The problem asks to evaluate the line integral:

$\int_C 5xy^2 \, ds,$

where $C$ is the right half of the circle $x^2 + y^2 = 9$.

Steps to Solve:

1. Parameterize the Curve $C$: The curve is the right half of the circle $x^2 + y^2 = 9$. A suitable parameterization for this half-circle is: $x = 3\cos(t), \quad y = 3\sin(t), \quad t \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right].$

2. Compute the Differential Arc Length $ds$: The arc length differential is given by: $ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt.$ From the parameterization: $\frac{dx}{dt} = -3\sin(t), \quad \frac{dy}{dt} = 3\cos(t),$ so: $ds = \sqrt{(-3\sin(t))^2 + (3\cos(t))^2} \, dt = \sqrt{9(\sin^2(t) + \cos^2(t))} \, dt = 3 \, dt.$

3. Substitute into the Integral: Substituting $x = 3\cos(t)$, $y = 3\sin(t)$, and $ds = 3 \, dt$, the integrand becomes: $5xy^2 \, ds = 5(3\cos(t))(3\sin(t))^2(3 \, dt).$ Simplify: $5xy^2 \, ds = 5(3\cos(t))(9\sin^2(t))(3) \, dt = 405\cos(t)\sin^2(t) \, dt.$

4. Evaluate the Integral: The integral becomes: $\int_C 5xy^2 \, ds = \int_{-\pi/2}^{\pi/2} 405\cos(t)\sin^2(t) \, dt.$ Use the identity $\sin^2(t) = 1 - \cos^2(t)$: $\int_{-\pi/2}^{\pi/2} 405\cos(t)\sin^2(t) \, dt = \int_{-\pi/2}^{\pi/2} 405\cos(t)(1 - \cos^2(t)) \, dt.$ Distribute and split the integral: $\int_{-\pi/2}^{\pi/2} 405\cos(t)(1 - \cos^2(t)) \, dt = \int_{-\pi/2}^{\pi/2} 405\cos(t) \, dt - \int_{-\pi/2}^{\pi/2} 405\cos(t)\cos^2(t) \, dt.$

   • The first term: $\int_{-\pi/2}^{\pi/2} 405\cos(t) \, dt = 405 [\sin(t)]_{-\pi/2}^{\pi/2} = 405 (\sin(\pi/2) - \sin(-\pi/2)) = 405 (1 - (-1)) = 810.$
   • The second term: $\int_{-\pi/2}^{\pi/2} 405\cos(t)\cos^2(t) \, dt = \text{(odd function over symmetric interval, so it evaluates to 0)}.$

   Thus, the integral simplifies to: $\int_C 5xy^2 \, ds = 810.$

Final Answer:

$\boxed{810}$

Would you like me to clarify any step or expand on specific concepts?

Related Questions:

1. What is the significance of symmetry in simplifying integrals over specific intervals?
2. How would the integral change if $C$ represented the full circle instead of the half-circle?
3. Why does the term involving $\cos(t)\cos^2(t)$ vanish in this case?
4. How can this problem be generalized to different radii or curves?
5. Can Green's Theorem or another vector calculus theorem be used here?

Tip:

When parameterizing curves, always double-check the domain and ensure your parameter