By Green's theorem we have \begin{align*} \int_{C} Mdx+Ndy = \iint_{D}(N'_{x} - M'_{y})dxdy \end{align*} where $D$ is the domain bounded by the curve $C$. So let's use this theorem: \begin{align*} \int_{C} 3ydx+2xdy = \iint_{D} (2-3)dxdy = - \iint_{D}dxdy = -\iint_{D}dydx = -\int_{0}^{\pi}\int_{0}^{\sin x} dydx = -\int_{0}^{\pi}\sin x dx = \cos \pi - \cos 0 = -2 \end{align*} Please take into account that we switched the order of integration. The point is that in this case calculating the double indegral becomes easier.

We know that for the graph given explicitly by $z=f(x,y)$ area can be computed by the following formula \begin{align*} \text{Area} = \iint_{D} \sqrt{(f'_{x})^{2}+(f'_{y})^{2}+1}dxdy \end{align*} where domain $D$ is the domain (belonging to the $OXY$ plane) of parametrization of the sruface in cartesian coordinate system. Since $1/4 \leq z\leq 1$ and $z=x^{2}+y^{2}$ therefore our domain $D$ is $D= \{ (x,y) \; 1/4 \leq x^{2}+y^{2} \leq 1\}$ So for $f(x,y) = x^{2}+y^{2}$ we have \begin{align*} &\text{Area} = \iint_{\frac{1}{4}\leq x^{2}+y^{2}\leq 1} \sqrt{1+(f'_{x})^{2} + (f'_{y})^{2}}dxdy = \iint_{\frac{1}{4}\leq x^{2}+y^{2}\leq 1} \sqrt{1+4x^{2}+4y^{2}}dxdy=\\ &\text{Polar coordinates} (x =r\cos \theta, \; y=r\sin \theta)=\int_{0}^{2\pi}\int_{\frac{1}{2}}^{1} \sqrt{1+4r^{2}\cos^{2}\theta+4 r^{2}\sin^{2}\theta} \; rdrd\theta=\\ &\int_{0}^{2\pi}\int_{\frac{1}{2}}^{1} \sqrt{1+4r^{2}} \; rdrd\theta=2\pi \int_{1/2}^{1}\sqrt{1+4r^{2}} \; rdr = \text{change of variables}=\\ &2\pi \cdot \frac{1}{12}\cdot \left((1+4r^{2})^{3/2} \right)|_{1/2}^{1}=\frac{\pi}{6}\left( 5^{3/2}-2^{3/2}\right) \end{align*}