Evaluate $$\int _{\Gamma}x^2(y+1)dx-xy^2dy$$ where $\Gamma = \{(x,y):x^2+y^2=1, y\geq 0\}$ counter clockwise.
So $\Gamma$ is the upper half of the unit circle, we use Green:
$$\int_{x=1}^{x=-1}\int_{y=0}^{y=\sqrt{1-x^2}}-(x^2+y^2)ydx$$
Using polar we get
$$\int_{0}^{\pi}\int_{0}^{1}(-r^2cos^2\theta-r^2sin^2\theta)r dr d\theta = \int_{0}^{\pi}\int_{0}^{1}-r^3drd\theta= -\frac{\pi}{4}$$
next we have to reduce the curve from $-1$ to $1$
$\gamma(t)=(t,0)$ where $t\in[-1,1]$
So $$\int_{-1}^{1} (t^2(0+1),-t*0^2)\cdot(1,0)=\int _{-1}^{1}t^2dt=\frac{2}{3}$$
But the answer can not $$-\frac{\pi}{4}-\frac{2}{3}$$ so it is obviously a mistake, where did I got it wrong?
We have a curve $\Gamma = \{(x, y) \space : \space x^2 + y^2 =1, \space y \ge 0 \}$, which is not a closed curve, it is the upper semi-circular arc (counter clockwise). In this case it is better to find a parametric equation for such curve, what is $$\Gamma=\cases{x=\cos t\\y=\sin t}, \space \space \space t \in \left[ 0, \pi \right].$$
Now we have: $$\int_{\Gamma} x^2(y+1)dx - xy^2dy=\int_0^{\pi}\left[\cos^2t(\sin t + 1)(-\sin t) - \cos t\sin^2t\cos t\right]dt=\int_0^{\pi}\left[\cos^2t(-\sin^2t - \sin t) - \cos^2t\sin^2t\right]dt=\int_0^{\pi}\left[\cos^2t\left(-2\sin^2t - \sin t \right) \right]dt=\dots=\boxed{-\frac{2}{3}-\frac{\pi}{4}}.$$