To calculate the line integral of the gradient of $f(x, y, z)$ along the path $l$ using the Fundamental Theorem of Line Integrals, we proceed as follows:

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Step 1: Restate the fundamental theorem of line integrals

If $\nabla f(x, y, z)$ is the gradient of a scalar field $f(x, y, z)$, then the line integral along a path $r(t)$ parametrized from $t = a$ to $t = b$ is given by:

$\int_{l} \nabla f(x, y, z) \cdot d\mathbf{r} = f(r(b)) - f(r(a))$

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Step 2: Identify the components of the problem

1. The scalar field:
   $f(x, y, z) = x^3 + z^2 \ln(y)$

2. The parametrization of the path $r(t)$:
   $r(t) = \langle t^2, t^3, 2t \rangle, \quad 1 \leq t \leq 2$

   This gives: $x = t^2, \quad y = t^3, \quad z = 2t$

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Step 3: Calculate $f(r(t))$

Substitute $x = t^2$, $y = t^3$, and $z = 2t$ into $f(x, y, z)$: $f(r(t)) = (t^2)^3 + (2t)^2 \ln(t^3)$ $f(r(t)) = t^6 + 4t^2 \ln(t^3)$ $f(r(t)) = t^6 + 12t^2 \ln(t)$

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Step 4: Evaluate $f(r(2)) - f(r(1))$

1. Compute $f(r(2))$: $f(r(2)) = (2)^6 + 12(2)^2 \ln(2)$ $f(r(2)) = 64 + 48 \ln(2)$

2. Compute $f(r(1))$: $f(r(1)) = (1)^6 + 12(1)^2 \ln(1)$ Since $\ln(1) = 0$, this simplifies to: $f(r(1)) = 1$

3. Subtract $f(r(1))$ from $f(r(2))$: $f(r(2)) - f(r(1)) = (64 + 48 \ln(2)) - 1$ $f(r(2)) - f(r(1)) = 63 + 48 \ln(2)$

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Final Answer

$\int_{l} \nabla f(x, y, z) \cdot d\mathbf{r} = 63 + 48 \ln(2)$

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Would you like additional details or clarification?
Here are five related questions for further exploration:

1. How do we derive the parametric representation of a path for a given scalar field?
2. Why does the fundamental theorem of line integrals simplify the calculation?
3. What conditions are necessary for the fundamental theorem of line integrals to hold?
4. How does the path parametrization affect the integral of scalar fields?
5. Can this method be extended to non-conservative fields, and if so, how?

Tip: Always confirm that the vector field is conservative by verifying $\nabla \times \mathbf{F} = 0$ when applying the fundamental theorem of line integrals!