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To evaluate the line integral \( \int_C y \, dx + z \, dy + x \, dz \), we need to parameterize the path \( C \) given by the equations \( x = \sqrt{t} \), \( y = t \), and \( z = t^2 \) for \( 1 \leq t \leq 4 \).

### Step 1: Parameterize \( dx \), \( dy \), and \( dz \)
Given:
\[
x = \sqrt{t}, \quad y = t, \quad z = t^2
\]
we find the differentials \( dx \), \( dy \), and \( dz \) in terms of \( dt \):

1. \( dx = \frac{d}{dt}(\sqrt{t}) \, dt = \frac{1}{2\sqrt{t}} \, dt \)
2. \( dy = \frac{d}{dt}(t) \, dt = 1 \, dt \)
3. \( dz = \frac{d}{dt}(t^2) \, dt = 2t \, dt \)

### Step 2: Substitute into the Integral
Now, substitute \( y = t \), \( z = t^2 \), and \( x = \sqrt{t} \) along with the differentials into the integral:
\[
\int_C y \, dx + z \, dy + x \, dz = \int_{1}^{4} \left( t \cdot \frac{1}{2\sqrt{t}} + t^2 \cdot 1 + \sqrt{t} \cdot 2t \right) \, dt
\]

### Step 3: Simplify the Integrand
Breaking down each term:
1. \( y \, dx = t \cdot \frac{1}{2\sqrt{t}} = \frac{t}{2\sqrt{t}} = \frac{\sqrt{t}}{2} \)
2. \( z \, dy = t^2 \cdot 1 = t^2 \)
3. \( x \, dz = \sqrt{t} \cdot 2t = 2t\sqrt{t} \)

Thus, the integrand becomes:
\[
\frac{\sqrt{t}}{2} + t^2 + 2t\sqrt{t}
\]

### Step 4: Evaluate the Integral
Now we evaluate:
\[
\int_{1}^{4} \left( \frac{\sqrt{t}}{2} + t^2 + 2t\sqrt{t} \right) \, dt
\]

Each term can be integrated individually:
1. For \( \int_{1}^{4} \frac{\sqrt{t}}{2} \, dt = \int_{1}^{4} \frac{t^{1/2}}{2} \, dt \):
   \[
   = \frac{1}{2} \cdot \frac{t^{3/2}}{3/2} \Big|_{1}^{4} = \frac{1}{3} t^{3/2} \Big|_{1}^{4} = \frac{1}{3} \left(4^{3/2} - 1^{3/2}\right) = \frac{1}{3} (8 - 1) = \frac{7}{3}
   \]

2. For \( \int_{1}^{4} t^2 \, dt \):
   \[
   = \frac{t^3}{3} \Big|_{1}^{4} = \frac{4^3}{3} - \frac{1^3}{3} = \frac{64 - 1}{3} = \frac{63}{3} = 21
   \]

3. For \( \int_{1}^{4} 2t\sqrt{t} \, dt = \int_{1}^{4} 2t^{3/2} \, dt \):
   \[
   = 2 \cdot \frac{t^{5/2}}{5/2} \Big|_{1}^{4} = \frac{4}{5} t^{5/2} \Big|_{1}^{4} = \frac{4}{5} \left(4^{5/2} - 1^{5/2}\right) = \frac{4}{5} (32 - 1) = \frac{4}{5} \cdot 31 = \frac{124}{5}
   \]

### Step 5: Combine Results
Now, we add up each evaluated term:
\[
\frac{7}{3} + 21 + \frac{124}{5}
\]

To add these, we find a common denominator, which is 15:
1. \( \frac{7}{3} = \frac{35}{15} \)
2. \( 21 = \frac{315}{15} \)
3. \( \frac{124}{5} = \frac{372}{15} \)

Thus,
\[
\frac{35 + 315 + 372}{15} = \frac{722}{15}
\]

### Final Answer
\[
\int_C y \, dx + z \, dy + x \, dz = \frac{722}{15}
\]

Would you like further details, or do you have any questions?

### Related Questions:
1. How would the result change if \( C \) had different parameter bounds?
2. How can this line integral be interpreted geometrically?
3. What is the significance of each term \( y \, dx \), \( z \, dy \), and \( x \, dz \) in the context of the integral?
4. How would we approach this if \( x \), \( y \), and \( z \) were different functions of \( t \)?
5. What are some alternative methods for evaluating line integrals?

### Tip:
When calculating line integrals, ensure each part of the parameterization matches with the differential terms to avoid miscalculation.

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

Learn how to evaluate the line integral \( \int_C y \, dx + z \, dy + x \, dz \) over a parameterized path with \( x = \sqrt{t} \), \( y = t \), and \( z = t^2 \) for \( 1 \leq t \leq 4 \). This solution covers parameterization techniques and integral calculations, suitable for college-level calculus.

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