Let $\overrightarrow V, \Gamma$, be a vectorial field and a path such that:
$$\overrightarrow V=-(x^2+y^2)\overrightarrow i-(x^2-y^2)\overrightarrow j$$
and
$$\Gamma=\{(x,y)\in\mathbb{R}^2 | x^2+y^2=4, y < 0\}\cup \{(x,y)\in\mathbb{R^2}| (x-1)^2+y^2=1, y \geq0\}$$
I know that: $\int_{\Gamma}\overrightarrow V d\overrightarrow r=\int_{\Gamma}Pdx+Qdy$
$P(x,y)=-x^2-y^2$
$Q(x,y)=y^2-x^2$
I'm going to use this parametrization to calculate this:
$x(t)=2\cos t$
$y(t)=2\sin t$
$t\in[\pi,2\pi]$
But that is only for the first part of the path can I calculate the first part of the path with a parametrization and the second one of the path with another parametrization?
Your path is in fact the union of two different ones: the lower semicircle $\;x^2+y^2=4\;$ and the upper semicircle $\;(x-1)^2+y^2=1\;$ . For the first one you can parametrize (in the positive direction)
$$\Gamma_1: r_1(t)=(2\cos t,\,2\sin t)\;,\;\;t\in[\pi,2\pi]\;$$
and the second one
$$\Gamma_2: r_2(t)=(1+\cos t,\,\sin t)\;,\;\;t\in[0,\pi]$$
Usually paths given are path connected, yet this one isn't....anyway, now you can integrate on each separatedly.