Green's Theorem #3

The vector field in the above integral is $F(x,y)= (y^2, 3xy)$ . We can compute this integral using Green's theorem to convert the line integral into a double integral. The integrand of the double integral must be $\begin{aligned} \frac{\partial{F_2}}{\partial{x}} -\frac{\partial{F_1}}{\partial{y}} = 3y -2y = y .\end{aligned}$

Since the line integral was over the boundary of the half disk, the region of integration for the double integral is the half-disk D itself. (Since C was oriented counterclockwise, the orientation matches; otherwise, we would have had to multiple by negative one to get the correct sign.) The region D is described by $\begin{aligned} -1 \le x \le 1,\\ 0 \le y \le \sqrt{1-x^2}.\end{aligned}$

Therefore, by Green's Theorem, $\begin{aligned} \oint_C y^2 dx + 3xy\, dy & = \iint_D \left(\frac{\partial{F_2}}{\partial{x}} -\frac{\partial{F_1}}{\partial{y}}\right)dA\\ &= \iint_D y dA\\ &= \int_{-1}^1 \int_0^{\sqrt{1-x^2}} y\, dy \, dx\\ &= \int_{-1}^1 \left( \left.\frac{y^2}{2} \right|_{y=0}^{y=\sqrt{1-x^2}}\right) dx\\ &= \int_{-1}^1 \frac{1-x^2}{2} dx\\ &= \frac{x}{2} - \left.\left.\frac{x^3}{6} \right|_{-1}^1\right. = \frac{2}{3}. \end{aligned}$

Hence, the answer is $2+3=5.$