Finding the Volume of the Intersection of Two Cylinders

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Homework Help Overview

The problem involves finding the volume of the intersection of two solid cylinders defined by the equations x² + y² ≤ 1 and y² + z² ≤ 1. Participants are exploring the integration setup necessary to solve this volume calculation.

Discussion Character

  • Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss various orders of integration and their implications on the complexity of the integral. There are attempts to set up the bounds for integration, with some questioning the appropriateness of the chosen bounds and the relationships between the variables involved.

Discussion Status

There is an ongoing exploration of different integration strategies, with some participants suggesting changes to the order of integration. Questions about the correctness of bounds and relationships between variables are being raised, indicating a productive dialogue around the problem setup.

Contextual Notes

Some participants express concern about potential loss of information in the bounds and the implications of the relationships between x, y, and z. There is also mention of the complexity of the integral and the need for clarity in the bounds used for integration.

TranscendArcu
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Homework Statement


Find the volume of the intersection of the two solid cylinders x2 + y2 ≤ 1 and y2 + z2 ≤ 1.


The Attempt at a Solution


Apparently this is done most easily by cartesian coordinates. I have the integral:

\int_{-1} ^1 \int_{-sqrt(1-x^2)} ^{sqrt(1-x^2)} \int_{-sqrt(1-y^2)} ^{sqrt(1-y^2)} 1 dzdydxBut, this is disgusting to integrate (as far as I can tell). I think I either a) have the wrong bounds, or b) have missed a clue to make this problem easier.
 
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Changing the order of integration would help a lot. Try integrating dy last instead of dx. Order of integration can make a BIG difference.
 
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So let's see if I can do this:

\int_{-1} ^1 \int_{-sqrt(1-y^2)} ^{sqrt(1-y^2)} \int_{-x} ^x 1 dzdxdy
\int_{-1} ^1 \int_{-sqrt(1-y^2)} ^{sqrt(1-y^2)} 2x dxdyThis gives an integrand of zero for the last integral. Hmm, did I use incorrect bounds?
 
TranscendArcu said:
So let's see if I can do this:

\int_{-1} ^1 \int_{-sqrt(1-y^2)} ^{sqrt(1-y^2)} \int_{-x} ^x 1 dzdxdy
\int_{-1} ^1 \int_{-sqrt(1-y^2)} ^{sqrt(1-y^2)} 2x dxdyThis gives an integrand of zero for the last integral. Hmm, did I use incorrect bounds?

Umm, yes, you used the wrong bounds. Shouldn't the z bounds be determined by y^2+z^2<=1?? Why would you think they should be -x to x?
 
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I have two possible problems with those bounds: first, and without doing any actual calculations, those bounds don't make the integral look very much nicer than what I had in #1; second, I would be concerned that we "lose" information about x by not including it somewhere in the bounds (that is, I see we have z's relationship to y and x's relationship to y, but not x's relationship to z). I used -x≤z≤x because that's what I got by setting,

x^2 +y^2 = y^2 + z^2, which suggests,

z = +/- x.
 
TranscendArcu said:
I have two possible problems with those bounds: first, and without doing any actual calculations, those bounds don't make the integral look very much nicer than what I had in #1; second, I would be concerned that we "lose" information about x by not including it somewhere in the bounds (that is, I see we have z's relationship to y and x's relationship to y, but not x's relationship to z). I used -x≤z≤x because that's what I got by setting,

x^2 +y^2 = y^2 + z^2, which suggests,

z = +/- x.

I have some problems with that. How does x^2+y^2<=1 and y^2+z^2<=1 imply that x^2+y^2=y^2+z^2?? I suggested integrating dy last exactly because then the x bounds depend only on y and the z bounds depend only on y. Why do think there is some other dependency?
 
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Okay. I think I understand this now. I did the calculation out and got 16/3, which I have marked as the right answer. I guess doing the integration in that order was easier.
 
TranscendArcu said:
Okay. I think I understand this now. I did the calculation out and got 16/3, which I have marked as the right answer. I guess doing the integration in that order was easier.

A LOT easier. And yes, it is 16/3.
 

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