Can someone tell me what the bounds of the triple integral would be? I am confused as to how to start the problem.
**The previous integral I had was wrong so I am editing the post. I now think the integral would go from 0 to 2 for the outermost integral dx, then 0 to 2 for dy, and 0 to square root of y for dz the innermost integral.
On
Note that the integration region in the $xy$-plane consists of three parts, as indicated by I, II and III in the diagram. Thus, the volume integral can be expressed as,
$$\int_0^1 \int_0^{2-y-\sqrt y} \sqrt y \> dxdy + \int_0^1\int_{2-y-\sqrt y}^{2-y}(2-x-y)dxdy + \int_1^2\int_{0}^{2-y}(2-x-y)dxdy$$
Quanto's posted answer illustrates the point I made in comments. One should choose the order of integration judiciously to make one's life easier. (Otherwise, one may have to split into two, three, or — gulp — more regions.) In this case, because the two determining surfaces are $y=z^2$ and $y=2-x-z$, it makes sense to project onto the $xz$-plane and integrate with $dy$ innermost.
The two surfaces intersect when $x=2-z-z^2$. Thus, this parabola and the $x$- and $z$-axes determine the region in the $xz$-plane over which the 3-D region lies. This parabola intersects the $x$-axis at $(2,0,0)$ and the $z$-axis at $(0,0,1)$. Thus, we obtain the iterated integral $$\int_0^1\int_0^{2-z-z^2}\int_{z^2}^{2-x-z} dy\,dx\,dz.$$ (You should convince yourself that to be in the region determined in the problem, a line parallel to the $y$-axis for fixed $(x,z)$ enters the region at $y=z^2$ and exits at $y=2-x-z$. I.e., thinking of $y$ as "height," the surface $y=z^2$ is the lower surface and the surface $y=2-x-z$ is the higher. In the picture drawn, the first is leftmost and the second is rightmost.)