<s>[INST] $\int_\gamma{(x-y)dx + (x+y)dy}, \quad \gamma : x^2 + 2y^2 = 1 , \quad 0 \leq y $ I'm asked to find 
$$\int_\gamma{(x-y)dx + (x+y)dy}$$
where 
$$\gamma : x^2 + 2y^2 = 1 , \quad 0 \leq y$$
(with positive direction) i.e the upper half of the ellipse $x^2 + 2y^2 = 1$. 
My attempt 
Let $\sigma = \gamma + \gamma_1$ where $\gamma_1 = (t,0) \quad , \quad 0\leq t\leq 1.$ 
Since $\sigma$ is both positive and closed, Greens Formula can be used with
$\frac{dQ}{dx} - \frac{dP}{dy} = 2.$
$$\int_\gamma(x-y)dx + (x+y)dy = \int\int_D2dxdy - \int_{-1}^1t\cdot dt$$
$$= 2\cdot\frac{1}{2}\cdot (\frac{1}{\sqrt{2}}\pi) - 2 = \frac{\pi}{\sqrt{2}} - 2$$
where I evaluate the double integral simply by getting half of the area of the ellipsoid $ = \frac{1\cdot\frac{1}{\sqrt{2}}\cdot\pi}{2}.$
However, the answer is supposed to be $\frac{\pi}{\sqrt{2}}.$
What am I doing wrong?
 [/INST] Here is an alternate approach using straight forward substitution.
Let $x=\cos\theta,\,y=\frac{\sqrt{2}}{2}\sin\theta,\,0\le\theta\le\pi$
\begin{eqnarray}
\int_\gamma{(x-y)dx + (x+y)dy}&=&\int_0^\pi-\left(\cos\theta+\frac{\sqrt{2}}{2}\sin\theta\right)\sin\theta+\left(\cos\theta+\frac{\sqrt{2}}{2}\sin\theta\right)\cdot\frac{\sqrt{2}}{2}\cos\theta\,d\theta\\
&=&\int_0^\pi-\frac{1}{4}\sin2\theta+\frac{\sqrt{2}}{2}\,d\theta\\
&=&\left[\frac{1}{8}\cos2\theta+\frac{\theta}{\sqrt{2}}\right]_0^\pi\\
&=&\frac{\pi}{\sqrt{2}}
\end{eqnarray}
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