Troubles on finding a triple integral bounds

dunkelheit

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I have a doubt about this triple integral:

\(\displaystyle \iiint_A ydxdydz\)

Where \(\displaystyle A=\{(x,y,z)\in\mathbb{R}^3: x^2+y^2-2x<0,0<z<x,x^2+y^2<1,y>0\}\)

After using cylindrical coordinates I end up with

\(\displaystyle \iiint_B \rho^2 \sin \theta d\rho d\theta dz\)

Where \(\displaystyle B=\{(\rho \cos \theta,\rho \sin \theta,z)\in\mathbb{R}^3:\rho^2-2\rho \cos \theta<0,0<z<\rho \cos \theta,\rho^2<1,\rho \sin \theta>0\}\)

I find out that \(\displaystyle 0<z<\rho \cos \theta\) and that \(\displaystyle 0<\rho<\min \{1,2 \cos \theta \}\); until now I don't have any problems, I know that I have to split the integral in a sum of two integrals due to the fact that because of \(\displaystyle \theta\) the values of the upper bound for \(\displaystyle \rho\) can change.

The doubt is about discussing \(\displaystyle 0<z<\rho \cos \theta\), I have done it by letting \(\displaystyle \rho \cos \theta>0\) and obtaining \(\displaystyle \theta \in \left[0,\frac{\pi}{2}\right] \cup \left[\frac{3\pi}{2},2\pi\right]\) and I know this is the correct way, but I'm not completely aware of why is it!

I mean that I'm not completely sure why I don't have to put \(\displaystyle \rho \cos \theta>z\) instead, because by the description of the domain I have the information that \(\displaystyle z>0\), so I have a lower bound on \(\displaystyle \rho \cos \theta\) that is \(\displaystyle \rho \cos \theta>z\) that implies \(\displaystyle \cos \theta>\frac{z}{\rho}\) and not simply \(\displaystyle \rho \cos \theta>0\).

Can someone help me understand why I can be sure to have \(\displaystyle \rho \cos \theta>0\) from \(\displaystyle 0<z<\rho \cos \theta\)?

Thanks for your time.
 
I have a doubt about this triple integral:

\(\displaystyle \iiint_A ydxdydz\)

Where \(\displaystyle A=\{(x,y,z)\in\mathbb{R}^3: x^2+y^2-2x<0,0<z<x,x^2+y^2<1,y>0\}\)

After using cylindrical coordinates I end up with

\(\displaystyle \iiint_B \rho^2 \sin \theta d\rho d\theta dz\)

Where \(\displaystyle B=\{(\rho \cos \theta,\rho \sin \theta,z)\in\mathbb{R}^3:\rho^2-2\rho \cos \theta<0,0<z<\rho \cos \theta,\rho^2<1,\rho \sin \theta>0\}\)

I find out that \(\displaystyle 0<z<\rho \cos \theta\) and that \(\displaystyle 0<\rho<\min \{1,2 \cos \theta \}\); until now I don't have any problems, I know that I have to split the integral in a sum of two integrals due to the fact that because of \(\displaystyle \theta\) the values of the upper bound for \(\displaystyle \rho\) can change.

The doubt is about discussing \(\displaystyle 0<z<\rho \cos \theta\), I have done it by letting \(\displaystyle \rho \cos \theta>0\) and obtaining \(\displaystyle \theta \in \left[0,\frac{\pi}{2}\right] \cup \left[\frac{3\pi}{2},2\pi\right]\) and I know this is the correct way, but I'm not completely aware of why is it!

I mean that I'm not completely sure why I don't have to put \(\displaystyle \rho \cos \theta>z\) instead, because by the description of the domain I have the information that \(\displaystyle z>0\), so I have a lower bound on \(\displaystyle \rho \cos \theta\) that is \(\displaystyle \rho \cos \theta>z\) that implies \(\displaystyle \cos \theta>\frac{z}{\rho}\) and not simply \(\displaystyle \rho \cos \theta>0\).

Can someone help me understand why I can be sure to have \(\displaystyle \rho \cos \theta>0\) from \(\displaystyle 0<z<\rho \cos \theta\)?

Thanks for your time.

The first question to ask is, why are you using cylindrical coordinates? Were you told to, or is this your choice? It does allow you to avoid square roots, so it does make sense.

Given that this is what you want or need to do, the next question is, have you paid attention to the order of the differentials? As written, \(\displaystyle \iiint_B ... d\rho d\theta dz\), you are integrating with respect to z last, so your limits on that outer integral can only be numbers, not functions of rho and theta.

