Calculate $$\int_Cxy^2\ \text{d}y-yx^2\ \text{d}x$$$C=\{(x,y)\in\mathbb R^2:x^2+(y-1)^2=1\}$ using Green's theorem: $\int_{C}P\ \text{d}x+Q\ \text{d}y=\int_D(\frac{\partial Q}{\partial y}-\frac{\partial P}{\partial x})\ \text{d}\mathbb R^2$
Now we have $P=-yx^2,\ \ Q=xy^2,\ \ \frac{\partial Q}{\partial y}=2xy, \ \ \frac{\partial P}{\partial x}=-2xy$, hence $$\int_Cxy^2\ \text{d}y-yx^2\ \text{d}x=\int_D4xy\ \text{d}\mathbb R^2$$ with $D=\{(x,y)\in\mathbb R^2:x^2+(y-1)^2\le1\}$.
I have trouble finding the right parameterization for $D$. Can I say that $D=\{(x,y)\in\mathbb R^2:0\le x\le 1, -\sqrt{1-x^2}+1\le y\le \sqrt{1-x^2}+1\}$ and integrate $$\int_0^1 2\Big(\int_0^{\sqrt{1-x^2}+1}xy\ \text{d}y\Big)\ \text{d}x\ \ ?$$ Any help is greatly appreciated!
The suggested integral expression isn't quite right. The $x$-values of the region vary from $0$ to $2$, and solving for $y$ in the equation of the circle gives $y = \pm \sqrt{1 - (x - 1)^2}$, and these facts give the limits of the two integrals ordered this way. As you've indicates, we can use symmetry to simplify the result.
Alternatively, the double integral is easier to evaluate in polar coordinates, in which the boundary $$x^2 + (y - 1)^2 = 1$$ of $C$ is given by $$(r \sin \theta)^2 + [(r \cos \theta) - 1]^2 = 1 .$$ Rearranging gives a simple formula for $r$ as a function of $\theta$.