Find line integral

C
F · dr
of vector field
F(x, y, z) = 3x2zi + z2j + (x3 + 2yz)k
along curve C parameterized by
r(t) =

ln(t)
ln(2)

i + t3/2j + t cos(𝜋t)k,

1 ≤ t ≤ 4.

To evaluate the line integral of the vector field \(\mathbf{F}(x, y, z) = 3x^2z \, \mathbf{i} + z^2 \, \mathbf{j} + (x^3 + 2yz) \, \mathbf{k}\) along the curve \(C\) parameterized by \(\mathbf{r}(t) = \ln(t) \, \mathbf{i} + \frac{t^{3/2}}{\ln(2)} \, \mathbf{j} + t \cos(\pi t) \, \mathbf{k}\) for \(1 \leq t \leq 4\), we proceed as follows:

### Step 1: Compute \(\mathbf{F}(\mathbf{r}(t))\)
First, we substitute \(x = \ln(t)\), \(y = \frac{t^{3/2}}{\ln(2)}\), and \(z = t \cos(\pi t)\) into \(\mathbf{F}(x, y, z)\):

1. The \(i\)-component of \(\mathbf{F}\) is \(3x^2z = 3(\ln(t))^2 \cdot t \cos(\pi t)\).
2. The \(j\)-component of \(\mathbf{F}\) is \(z^2 = (t \cos(\pi t))^2 = t^2 \cos^2(\pi t)\).
3. The \(k\)-component of \(\mathbf{F}\) is \(x^3 + 2yz = (\ln(t))^3 + 2 \cdot \frac{t^{3/2}}{\ln(2)} \cdot t \cos(\pi t)\).

So,
\[
\mathbf{F}(\mathbf{r}(t)) = 3(\ln(t))^2 \, t \cos(\pi t) \, \mathbf{i} + t^2 \cos^2(\pi t) \, \mathbf{j} + \left((\ln(t))^3 + 2 \frac{t^{5/2} \cos(\pi t)}{\ln(2)}\right) \, \mathbf{k}.
\]

### Step 2: Compute \(\frac{d\mathbf{r}}{dt}\)
Differentiate each component of \(\mathbf{r}(t)\) with respect to \(t\):

1. For the \(i\)-component, \(\frac{d}{dt} \ln(t) = \frac{1}{t}\).
2. For the \(j\)-component, \(\frac{d}{dt} \frac{t^{3/2}}{\ln(2)} = \frac{3}{2} \frac{t^{1/2}}{\ln(2)}\).
3. For the \(k\)-component, \(\frac{d}{dt} (t \cos(\pi t)) = \cos(\pi t) - \pi t \sin(\pi t)\).

Thus,
\[
\frac{d\mathbf{r}}{dt} = \frac{1}{t} \, \mathbf{i} + \frac{3}{2} \frac{t^{1/2}}{\ln(2)} \, \mathbf{j} + \left(\cos(\pi t) - \pi t \sin(\pi t)\right) \, \mathbf{k}.
\]

### Step 3: Compute \(\mathbf{F}(\mathbf{r}(t)) \cdot \frac{d\mathbf{r}}{dt}\)
The dot product \(\mathbf{F}(\mathbf{r}(t)) \cdot \frac{d\mathbf{r}}{dt}\) is given by:
\[
\mathbf{F}(\mathbf{r}(t)) \cdot \frac{d\mathbf{r}}{dt} = \left(3(\ln(t))^2 \, t \cos(\pi t)\right) \cdot \frac{1}{t} + \left(t^2 \cos^2(\pi t)\right) \cdot \frac{3}{2} \frac{t^{1/2}}{\ln(2)} + \left((\ln(t))^3 + 2 \frac{t^{5/2} \cos(\pi t)}{\ln(2)}\right) \cdot \left(\cos(\pi t) - \pi t \sin(\pi t)\right).
\]

Simplify each term individually:

1. The first term: \[3(\ln(t))^2 \cos(\pi t).\]
2. The second term: \[\frac{3}{2} \frac{t^{5/2} \cos^2(\pi t)}{\ln(2)}.\]
3. The third term: \[\left((\ln(t))^3 + 2 \frac{t^{5/2} \cos(\pi t)}{\ln(2)}\right) \left(\cos(\pi t) - \pi t \sin(\pi t)\right).\]

### Step 4: Integrate from \(t = 1\) to \(t = 4\)
Now we need to integrate the resulting expression with respect to \(t\) over the interval \([1, 4]\):

\[
\int_{1}^{4} \left(3(\ln(t))^2 \cos(\pi t) + \frac{3}{2} \frac{t^{5/2} \cos^2(\pi t)}{\ln(2)} + \left((\ln(t))^3 + 2 \frac{t^{5/2} \cos(\pi t)}{\ln(2)}\right) \left(\cos(\pi t) - \pi t \sin(\pi t)\right)\right) \, dt.
\]

This integral is quite complex and would generally be evaluated using numerical integration techniques.

Would you like a detailed breakdown of the steps involved in computing this numerically, or do you have any other questions?

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**Tip:** When solving line integrals involving complex functions, substituting and simplifying before integrating can sometimes reveal easier ways to handle the calculation.

Find line integral ∫C F · dr of vector field F(x, y, z) = 3x^2zi + z^2j + (x^3 + 2yz)k along curve C parameterized by r(t) = ln(t)i + (t^3/2)/ln(2)j + t cos(πt)k, 1 ≤ t ≤ 4.

Find the line integral of vector field F(x, y, z) = 3x^2zi + z^2j + (x^3 + 2yz)k along the parameterized curve r(t) = ln(t)i + (t^3/2)/ln(2)j + t cos(πt)k for 1 ≤ t ≤ 4. Includes detailed steps on evaluating F(r(t)) and computing the dot product with r'(t), suitable for advanced calculus students.

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