Finding the area between 2 curves using Green's Theorem Find the area bounded by $y=x^2$ and $y=x$ using Green's Theorem.

I know that I have to use the relationship $\int_c Pdx + Qdy = \int\int_D1dA$. But I don't know what my boundaries for the integral would be since it consists of two curves. 

I know that if it were a circle, I would simply parametrize the circle equation, which I understand since the curve consists of only one equation. However, since there are two equations, I'm not too sure how to approach this.

Thanks
$$Area=\frac{1}{2}\oint_C xdy-ydx=\frac{1}{2}\oint_{C_1} xdy-ydx+\frac{1}{2}\oint_{C_2} xdy-ydx$$ where $C$ is the curve consisting of the curve $y=x^2$ from $0$ to $1$ and the line $y=x$ from $1$ to $0$, say $C_1$ and $C_2$ respectively.then $$\frac{1}{2}\oint_C xdy-ydx=\frac{1}{2}\int_0^1 x(2xdx)-x^2dx+\frac{1}{2}\int_{1}^0 xdx-xdx=\frac{1}{2}\int_0^1x^2dx=\frac{1}{6}$$
Given $y=x^2$ and $y=x$, we can have a line integral $1/2\int(xdy-ydx)$.
By Green's theorem, $\int_C(Mdx+Ndy) =\int_R[
∂N/∂x-∂M/∂y]dxdy$.
Now, $M=-y/2$ and $N=x/2$.

$∂N/∂x=1/2$ and
$∂M/∂y=-1/2$
so,$[
∂N/∂x-∂M/∂y]=1/2-(-1/2)=1$.

Thus, the area $$A=\int_C(Mdx+Ndy) =\int_R[
∂N/∂x-∂M/∂y]dxdy=\int_D dxdy,$$ 
where $D$ is defined as the region where $y$ goes from $y=x^2$ to $y=x$ and $x$ from 0 to 1. Then,the integral equals $1/6$.