Triple integral evaluation

[Ozma]

Given $\displaystyle \Omega = \left\{(x,y,z) \in \mathbb{R}^3 \ | \ 1 < x^2+y^2 < 2x, 0 < z < \frac{x^2+y^2}{x^2} \right\}$, evaluate $\displaystyle \iiint_{\Omega} \frac{y^2}{x^2+y^2} dxdydz$.
I used cylindrical coordinates $\displaystyle x=r \cos t, y= r \sin t, z=z$ with $\displaystyle 0 \leq t \leq 2\pi$ and $\displaystyle r \geq 0$: so the integral becomes
$\displaystyle \iiint_{D} r \sin^2 t drdtdz$
With $\displaystyle D=\left\{(r,t,z) \in \mathbb{R}^3 \ | \ 1 < r^2 < 2 r \cos t,0 < z < \frac{1}{\cos^2 t}\right\}$.
It is then $\displaystyle 1 < r^2 \iff r>1$ (because $\displaystyle r \geq 0$) and it is $\displaystyle r^2 <2r \cos t \iff r < 2 \cos t$ so it is $\displaystyle 1< r < 2 \cos t$. The condition on $\displaystyle z$ is already determined, while $\displaystyle 0 < z < \frac{1}{\cos^2 t}$ implies $\displaystyle \frac{1}{\cos^2 t} >0$ which is always verified.
I am not sure how to find other conditions on $\displaystyle t$ from the inequalities $\displaystyle 1<r^2<2r \cos t$: maybe from $\displaystyle 1< r < 2 \cos t$ I can deduce that $\displaystyle \cos t>\frac{1}{2}$ and so $\displaystyle 0< t < \pi/3 \lor 5\pi/3 < t < 2\pi$? Could this work?

Another question: the inequalities like $\displaystyle a<b<c$ mean in logic "$\displaystyle a<b$ and $\displaystyle b<c$ and $\displaystyle a<c$"? I believe it is something like "$\displaystyle a<b$ and $\displaystyle b<c$" and this implies $\displaystyle a<c$, because if it was "and $\displaystyle a<c$" instead of the implicatuon then I couldn't deduce $\displaystyle \cos t>\frac{1}{2}$ from $\displaystyle 1< r < 2 \cos t$.

 

[Dr.Peterson]

Another question: the inequalities like $\displaystyle a<b<c$ mean in logic "$\displaystyle a<b$ and $\displaystyle b<c$ and $\displaystyle a<c$"? I believe it is something like "$\displaystyle a<b$ and $\displaystyle b<c$" and this implies $\displaystyle a<c$, because if it was "and $\displaystyle a<c$" instead of the implicatuon then I couldn't deduce $\displaystyle \cos t>\frac{1}{2}$ from $\displaystyle 1< r < 2 \cos t$.

I'll focus on this part for the moment.

I don't see a difference. Can you explain more? As I read it, $\displaystyle a<b<c$ means "$\displaystyle a<b$ and $\displaystyle b<c$", but from that you can conclude that $\displaystyle a<c$. Why would anything be different if you took the latter as part of the meaning?

If you are told that $\displaystyle 1< r < 2 \cos t$, then you can certainly conclude that $\displaystyle 1< r$, and that $\displaystyle r < 2 \cos t$, and that $\displaystyle 1< 2 \cos t$, whether you think of the last as directly stated or merely implied. The latter certainly does imply that $\displaystyle \cos t>\frac{1}{2}$, in any case. Why couldn't you deduce it?

 

[Ozma]

@Dr. Peterson: As you say, if I read $\displaystyle a<b<c$ only as "$\displaystyle a<b$ and $\displaystyle b<c$" then of course I can conclude $\displaystyle a<c$ because of transitivity. However, this is an implication "$\displaystyle a<b$ and $\displaystyle b<c$ implies $\displaystyle a<c$" and not a conjunction "$\displaystyle a<b$ and $\displaystyle b<c$ and $\displaystyle a<c$". I said that I can't conclude that $\displaystyle a<c$ because it seems to me that, for $\displaystyle 1<r^2<2r \cos t$ in particular, if we read $\displaystyle a<b<c$ as "$\displaystyle a<b$ and $\displaystyle b<c$ and $\displaystyle a<c$" I can't "isolate" the variable $\displaystyle t$ because then from $\displaystyle 1<r^2<2r \cos t$ it must be true as well that $\displaystyle 1<2r \cos t$ (because of the conjunction) and so I get $\displaystyle \cos t > \frac{1}{2r}$ while, from the implication, I can "separate" the inequalities as I've done in #1. Am I understanding something wrong? Thank you.

