\( \begin{array}{llrl}\text { b) } & \text { Evaluate } \oint_{C} y^{2} d x+3 x y d y, \text { where the region } D \text { enclosed by } C & \text { 11. Give } \\ & \text { is given by } D=\{(r, \theta): 1 \leq r \leq 2,0 \leq \theta \leq \pi\} . & {[8 \text { marks }]} & \text { A. } \\ & & & \text { B. }\end{array} \)

To evaluate the line integral \( \oint_{C} y^{2} dx + 3xy dy \) over the closed curve \( C \) which bounds the region \( D=\{(r, \theta): 1 \leq r \leq 2,0 \leq \theta \leq \pi\} \), we can employ Green's Theorem. This theorem connects line integrals around a simple curve to a double integral over the region bounded by the curve. By converting the given integral to a double integral using Green’s Theorem, we identify the components \( P = y^2 \) and \( Q = 3xy \). Thus, we need to calculate: \[ \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = \frac{\partial}{\partial x}(3xy) - \frac{\partial}{\partial y}(y^2) \] Calculating these partial derivatives gives: \[ \frac{\partial Q}{\partial x} = 3y \quad \text{and} \quad \frac{\partial P}{\partial y} = 2y \] So we combine them: \[ 3y - 2y = y \] Now, we can set up the double integral over the region \( D \): \[ \iint_{D} y \, dA \] In polar coordinates, where \( y = r\sin(\theta) \) and \( dA = r \, dr \, d\theta \), the integral becomes: \[ \int_0^{\pi} \int_1^{2} (r\sin(\theta)) r \, dr \, d\theta = \int_0^{\pi} \sin(\theta) \, d\theta \int_1^{2} r^2 \, dr \] Calculating the inner integral yields: \[ \int_1^{2} r^2 \, dr = \left[\frac{r^3}{3}\right]_1^2 = \frac{8}{3} - \frac{1}{3} = \frac{7}{3} \] The outer integral: \[ \int_0^{\pi} \sin(\theta) \, d\theta = [-\cos(\theta)]_0^{\pi} = -(-1 - 1) = 2 \] Thus, the overall integral evaluates to: \[ \frac{7}{3} \cdot 2 = \frac{14}{3} \] Finally, we conclude that: \[ \oint_{C} y^{2} dx + 3xy dy = \frac{14}{3} \] So the required value of the line integral is: \[ \boxed{\frac{14}{3}} \]