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We are tasked with calculating a line integral of a vector field \(\mathbf{F}\) along the curve \(\mathbf{C}\). The problem is defined as follows:

Given:
- The vector field \(\mathbf{F} = (x + y, z, 2x - y)\),
- The curve \(\mathbf{r}(t) = (3t, t^2, t^3)\) for \( t \in [0, 2] \),

The line integral is:

\[
\int_C \mathbf{F} \cdot d\mathbf{r}
\]

### Step 1: Parameterizing the Vector Field
The curve \(\mathbf{r}(t) = (x(t), y(t), z(t)) = (3t, t^2, t^3)\). This gives:
- \( x(t) = 3t \),
- \( y(t) = t^2 \),
- \( z(t) = t^3 \).

We substitute these into the vector field \(\mathbf{F}(x,y,z) = (x + y, z, 2x - y)\) to express \(\mathbf{F}\) as a function of \(t\):
\[
\mathbf{F}(t) = (x(t) + y(t), z(t), 2x(t) - y(t)).
\]

Substituting \( x(t) = 3t \), \( y(t) = t^2 \), and \( z(t) = t^3 \) into \(\mathbf{F}\):
\[
\mathbf{F}(t) = (3t + t^2, t^3, 2(3t) - t^2) = (3t + t^2, t^3, 6t - t^2).
\]

### Step 2: Finding \( d\mathbf{r}(t) \)
The differential \( d\mathbf{r}(t) \) is the derivative of the curve \(\mathbf{r}(t)\) with respect to \(t\):
\[
\frac{d}{dt} \mathbf{r}(t) = \frac{d}{dt} (3t, t^2, t^3) = (3, 2t, 3t^2).
\]
Thus, \( d\mathbf{r}(t) = (3, 2t, 3t^2) dt \).

### Step 3: Computing the Dot Product
Now, compute the dot product \(\mathbf{F}(t) \cdot d\mathbf{r}(t)\):
\[
\mathbf{F}(t) \cdot d\mathbf{r}(t) = (3t + t^2, t^3, 6t - t^2) \cdot (3, 2t, 3t^2).
\]
This gives:
\[
(3t + t^2)(3) + (t^3)(2t) + (6t - t^2)(3t^2).
\]
Expanding each term:
\[
(3t + t^2)(3) = 9t + 3t^2,
\]
\[
(t^3)(2t) = 2t^4,
\]
\[
(6t - t^2)(3t^2) = 18t^3 - 3t^4.
\]
Adding these up:
\[
9t + 3t^2 + 2t^4 + 18t^3 - 3t^4 = 9t + 3t^2 + 18t^3 - t^4.
\]

### Step 4: Integrating Over \( t \in [0, 2] \)
Now, we integrate the expression \( 9t + 3t^2 + 18t^3 - t^4 \) over the interval \( [0, 2] \):
\[
\int_0^2 (9t + 3t^2 + 18t^3 - t^4) dt.
\]
We compute the integral term by term:
\[
\int_0^2 9t \, dt = \left[ \frac{9t^2}{2} \right]_0^2 = \frac{9(2)^2}{2} = 18,
\]
\[
\int_0^2 3t^2 \, dt = \left[ t^3 \right]_0^2 = 8,
\]
\[
\int_0^2 18t^3 \, dt = \left[ \frac{18t^4}{4} \right]_0^2 = \frac{18(2)^4}{4} = 72,
\]
\[
\int_0^2 -t^4 \, dt = \left[ \frac{-t^5}{5} \right]_0^2 = \frac{-32}{5}.
\]
Summing these values:
\[
18 + 8 + 72 - \frac{32}{5} = 98 - \frac{32}{5} = \frac{490}{5} - \frac{32}{5} = \frac{458}{5} = 91.6.
\]

Thus, the value of the line integral is **91.6**.

Would you like more details or have any questions?

Here are 5 related questions you might find interesting:
1. How do parameterizations of curves affect line integrals?
2. What physical interpretations can line integrals have in vector fields?
3. How does Green's Theorem relate to line integrals?
4. Can line integrals be used to compute work in a force field?
5. What are the conditions for a vector field to be conservative?

**Tip:** Line integrals often have physical interpretations, such as computing the work done by a force field along a curve. Understanding these can deepen your grasp of the concept.

Calculate the line integral \( \int_C \mathbf{F} \cdot d\mathbf{r} \) where \( \mathbf{F} = (x + y, z, 2x - y) \) and \( C \) is a curve given by \( \mathbf{r}(t) = (3t, t^2, t^3) \) for \( t \in [0, 2] \).

Learn how to calculate the line integral of the vector field \( \mathbf{F} = (x + y, z, 2x - y) \) along the curve \( \mathbf{r}(t) = (3t, t^2, t^3) \) for \( t \in [0, 2] \). This solution provides a step-by-step approach, covering concepts in vector fields, line integrals, and parametric curves, suitable for undergraduate calculus students.

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