I am trying to evaluate the surface integral of a vector function u (whose definition is irrelevant to this problem) over the surface $$z=x+y^2 $$ with $z<0$ and $x>-1$.
Initially I put $$x+y^2<0 $$ $$-\sqrt{-x}<y<\sqrt{-x} $$ $$-1<x<0 $$ which gave $$\int\int_S\bf{u} \cdot{\bf{n}}\textit{dS}=\int_{x=-1}^{x=0}\int_{y=-\sqrt{-x}}^{y=\sqrt{-x}}\bf{u} \cdot{\frac{\partial r}{\partial \textit{x}}}\times\frac{\partial r}{\partial \textit{y}}\textit{dydx} $$ and yielded the correct answer.
But I then tried to reverse the order of integration for $x$ and $y$, so again $$x+y^2<0$$ $$-1<x<-y^2$$ but now I'm struggling to find the integration limits for $y$ (setting $x$ to zero also makes no sense in this case).
Is there a way to find such limits (for $y$) without having to plot the whole 3d graph for $z=x+y^2$?