We are tasked with evaluating the line integral of a vector field $\mathbf{F}$ along a curve $C$ parameterized by $\mathbf{r}(t)$. The integral is given by:

$\int_C \mathbf{F} \cdot \mathbf{T} \, ds$

where $\mathbf{F} = \langle -6y, 6x \rangle$ and the curve $C$ is the semicircle parameterized by:

$\mathbf{r}(t) = \langle 4\cos t, 4\sin t \rangle$

for $t \in [0, \pi]$.

Step 1: Find the expression for $\mathbf{T}$ (the unit tangent vector).

The unit tangent vector $\mathbf{T}(t)$ is the derivative of the position vector $\mathbf{r}(t)$ normalized:

$\mathbf{r}(t) = \langle 4 \cos t, 4 \sin t \rangle$

The derivative of $\mathbf{r}(t)$ is:

$\mathbf{r}'(t) = \langle -4 \sin t, 4 \cos t \rangle$

The magnitude of $\mathbf{r}'(t)$ is:

$|\mathbf{r}'(t)| = \sqrt{(-4 \sin t)^2 + (4 \cos t)^2} = \sqrt{16 \sin^2 t + 16 \cos^2 t} = 4$

Thus, the unit tangent vector is:

$\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|} = \langle -\sin t, \cos t \rangle$

Step 2: Evaluate the vector field $\mathbf{F}$ along the curve $C$.

The components of the vector field $\mathbf{F} = \langle -6y, 6x \rangle$ must be evaluated along the curve $C$. Since $\mathbf{r}(t) = \langle 4 \cos t, 4 \sin t \rangle$, we substitute these values into $\mathbf{F}$:

$\mathbf{F}(\mathbf{r}(t)) = \langle -6(4 \sin t), 6(4 \cos t) \rangle = \langle -24 \sin t, 24 \cos t \rangle$

Step 3: Compute the dot product $\mathbf{F} \cdot \mathbf{T}$.

Now, we compute the dot product $\mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{T}(t)$:

$\mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{T}(t) = \langle -24 \sin t, 24 \cos t \rangle \cdot \langle -\sin t, \cos t \rangle$

The dot product is:

$\mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{T}(t) = (-24 \sin t)(-\sin t) + (24 \cos t)(\cos t) = 24 \sin^2 t + 24 \cos^2 t = 24(\sin^2 t + \cos^2 t) = 24$

Step 4: Set up the integral.

The differential $ds$ is given by $ds = |\mathbf{r}'(t)| \, dt = 4 \, dt$. Thus, the line integral becomes:

$\int_C \mathbf{F} \cdot \mathbf{T} \, ds = \int_0^\pi 24 \cdot 4 \, dt = 96 \int_0^\pi dt$

Step 5: Compute the integral.

The integral is straightforward:

$96 \int_0^\pi dt = 96 \cdot \pi = 96\pi$

Final Answer:

$\int_C \mathbf{F} \cdot \mathbf{T} \, ds = 96\pi$

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Would you like further clarification on any step or additional questions? Here are some related questions you might find helpful:

1. How would the result change if the curve were a full circle rather than a semicircle?
2. What would the integral be if the vector field $\mathbf{F}$ were $\langle 6y, -6x \rangle$?
3. How do you compute a line integral of a vector field over a parametric curve in general?
4. What is the physical interpretation of the line integral of a vector field along a curve?
5. How can you use Green's Theorem to evaluate a line integral over a closed curve?

Tip: When evaluating line integrals, always ensure that the vector field is correctly parameterized and that the differential $ds$ is appropriately factored in.