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To evaluate the line integral \(\int_C 3xy^2 \, ds\), where \(C\) is the right half of the circle \(x^2 + y^2 = 16\), parametrized by \(x = 4 \cos(t)\) and \(y = 4 \sin(t)\), we can proceed with the following steps:

1. **Set up the parametric equations:** The given parametrization of \(C\) is:
   \[
   x = 4 \cos(t), \quad y = 4 \sin(t)
   \]
   for \(t\) ranging from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\) (since we are only considering the right half of the circle).

2. **Calculate \(ds\):** The element \(ds\) represents the differential arc length, given by:
   \[
   ds = \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt
   \]
   Let's compute \(dx/dt\) and \(dy/dt\):
   \[
   \frac{dx}{dt} = -4 \sin(t), \quad \frac{dy}{dt} = 4 \cos(t)
   \]
   Then,
   \[
   ds = \sqrt{(-4 \sin(t))^2 + (4 \cos(t))^2} \, dt = \sqrt{16 \sin^2(t) + 16 \cos^2(t)} \, dt
   \]
   Using the Pythagorean identity \(\sin^2(t) + \cos^2(t) = 1\), we have:
   \[
   ds = \sqrt{16} \, dt = 4 \, dt
   \]

3. **Substitute into the integral:** We are given the integrand \(3xy^2\), so we need to express \(xy^2\) in terms of \(t\):
   \[
   x = 4 \cos(t), \quad y = 4 \sin(t) \Rightarrow y^2 = (4 \sin(t))^2 = 16 \sin^2(t)
   \]
   Thus,
   \[
   xy^2 = (4 \cos(t))(16 \sin^2(t)) = 64 \cos(t) \sin^2(t)
   \]
   Therefore, the integral becomes:
   \[
   \int_C 3xy^2 \, ds = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 3 \cdot 64 \cos(t) \sin^2(t) \cdot 4 \, dt
   \]
   Simplify the constants:
   \[
   = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 768 \cos(t) \sin^2(t) \, dt
   \]

4. **Evaluate the integral:** To simplify the integral \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 768 \cos(t) \sin^2(t) \, dt\), we can use the identity \(\sin^2(t) = \frac{1 - \cos(2t)}{2}\):
   \[
   = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 768 \cos(t) \cdot \frac{1 - \cos(2t)}{2} \, dt
   \]
   \[
   = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 384 \cos(t) (1 - \cos(2t)) \, dt
   \]
   Now distribute \(\cos(t)\):
   \[
   = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 384 \cos(t) \, dt - \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 384 \cos(t) \cos(2t) \, dt
   \]
   
Since \(\cos(t)\) is an odd function over a symmetric interval around zero, both integrals evaluate to zero:
   \[
   \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 384 \cos(t) \, dt = 0, \quad \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 384 \cos(t) \cos(2t) \, dt = 0
   \]

Thus, the final answer is:
\[
\int_C 3xy^2 \, ds = 0
\]

Would you like further details on any part of the solution?

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Here are some related questions to explore further:
1. How would this change if \(C\) were the entire circle instead of just the right half?
2. What if we parametrized \(C\) differently? How would that affect the solution?
3. Could we evaluate this line integral using Green's Theorem?
4. How would the process change if the integral were \(\int_C (x^2 + y^2) \, ds\) instead?
5. What role does the symmetry of the function play in determining the value of the integral?

**Tip:** Always check for symmetry in line integrals, as it can often simplify calculations.

Evaluate the line integral \(\int_C 3xy^2 \, ds\), where \(C\) is the right half of the circle \(x^2 + y^2 = 16\), parametrized by \(x = 4 \cos(t)\), \(y = 4 \sin(t)\).

This page provides a step-by-step solution for evaluating the line integral \(\int_C 3xy^2 \, ds\), where \(C\) represents the right half of the circle \(x^2 + y^2 = 16\), parametrized by \(x = 4 \cos(t)\) and \(y = 4 \sin(t)\). Learn about line integrals, arc length differentials, and trigonometric identities in this calculus problem.

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