Changing the order of integration of triple integral We have a solid in a first octant, $x,y,z \ge 0$, which is bounded by the coordinate planes, the plane $x + y = 2$ and the parabolic cylinder $z + {y^2} = 1$. The task is to change the order of integration and calculate: 


*

*$\int {\left( {\int { \ldots \,\partial x} } \right)} \,\partial y$

*$\int {\left( {\int { \ldots \,\partial y} } \right)} \,\partial z$


The following graph helps to visualize the problem: 

I have successfully managed to calculate with orders:
$$\partial z\,\partial y\,\partial x{\text{   and   }}\partial x\,\partial z\,\partial y{\text{ }}$$
Both gave the result $\frac{{13}}{{12}}$
I'm having troubles integrating in order $\partial y\,\partial x\,\partial z$. It is clear to me that we first have to integrate with respect to the height $y$ and base $xz$.
The base $xz$ projected onto $xz$ plane has a rectangular shape and the projected solid should look like this 
I set the following limits of integration:
$$\int\limits_0^1 {\int\limits_0^2 {\int\limits_0^{\sqrt {1 - z} } {1\,\,\partial y\partial x\partial z = \frac{4}{3}} } } $$
And the result does not match with the result i obtained from other two orders. Can someone show me the mistake?

The problem is with the upper boundary of the region (where "up" here is taken to be in the positive direction of the $y$-axis, away from the projection onto the $xz$-plane). Notice that if $x$ is closer to $0$, then the the upper boundary is the (half) cylinder $y = \sqrt{1 - z}$. Otherwise, if $x$ is closer to $2$, then the upper boundary is the plane $y = 2 - x$. To figure out the exact threshold, we need to figure out where the two boundaries intersect and express the result in terms of $x$:
$$
2 - x = y = \sqrt{1 - z} \iff x = 2 - \sqrt{1 - z}
$$
Hence, we split up into two cases:
$$
\int_0^1 \int_0^{2 - \sqrt{1 - z}} \int_0^\sqrt{1 - z} dy \, dx \, dz + \int_0^1 \int_{2 - \sqrt{1 - z}}^2 \int_0^{2 - x} dy \, dx \, dz
= \frac{13}{12}
$$

Another way to think about this is via the cross-section method. Suppose that you cut the region with a plane at some fixed (constant) value of $z$, so that the slice is parallel to the $xy$-plane. Then this cross-section is the trapezoid bounded by the vertical line $x = 0$, the horizontal line $y = \sqrt{1 - z}$, the diagonal line $y = 2 - x$, and the horizontal line $y = 0$. This will yield the same result as above.
