To evaluate the line integral

$\int_C y \, dx + z \, dy + x \, dz$

where the path $C$ is parametrized by the equations:

• $x = \sqrt{t}$
• $y = t$
• $z = t^2$

with $t$ ranging from 1 to 4, we can proceed as follows:

1. Find $dx$, $dy$, and $dz$ in terms of $dt$:

   • Given $x = \sqrt{t}$, then: $dx = \frac{1}{2\sqrt{t}} \, dt$
   • Given $y = t$, then: $dy = 1 \, dt$
   • Given $z = t^2$, then: $dz = 2t \, dt$

2. Substitute into the integral: Substitute $y = t$, $z = t^2$, $x = \sqrt{t}$, and the expressions for $dx$, $dy$, and $dz$ in terms of $dt$: $\int_C y \, dx + z \, dy + x \, dz = \int_{1}^{4} \left( t \cdot \frac{1}{2\sqrt{t}} + t^2 \cdot 1 + \sqrt{t} \cdot 2t \right) dt$

3. Simplify each term:

   • The first term $t \cdot \frac{1}{2\sqrt{t}} = \frac{t}{2\sqrt{t}} = \frac{\sqrt{t}}{2}$
   • The second term is $t^2$
   • The third term is $\sqrt{t} \cdot 2t = 2t\sqrt{t}$

   Thus, the integral becomes: $\int_{1}^{4} \left( \frac{\sqrt{t}}{2} + t^2 + 2t\sqrt{t} \right) dt$

4. Evaluate each component separately:

   • For $\int_{1}^{4} \frac{\sqrt{t}}{2} \, dt$: Substitute $u = t^{1/2} \Rightarrow t = u^2 \Rightarrow dt = 2u \, du$: [ \int \frac{\sqrt{t}}{2} , dt = \int u , du = \frac{u^2}{2} + C