[Math] Solid region $S$ bounded by parabolic cylinder $z+x^2 = 2$ and planes $z=0$, $y=0$ and $y=x$.

calculusintegrationmultivariable-calculus

The problem

There exists a (unique, I guess) solid region $S$ bounded by parabolic cylinder $z+x^2 = 2$ and planes $z=0$, $y=0$ and $y=x$.

  1. Sketch the part of $S$ in the first octant of $\mathbb R^3$.

  2. Set up the bounds for the triple integral $$I_S = \int\int\int _S f dz dy dx$$

for any suitable $f: \mathbb R^3 \to \mathbb R$ (uniformly continuous, Riemann integrable, Lebesgue integrable or whatever you need).

Questions:

  1. Question: For 1, how is $S$ not completely in the first octant (except possibly for some of its boundary)?

I think $S$ is a cut out of an infinite loaf of bread, with the bottom as $z=0$ cut and the top as the $z+x^2=2$, where the infinite loaf is cut by $y=0$ and $y=x$. Looks like first octant to me.

  1. Question: For 2, this is what I got at least for $S$ in the first octant, denoted $S_1$. Is this correct?

$z$ is from $0$ to $2-x^2$. So we get

$$I_{S_1} = \int\int_{R_1} \int_{z=0}^{z=2-x^2} f dy dx$$

for some region $R_1$.

Afterwards, I let $g = \int_{z=0}^{z=2-x^2} f dz$, $g: \mathbb R^2 \to \mathbb R$ and pretend we're in the $xy$-plane ($z=0$) and then set up the bounds for the double integral

$$I_{S_1} = \int\int_{R_1} g dy dx$$

over $R_1$ which appears to be a triangle bounded by $y=x$, $x=\sqrt{2}$ and $y=0$. Therefore,

$$I_{S_1} = \int_{x=0}^{x=\sqrt{2}} \int_{y=0}^{y=x} g dy dx = \int_{x=0}^{x=\sqrt{2}} \int_{y=0}^{y=x} \int_{z=0}^{z=2-x^2} f dz dy dx$$

  1. Question: If $S$ is not completely in the first octant (except possibly for some of its boundary), then why, and what's the integral?

    • See below.

Update

Based on Quanto's answer, in particular the term 'third octant' (which I hope means negative y, negative x and positive z based on the '2' in the 'gray code' here), I realised my (implicit) mistake in extending my answer to Question (2). Here's what I think to do:

Forget $y=x$ and $y=0$ for now.

The solid region bounded by $z=0$ and $z+x^2=2$ is $S^\# = \{(x,y,z) \in \mathbb R^3| z \ge 0, z=2-x^2\}$. I see this as an infinite hollow loaf of bread where the bread is shaped parabolically and where the $y$-axis is the centre of the base of the infinite loaf.

Observe $S^\#$ can also be written $S^\# = \{(x,y,z) \in \mathbb R^3| z=2-x^2, x^2 \le 2 \}$, i.e. the condition '$z \ge 0$' is equivalent to the condition '$x^2 \le 2$'.

Flattening $S^\#$, in the sense of projecting $S^\#$ onto $z=0$, gives us $\tilde A = \{(x,y,z) \in \mathbb R^3| z=0, x^2 \le 2 \}$ which is pretty much the same as (In higher maths: 'pretty much the same as' means 'homeomorphic, as topological manifolds with boundary, to' or something) the set $A = \{(x,y) \in \mathbb R^2|x^2 \le 2 \}$ of 2 parallel vertical lines and the space in between them.

Now we bring back $y=x$ and $y=0$. It appears my mistake was that I implicitly assumed 'bounded by $y=0$' meant 'is contained in $y \ge 0$'.

Observe that $A$ is cut up by $y=x$ and $y=0$ into 4 spaces (Let these spaces include the boundaries), 2 of which are bounded. Denote these bounded spaces as $A_1$ and $A_2$. One of $A_1$ and $A_2$ without its boundary is completely in the first quadrant. The other without its boundary is completely in the third quadrant. Let $A_1$ be the former and $A_2$ be the latter.

The part $S_1$ (without boundary) of $S$ in the first octant projects to $A_1$ (without boundary). $A_2$ (without boundary) indeed corresponds to a non-empty part $S_2$ (without boundary) of $S$ not in the first octant. Just as $A_2$ (without boundary) is in the third quadrant, so do we have $S_2$ (without boundary) to be in the third octant.

Now, the bounds for $z$ in $I_{S_2}$ are the same as in $I_{S_1}$, I think: $$I_{S_2} = \int\int_{R_2}\int_{z=0}^{z=2-x^2} f dy dx = \int\int_{R_2} g dy dx$$

Now, $R_2$ is the region in $\mathbb R^2$ which appears to be a triangle bounded by $y=x$, $x=-\sqrt{2}$ (rather than $x=\sqrt{2}$ !) and $y=0$. Therefore, $$I_{S_2} = \int_{x=-\sqrt{2}}^{x=0} \int_{y=x}^{y=0} g dy dx = \int_{x=-\sqrt{2}}^{x=0} \int_{y=x}^{y=0} \int_{z=0}^{z=2-x^2} f dz dy dx$$

Finally, I believe that $S = S_1 \cup S_2$ and that $S_1 \cap S_2$ is empty except for boundary points. Therefore, $$I_S = I_{S_1} + I_{S_2} = \int \int \int_{S_1} f dz dy dx + \int \int \int_{S_2} f dz dy dx$$

Best Answer

Note that the region enclosed by $z=0$ and $z=-x^2+2$ is a tube with a parabolic segment as its cross section and oriented along the $y$-axis.

1) The vertical planes $y=0$ and $x=y$ divide the tube into four sections. Only one of the sections is in the first octant.

2) The integral below for the first octant is correct $$I =\int_{x=0}^{x=\sqrt{2}} \int_{y=0}^{y=x} \int_{z=0}^{z=2-x^2} f dz dy dx$$

3) Of the other three sections, only the one in the third octant is enclosed and its integral could be set up similarly as that in the first octant. The other two sections are open-ended, extending into both directions along the $y$-axis.

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