I'm having a difficult time answering this problem with a path integral.

Let $F(x,y) = (x, 3y)$ be a fluid flow. Compute the net flow out of the region below $y = 1 - x^2$ and above $y = 0$ using a line integral.

Let $D$ be the above mentioned region and $\partial D$ be this region's boundary. We have: $$\iint_D \vec{\nabla} \cdot \vec{F}(x,y)~dA = \oint_{\partial D}\vec{F}(x,y) \cdot \hat{n} ~ d(\partial D) = \Phi_{\text{flux}}$$ So: $$\iint \vec{\nabla} \cdot \vec{F}(x,y)~dA = 4\int_{-1}^{1}\int_{0}^{1-x} dydx = \frac{16}{3}$$ Now, here's where I have a hard time. I know that it will be split up into two path integrals (where one is around the parabolic path, and the other is along the $x$ axis from $-1$ to $1$, where $-\hat{y}$ is the normal vector). How do I find the normal vector of the parabola?

I've began with parametrizing the parabola with... $$\begin{aligned} x & = t \\ y & = 1 - t^2 \end{aligned}$$ ... so our vector is $(t, 1-t^2)$. The normal to this $(x, y)$ is $(-dy, dx)$, so the normal of this vector is $(-2t, 1)$. I'm... not sure what to do here. Do I dot the two and integrate? (This did not give me the correct answer.) Do I normalize $n$? Do I take the Jacobian determinant... or...

... I'm just confused.


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