The limits you list suggest that you are actually integrating with respect to z, then rho, then theta, that is, \(\displaystyle \iiint_B ... dz d\rho d\theta\), which is probably what I would do.

But having figured that out, I don't understand your question. You seem to be thinking that what I am interpreting as limits on z, \(\displaystyle 0<z<\rho \cos \theta\), is a condition not on z but on \(\displaystyle \rho \cos \theta\). Can you explain your thinking there? As I see it, you are first integrating from 0 to \(\displaystyle \rho \cos \theta\); and you can be sure that \(\displaystyle \rho \cos \theta > 0\) because that is x, which you know is positive.
 
I have a doubt about this triple integral:

\(\displaystyle \iiint_A ydxdydz\)

Where \(\displaystyle A=\{(x,y,z)\in\mathbb{R}^3: x^2+y^2-2x<0,0<z<x,x^2+y^2<1,y>0\}\)

After using cylindrical coordinates I end up with

\(\displaystyle \iiint_B \rho^2 \sin \theta d\rho d\theta dz\)

Where \(\displaystyle B=\{(\rho \cos \theta,\rho \sin \theta,z)\in\mathbb{R}^3:\rho^2-2\rho \cos \theta<0,0<z<\rho \cos \theta,\rho^2<1,\rho \sin \theta>0\}\)

I find out that \(\displaystyle 0<z<\rho \cos \theta\) and that \(\displaystyle 0<\rho<\min \{1,2 \cos \theta \}\); until now I don't have any problems, I know that I have to split the integral in a sum of two integrals due to the fact that because of \(\displaystyle \theta\) the values of the upper bound for \(\displaystyle \rho\) can change.

The doubt is about discussing \(\displaystyle 0<z<\rho \cos \theta\), I have done it by letting \(\displaystyle \rho \cos \theta>0\) and obtaining \(\displaystyle \theta \in \left[0,\frac{\pi}{2}\right] \cup \left[\frac{3\pi}{2},2\pi\right]\) and I know this is the correct way, but I'm not completely aware of why is it!

I mean that I'm not completely sure why I don't have to put \(\displaystyle \rho \cos \theta>z\) instead, because by the description of the domain I have the information that \(\displaystyle z>0\), so I have a lower bound on \(\displaystyle \rho \cos \theta\) that is \(\displaystyle \rho \cos \theta>z\) that implies \(\displaystyle \cos \theta>\frac{z}{\rho}\) and not simply \(\displaystyle \rho \cos \theta>0\).

Can someone help me understand why I can be sure to have \(\displaystyle \rho \cos \theta>0\) from \(\displaystyle 0<z<\rho \cos \theta\)?

Thanks for your time.

Can someone help me understand why I can be sure to have \(\displaystyle \rho \cos \theta>0\) from \(\displaystyle 0<z<\rho \cos \theta\)?

I'm not sure if a follow, but if cos is +, then \(\displaystyle \rho \cos \theta>0\)
obviously, you want to integrate \(\displaystyle \theta\) after z, since z=\(\displaystyle \rho \cos \theta\).
 
The first question to ask is, why are you using cylindrical coordinates? Were you told to, or is this your choice? It does allow you to avoid square roots, so it does make sense.

It is a choice of mine.

Given that this is what you want or need to do, the next question is, have you paid attention to the order of the differentials? As written, \(\displaystyle \iiint_B ... d\rho d\theta dz\), you are integrating with respect to z last, so your limits on that outer integral can only be numbers, not functions of rho and theta.

I'm not sure if it is right, but with that I mean that I'm still not applying the Fubini's reduction theorem; I've just substituted the new coordinates and the order of differentials will be decided by how I will describe the new domain in the new coordinates (is this wrong?).

The limits you list suggest that you are actually integrating with respect to z, then rho, then theta, that is, \(\displaystyle \iiint_B ... dz d\rho d\theta\), which is probably what I would do.

That's exactly what I want to do, but I got stuck by that doubt :(
I would write it like

\(\displaystyle \int_{0}^{\frac{\pi}{3}} \left(\int_{0}^{1} \left(\int_{0}^{\rho \cos \theta} \rho^2 \sin \theta dz \right) d \rho \right) d \theta + \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \left(\int_{0}^{2 \cos \theta} \left(\int_{0}^{\rho \cos \theta} \rho^2 \sin \theta dz \right) d \rho \right) d \theta \)

But I've chosen the interval of \(\displaystyle \theta\) just by intuition, I haven't a full awareness of it; I will explain better what I mean in the next part of this message.

