I was working on practice problems in the textbook and got stuck on this question. Any help would be greatly appreciated.
Set up two triple integrals with two different orders of integration that represent the volume with the following bounds:
$x=y^2$, $z=0$, $x+z=1$
Edit: I know that the figure has a shape of a parabola that is extruded up into the $x+z=1$ plane. I got integral from $0$ to $1$, $-\sqrt{x}$ to $\sqrt{x}$, $0$ to $1-x$ of $1\ dzdydx$. I also got $0$ to $1$, $0$ to $1-x$, $-\sqrt{x}$ to $\sqrt{x}$ of $1\ dydzdx$. Is this right?
I don't think it makes sense to move $z$ around in the order of integration.
Looking at the parabola, how do you describe the region in the $(x,y)$-plane?
Describe y in terms of $x$. and find the limits of $x$.
$\int_{0}^{1}\int_{-\sqrt x}^{\sqrt x} \int_0^{1-x} \ dz\ dy\ dx$
Or, describe $x$ in terms of $y$ and find the limits of $y$.
$\int_{-1}^{1}\int_{y^2}^{1} \int_0^{1-x} \ dz\ dx\ dy$