<s>[INST] Green's Theorem in the plane When calculating  $\oint _{ C }^{  }{ \frac { { x }^{ 2 }-{ y }^{ 2 } }{ 2 } dx+\left( \frac { { x }^{ 2 } }{ 2 } +{ y }^{ 4 } \right) dy } $  where $C$ Is the boundary of the region $D=\left\{ \left( x,y \right) \in \mathbb{R}^{ 2 }: 1\le { x }^{ 2 }+{ y }^{ 2 }\le 4,x\ge 0,y\ge 0 \right\} $, 
${ C }_{ 1 }:\alpha \left( t \right) =\left( t,0 \right) ,t\in \left[ 1,2 \right] $,
${ C }_{ 2 }:\alpha \left( t \right) =\left( 2\cos { t } ,2\sin { t }  \right) ,t\in \left[ 0,\frac { \pi  }{ 2 }  \right] $,
${ C }_{ 3 }:\alpha \left( t \right) =\left( 0,4-2t \right) ,t\in \left[ 0,1 \right] $, and 
${ C }_{ 4 }:\alpha \left( t \right) =\left( \sin { t } ,\cos { t }  \right) ,t\in \left[ 0,\frac { \pi  }{ 2 }  \right] $
are the curves bordering the region $D$. Differs by using Green's theorem, which is the reason for it, unless it calculates the integrals badly. Thanks for your help. 
 [/INST] Let $P(x,y)= \frac { { x }^{ 2 }-{ y }^{ 2 } }{ 2 }$  and $Q(x,y)= \frac { { x }^{ 2 } }{ 2 } +{ y }^{ 4 }$. 
Then
$$
\begin{align*}
\oint _{ C }^{  }{ \frac { { x }^{ 2 }-{ y }^{ 2 } }{ 2 } dx+\left( \frac { { x }^{ 2 } }{ 2 } +{ y }^{ 4 } \right) dy } 
&= \iint_D ( Q_x-P_y ) dA \\ 
&=  \iint_D x+y \: dA \\ 
&= \int_{0}^{\pi/2} \int_{1}^{2} (r \cos \theta + r\sin\theta)\; rdrd\theta \\ 
&= \int_{0}^{\pi/2} \int_{1}^{2} r^2 (\cos \theta + \sin\theta) drd\theta \\ 
&= \int_{0}^{\pi/2}  \dfrac{r^3}{3} (\cos \theta + \sin\theta) \Bigg|_{1}^{2} d\theta \\
&= \int_{0}^{\pi/2}  \dfrac{7}{3} (\cos \theta + \sin\theta) d\theta \\
&= \dfrac{7}{3} (\sin \theta - \cos\theta) \Bigg|_{0}^{\pi/2} \\
&= \dfrac{7}{3}(2) \\
&= \frac{14}{3}. \\
\end{align*}
$$ 
</s>