Set up, but do not compute, a triple integral to that gives the volume of the region in the first octant bounded by the plane $y+z=2$ and the cylinder $x=4-y^2$. Please use $dz\, dy\, dx$ as the order of integration:
Is this correct? $$\int_{-2}^2 \int_{-\sqrt{4-x}}^{\sqrt{4-x}} \int_0^{2-y} 1dzdydx$$
In the first octant $x,y,z$ are positive. $$\int_{0}^2 \int_0^{\sqrt{4-x}} \int_0^{2-y} 1dzdydx$$ is correct.