Evaluate the line integral, where C is the given plane curve. int _(C)xy^2ds C is the right half of the circle x^2+y^2=16 2=16 oriented counterclockwise

1. Parameterize the Curve<br /> The right half of the circle $x^2 + y^2 = 16$ can be parameterized as $x = 4\cos(t)$, $y = 4\sin(t)$ for $t \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.<br />2. Compute $ds$<br /> $ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt$. Here, $\frac{dx}{dt} = -4\sin(t)$ and $\frac{dy}{dt} = 4\cos(t)$. Thus, $ds = \sqrt{(-4\sin(t))^2 + (4\cos(t))^2} \, dt = 4 \, dt$.<br />3. Substitute into Integral<br /> Substitute $x = 4\cos(t)$, $y = 4\sin(t)$, and $ds = 4 \, dt$ into the integral: $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (4\cos(t))(4\sin(t))^2 \cdot 4 \, dt$.<br />4. Simplify and Integrate<br /> Simplify to $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 256 \cos(t) \sin^2(t) \, dt$. Use the identity $\sin^2(t) = \frac{1 - \cos(2t)}{2}$ to get $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 128 \cos(t) (1 - \cos(2t)) \, dt$.<br /> Split into two integrals: $128 \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos(t) \, dt - 128 \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos(t)\cos(2t) \, dt$.<br /> The first integral evaluates to zero because $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos(t) \, dt = [\sin(t)]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = 0$.<br /> For the second integral, use the product-to-sum formula: $\cos(t)\cos(2t) = \frac{1}{2}[\cos(t + 2t) + \cos(t - 2t)] = \frac{1}{2}[\cos(3t) + \cos(-t)]$.<br /> This becomes $-64 \left(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos(3t) \, dt + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos(t) \, dt\right)$.<br /> Both integrals evaluate to zero, so the entire expression is zero.