How do I express the region bounded by $z=x^2, z+y=1, z-y=1$ as a z-simple triple integral?

Question

I've been able to figure out the y-simple and x-simple triple integrals, but I'm having trouble with the z-simple one. I split the region in half over the x-axis, from what I can see, the projection onto the xy-plane of this is a rectangle of $[-1,1] \times [0,1]$ and $x^2 \leq z \leq 1-y$.

Answer 1

Not an answer. Just posting an image that hopefully makes it easier to grasp which region exactly is supposed to be covered. Normally I would add a picture to the question body, but this time I use a CW answer, because the region may still be in question (and hence my picture misleading).

More often than not a bounded region is intended. My guess (looking at the picture) is the region above the surface $z=x^2$ and below both the planes $z=1\pm y$. I might write the integral of the function $f(x,y,z)$ over that region as the iterated integral $$ \int_{x=-1}^1\int_{z=x^2}^1\int_{y=z-1}^{1-z}f(x,y,z)\,dy\,dz\,dx, $$ but I don't know what the OP means by $z$-simple?

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If we want an iterated integral with $z$-integration being the first (= the innermost), then I think it would be $$ \int_{x=-1}^1\int_{y=x^2-1}^{1-x^2}\int_{z=x^2}^{1-|y|}f(x,y,z)\,dz\,dy\,dx. $$ The reasoning is that the projection $(x,y)$ on the plane $z=0$ of any point in the region must be between the parabolas $y=x^2-1$ and $y=1-x^2$. Please check :-)

If you want to calculate the last iterated integral, you may need to split it into two regions according to the sign of the $y$-coordinate. Business as usual.

Answer 2

The planes $z+y=1$ and $z-y=1$ intersect at $\langle x, 0, 1\rangle$; a line parallel to the $x$-axis, 1 unit distant in the $z$ direction.

The $xz$ plane parallel cross-sections of curve $z=x^2$ are parabola with minima along the line $\langle 0,y,0\rangle$; the $y$-axis.

Therefore the shape which these planes and curve bound has an edge parallel to the $x$-axis at height $z=1$, two flat sides sloping down at $45^\circ$ angles to meet the parabolic-curved base which has a keel along the $y$-axis.

$$\{\langle x,y,z\rangle: -1\leq x\leq 1~,~x^2\leq z\leq 1~,~z-1\leq y\leq 1-z\}$$

The projection of this 'lemon-wedge' onto the $xy$ plane is not a rectangle.

Answer 3

It is easy to figure out that we want the parabolic sheet $z = x^2$ to be below the "roof-like" double-planes $ z = 1 - y $ and $ z = 1 + y$.

Since $ z = x^2 $, then we must have

$ z \le 1 - y $

$ z \le 1 + y $

i.e.

$ x^2 \le 1 - y $

$ x^2 \le 1 + y $

The first one gives us

$ y \le 1 - x^2 $

The second one give sus

$ y \ge x^2 - 1 $

Hence the boundaries are

$ x^2 - 1 \le y \le 1 - x^2 $

Since we must have $(1 - x^2) \ge (x^2 - 1)$, then $x^2 \le 1 $ i.e. $x \in [-1, 1] $

So, to calcualte the volume, we have the following

$ V = \displaystyle \int_{x=-1}^{1} \int_{y=x^2 - 1}^{1 - x^2} \int_{z = x^2 }^{1 - |y| } \ dz \ dy \ dx $