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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
    

Evaluate \( \int_C y \, dx + z \, dy + x \, dz \), where \( C \) is given by the equations \( x = \sqrt{t} \), \( y = t \), \( z = t^2 \), \( 1 \leq t \leq 4 \).

Explore how to evaluate the line integral \( \int_C y \, dx + z \, dy + x \, dz \) where the path \( C \) is defined by the parametric equations \( x = \sqrt{t} \), \( y = t \), and \( z = t^2 \) for \( 1 \leq t \leq 4 \). This solution demonstrates calculus concepts involving line integrals, parametric equations, and differentiation.

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