $f(x, y, z) = x + \sqrt{y} - z^2$ along the path from $(0, 0, 0)$ to $(2, 4, 2)$, which is split into two segments $C_1$ and $C_2$, we proceed as follows:

Step 1: Evaluate the integral along $C_1$

The path $C_1$ is given by: $\vec{r}_1(t) = t\vec{i} + t^2\vec{j}, \quad 0 \leq t \leq 2.$ This gives:

• $x = t$
• $y = t^2$
• $z = 0$ (since there is no $k$-component in $C_1$).

The differential $d\vec{r}_1$ is: $\frac{d\vec{r}_1}{dt} = \frac{d}{dt}(t\vec{i} + t^2\vec{j}) = \vec{i} + 2t\vec{j}.$ Thus, $d\vec{r}_1 = (\vec{i} + 2t\vec{j}) \, dt$.

Substituting into $f(x, y, z) = x + \sqrt{y} - z^2$: f(x, y, z) = t + \sqrt{t^2} - 0^2 = t + |t| = 2t \quad \text{(since t \geq 0)}.

The integral along $C_1$ is: $\int_{C_1} f(x, y, z) \, d\vec{r}_1 = \int_0^2 (2t) \cdot \sqrt{1 + (2t)^2} \, dt.$