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?

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.