Changing the order of integration of triple integral

Question

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:

1. $\int {\left( {\int { \ldots \,\partial x} } \right)} \,\partial y$
2. $\int {\left( {\int { \ldots \,\partial y} } \right)} \,\partial z$

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?

Answer 1

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: $$ \int_0^1 \int_0^{2 - \sqrt{1 - z}} \int_0^\sqrt{1 - z} dy \, dx \, dz + \int_0^1 \int_{2 - \sqrt{1 - z}}^2 \int_0^{2 - x} dy \, dx \, dz = \frac{13}{12} $$

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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.