 

[Dr.Peterson]

@Dr. Peterson: As you say, if I read $\displaystyle a<b<c$ only as "$\displaystyle a<b$ and $\displaystyle b<c$" then of course I can conclude $\displaystyle a<c$ because of transitivity. However, this is an implication "$\displaystyle a<b$ and $\displaystyle b<c$ implies $\displaystyle a<c$" and not a conjunction "$\displaystyle a<b$ and $\displaystyle b<c$ and $\displaystyle a<c$". I said that I can't conclude that $\displaystyle a<c$ because it seems to me that, for $\displaystyle 1<r^2<2r \cos t$ in particular, if we read $\displaystyle a<b<c$ as "$\displaystyle a<b$ and $\displaystyle b<c$ and $\displaystyle a<c$" I can't "isolate" the variable $\displaystyle t$ because then from $\displaystyle 1<r^2<2r \cos t$ it must be true as well that $\displaystyle 1<2r \cos t$ (because of the conjunction) and so I get $\displaystyle \cos t > \frac{1}{2r}$ while, from the implication, I can "separate" the inequalities as I've done in #1. Am I understanding something wrong? Thank you.

I still have no idea what you are thinking. If you infer something, then it is just as true as if it were directly stated; in both cases the conjunction of these statements is still true. That is, knowing that $\displaystyle a<b$ and $\displaystyle b<c$, it is therefore true that $\displaystyle a<b$ and $\displaystyle b<c$ and $\displaystyle a<c$. And that (whether you take it as the literal meaning of what you are told, or as an inference) includes the fact that $\displaystyle a<c$.

But maybe my confusion (or yours) is partly because you didn't say what you meant at the start. In #1 you said

... then I couldn't deduce $\displaystyle \cos t>\frac{1}{2}$ from $\displaystyle 1< r < 2 \cos t$.

while now you are saying

... from $\displaystyle 1<r^2<2r \cos t$ it must be true as well that $\displaystyle 1<2r \cos t$ (because of the conjunction) and so I get $\displaystyle \cos t > \frac{1}{2r}$ while, from the implication, I can "separate" the inequalities as I've done in #1. Am I understanding something wrong? Thank you.

You are starting from different inequalities in these two questions. I'm now looking more closely at your first paragraph to figure out what you are talking about; I was looking only at your question about the logic, which I still think is separate.

From $\displaystyle 1<r^2<2r \cos t$, which is more or less directly stated by the problem, you can conclude that $\displaystyle 1<2r \cos t$ and therefore that $\displaystyle \cos t > \frac{1}{2r}$.

From $\displaystyle 1<r<2 \cos t$, which you obtained indirectly, you can conclude that $\displaystyle 1<2 \cos t$, and therefore that $\displaystyle \cos t > \frac{1}{2}$.

Are you thinking that these two different conclusions shouldn't both be true? I'm still not following you after spending a long time (too long) trying to see what you are saying.

 

[Ozma]

@Dr.Peterson: Sorry for the time loss and thanks for your help, I think I am expressing badly my doubts because I'm a lot confused as well.

I don't think that the two conclusions $\displaystyle \cos t > \frac{1}{2r}$ and $\displaystyle \cos t > \frac{1}{2}$ shouldn't be both true, but since I'm looking for an interval like $\displaystyle t_1 < t < t_2$ to be able to integrate, why should I "prefer" $\displaystyle \cos t > \frac{1}{2}$? What assures that this is the correct interval on $\displaystyle t$ instead of, for example, the interval $\displaystyle 0 < t < \arccos \left(\frac{1}{2r}\right)$? While I know that this would "trap" me because I will not have a numerical interval like the one I get from $\displaystyle \cos t > \frac{1}{2}$ and so I couldn't integrate getting a number at the end, I don't feel safe on "ignoring" the fact that $\displaystyle \cos t > \frac{1}{2r}$ is another condition that I'm not discussing. Sorry if this is confused, but I don't know how to explain my doubt in another way. I hope that this helps to understand why I had the doubt on the conjunction/implication thing, it comes from this.

 

[Dr.Peterson]

... why should I "prefer" $\displaystyle \cos t > \frac{1}{2}$? What assures that this is the correct interval on $\displaystyle t$ instead of, for example, the interval $\displaystyle 0 < t < \arccos \left(\frac{1}{2r}\right)$? While I know that this would "trap" me because I will not have a numerical interval like the one I get from $\displaystyle \cos t > \frac{1}{2}$ and so I couldn't integrate getting a number at the end, I don't feel safe on "ignoring" the fact that $\displaystyle \cos t > \frac{1}{2r}$ is another condition that I'm not discussing.

You're looking for an interval between numbers, involving only the one variable you want; so when you find it, you use it! That's all there is to it.

The other "interval" can be used in graphing the curve (or in writing limits of an inner integral). In general, the outer integral will have numerical limits; the middle integral can have limits in terms of one variable; and the inner integral will have limits that may involve both "other" variables. I would likely make the inner integral use dz and the outer one use dt (i.e. $d\theta$). To some extent, that choice depends on which limits can most easily be expressed in terms of fewest variables.

If you hadn't tried to broaden your question into one about logic, this could have been worked out more quickly. Please take the thinking you showed initially, and make an attempt at writing out the actual triple integral. In doing that, you will get a better idea of your needs, and may solve your own problem. At least there will be more to talk about.