To evaluate the line integral $\int_C \mathbf{F} \cdot d\mathbf{r}$, we proceed as follows:

Given:

$\mathbf{F} = \langle 4\sin x, -\cos y, 10xz \rangle,$ and the path $\mathbf{r}(t) = (-3t^3, 3t^2, 2t)$ for $0 \leq t \leq 1$.

Step 1: Parametrize $\mathbf{F}$

Substitute $\mathbf{r}(t)$ into $\mathbf{F}$: $x = -3t^3, \quad y = 3t^2, \quad z = 2t.$

Thus, $\mathbf{F}(\mathbf{r}(t)) = \langle 4\sin(-3t^3), -\cos(3t^2), 10(-3t^3)(2t) \rangle.$

Simplify: $\mathbf{F}(\mathbf{r}(t)) = \langle 4\sin(-3t^3), -\cos(3t^2), -60t^4 \rangle.$

Step 2: Compute $d\mathbf{r}$

The derivative of $\mathbf{r}(t)$ with respect to $t$ is: $\mathbf{r}'(t) = \frac{d}{dt}(-3t^3, 3t^2, 2t) = \langle -9t^2, 6t, 2 \rangle.$

Thus, $d\mathbf{r} = \langle -9t^2, 6t, 2 \rangle \, dt.$

Step 3: Compute $\mathbf{F} \cdot d\mathbf{r}$

The dot product $\mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t)$ is: $\mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) = \langle 4\sin(-3t^3), -\cos(3t^2), -60t^4 \rangle \cdot \langle -9t^2, 6t, 2 \rangle.$

Perform the dot product: $\mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) = (4\sin(-3t^3))(-9t^2) + (-\cos(3t^2))(6t) + (-60t^4)(2).$

Simplify each term: $-36t^2 \sin(-3t^3) - 6t\cos(3t^2) - 120t^4.$

Step 4: Integrate over $t \in [0, 1]$

The integral becomes: $\int_C \mathbf{F} \cdot d\mathbf{r} = \int_0^1 \big( -36t^2 \sin(-3t^3) - 6t\cos(3t^2) - 120t^4 \big) \, dt.$

Split the integral:

$\int_C \mathbf{F} \cdot d\mathbf{r} = \int_0^1 -36t^2 \sin(-3t^3) \, dt - \int_0^1 6t\cos(3t^2) \, dt - \int_0^1 120t^4 \, dt.$

Solve each term:

1. First term: Let $u = -3t^3$, so $du = -9t^2 \, dt$: $\int_0^1 -36t^2 \sin(-3t^3) \, dt = \int_0^{-3} -36 \sin(u) \frac{du}{-9} = \int_0^{-3} 4 \sin(u) \, du.$

The integral of $\sin(u)$ is: $\int \sin(u) \, du = -\cos(u).$

Evaluate: [ \int_0^{-3} 4 \sin(u) , du = -4\cos(u) \Big|_0^{-3} = -4[\cos(-3) - \cos(0)] = -4[\cos(-3) - 1]. ]

2. Second term: Let $v = 3t^2$, so $dv = 6t \, dt$: $\int_0^1 6t \cos(3t^2) \, dt = \int_0^3 \cos(v) \, dv.$

The integral of $\cos(v)$ is: $\int \cos(v) \, dv = \sin(v).$

Evaluate: $\int_0^3 \cos(v) \, dv = \sin(v) \Big|_0^3 = \sin(3) - \sin(0) = \sin(3).$

3. Third term: $\int_0^1 120t^4 \, dt = 120 \int_0^1 t^4 \, dt.$

The integral of $t^4$ is: $\int t^4 \, dt = \frac{t^5}{5}.$

Evaluate: $120 \int_0^1 t^4 \, dt = 120 \left[\frac{t^5}{5}\right]_0^1 = 120 \cdot \frac{1^5}{5} = 24.$

Step 5: Combine the results

$\int_C \mathbf{F} \cdot d\mathbf{r} = -4[\cos(-3) - 1] - \sin(3) - 24.$

Simplify further if needed.

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Do you want detailed explanations for any specific step or further simplification? Here are some related questions for further exploration:

1. What is the physical interpretation of a line integral in vector fields?
2. How can we verify line integral results using computational tools?
3. What are the advantages of parameter substitution in line integrals?
4. How does the orientation of the curve affect the line integral?
5. What happens if the vector field is conservative?

Tip: Always check if the vector field is conservative before solving; it can simplify the computation significantly.