But having figured that out, I don't understand your question. You seem to be thinking that what I am interpreting as limits on z, \(\displaystyle 0<z<\rho \cos \theta\), is a condition not on z but on \(\displaystyle \rho \cos \theta\). Can you explain your thinking there? As I see it, you are first integrating from 0 to \(\displaystyle \rho \cos \theta\); and you can be sure that \(\displaystyle \rho \cos \theta > 0\) because that is x, which you know is positive.

I will try to explain better my doubt, sorry if I wasn't clear enough and I hope I will be now: my doubt is that, as far as I've understood about these integrals (maybe I still haven't understood to find the interval where \(\displaystyle \theta\) vary I have to solve \(\displaystyle \rho \sin \theta>0\) from the previous \(\displaystyle y>0\) and I have to solve \(\displaystyle 0<z<\rho \cos \theta\); the point is that how can I be sure to solve directly \(\displaystyle 0<\rho \cos \theta\) and "ignoring" \(\displaystyle z<\rho \cos \theta\)?
How can I know that I'm not integrating on a "longer interval" by letting directly
\(\displaystyle 0<\rho \cos \theta\) since \(\displaystyle z>0\) and so "adding some parts of the integral" due to the fact that there is a gap between \(\displaystyle z\) and \(\displaystyle 0\)?
 
I would write it like

\(\displaystyle \int_{0}^{\frac{\pi}{3}} \left(\int_{0}^{1} \left(\int_{0}^{\rho \cos \theta} \rho^2 \sin \theta dz \right) d \rho \right) d \theta + \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \left(\int_{0}^{2 \cos \theta} \left(\int_{0}^{\rho \cos \theta} \rho^2 \sin \theta dz \right) d \rho \right) d \theta \)

But I've chosen the interval of \(\displaystyle \theta\) just by intuition, I haven't a full awareness of it; I will explain better what I mean in the next part of this message.

I will try to explain better my doubt, sorry if I wasn't clear enough and I hope I will be now: my doubt is that, as far as I've understood about these integrals (maybe I still haven't understood to find the interval where \(\displaystyle \theta\) vary I have to solve \(\displaystyle \rho \sin \theta>0\) from the previous \(\displaystyle y>0\) and I have to solve \(\displaystyle 0<z<\rho \cos \theta\); the point is that how can I be sure to solve directly \(\displaystyle 0<\rho \cos \theta\) and "ignoring" \(\displaystyle z<\rho \cos \theta\)?
How can I know that I'm not integrating on a "longer interval" by letting directly
\(\displaystyle 0<\rho \cos \theta\) since \(\displaystyle z>0\) and so "adding some parts of the integral" due to the fact that there is a gap between \(\displaystyle z\) and \(\displaystyle 0\)?

The way I would think of it is to visualize the region over which you are integrating. It is a sort of wedge between the planes z=0 and z=x, cut out from a cylinder whose base is the intersection of two circles in the x-y plane.

You are first integrating along a vertical "column" from the x-y plane to the plane z=y. Then you are moving this column along a given ray (determined by θ) in the x-y plane, from the origin to the appropriate circle. Finally, you are moving that ray from the x-axis, to bounds determined by the shapes of the two circles (whose equations, after simplifying, are ρ=1 and ρ = 2 cos(θ)).

First rho goes out to the circle ρ=1, from θ=0 until the angle where the two circles meet. To find this angle, we solve 2 cos(θ) = 1, finding the angle π/3. So this is the upper limit of the first integral and the lower limit of the second. The upper limit of the second integral is determined by the circle ρ = 2 cos(θ) alone. What is the greatest value of θ when we make one loop around the circle? That is determined by when ρ returns to 0: 2 cos(θ) = 0, whose solution is θ = π/2. So that is the upper limit of the second integral. This happens to coincide with the restriction z>0, which, since the upper plane is z=x, implies that x>0, and so θ < π/2.

I'm still not sure why you are feeling so confused. It may be that you are trying to follow some mechanical method based on the various inequalities in the abstract, rather than merely looking at a graph of the region over which you are integrating, and solving equations only as needed to find details. Ultimately, I think the way to answer you questions is just to master graphing in cylindrical coordinates.
 
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