Volume inside loop using Green's theorem. Let $\mathcal{C}$ be the curve defined by the vector function $\vec r(t)=(1-t^2)\vec i+(t-t^3)\vec j$ with $t\in \Bbb R$. I need to find the area confined in the closed loop $\gamma$ formed by $\mathcal{C}$, using Green's theorem.
$\gamma$ is smooth, closed, simple and arbitrarily oriented counter clockwise, so :
$$\int\int_DdA=\frac{1}{2}\oint_\gamma xdy-ydx$$
We have : $\begin{cases}
x=1-t^2\\
y=t-t^3
\end{cases}$
And by the figure I have of the curve (I don't know how we could plot it otherwise), I can tell that the loop starts at $(0,0)$ and ends at $(1,0)$ so that $t\in[0,1]$.
Which leads to :
$$\int\int_DdA=\frac{1}{2}\oint_{0}^{1} ((1-t^2)(1-3t^2)-(t-t^3)(-2t))dt$$
$$=\frac{1}{2}\oint_{0}^{1} (t^4-2t^2+1)dt=4/15$$
I think this was simple enough and that I got it right. However, now, knowing this result, I must calculate the following integral :
$$\oint_{\gamma}\sqrt{1+x^3}dx+2xdy$$
We know that, as per Green's theorem, it can be rewritten as :
$$\int\int_D\Big(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\Big)dA=\int\int_D(2-0)dA$$
Sweet, the root vanishes ! But I don't know how to define domain $D$.
Here is a figure of domain $D$ :

Any help would be much appreciated. Thanks !
 A: There are two values of t that give the origin for that parameterization: t =-1 and t =1. I believe if you just evaluate the line integral with those limits you'll get the area inside the closed part of the curve.
A: We have the curve $C$ defined parametrically as
$$\begin{align}
\vec r(t)&=\hat xx(t)+\hat yy(t)\\\\
&=\hat x(1-t^2)+\hat yt(1-t^2)\\\\
&\implies x(t)=1-t^2\,\,\text{and}\,\,\,y(t)=t(1-t^2)\tag 1
\end{align}$$
for $t\in [0,1]$.


NOTE:
For $t\in[0,1]$, we see that $y$ is greater than $0$.  The closed curve $\gamma$ is formed by two parts; (i) the curve $C$ defined by the aforementioned parameterization and (ii) the line segment $C_x$ along the x-axis from $(0,0)$ to $(1,0)$.  However, we note that on the line segment, $y=0$ and $dy=0$ and the contribution from the integration over $C_x$ is zero.  We therefore have$$\oint_{\gamma}(xdy-ydx)=\int_{C}(xdy-ydx)+\int_{C_x}(xdy-ydx)=\int_{C}(xdy-ydx)$$


Now, using the first relationship in $(1)$ we find that
$$t^2=1-x \tag 2$$
while using both relationships in $(1)$ simultaneously we find that 
$$t^2=(y/x)^2 \tag 3$$
Noting that the right-hand sides of $(2)$ and $(3)$ are equal reveals
$$y^2=x^2(1-x)$$
We recall that $t\in[0,1] \implies x\in[0,1]\,\,\text{and}\,\,y\ge0$.
Thus, the area $A$ inside the region bounded by $y=0$ and $y= x\sqrt{1-x}$ is given by
$$\begin{align}
A&=\int_0^1\int_{0}^{x\sqrt{1-x}}dydx\\\\
&=\int_0^1\,x\sqrt{1-x}\,dx\\\\
&=-\frac{2}{3}\left.x(1-x)\right|_0^1+\frac23\int_0^1\,(1-x)^{3/2}\,dx\\\\
&=-\frac23\,\frac{2}{5}\,\left.(1-x)^{5/2}\right|_0^1\\\\
&=\frac{4}{15}
\end{align}$$
as expected!!
