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.