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Homework Help: Intersection of Two Cylinders

  1. Dec 15, 2017 #1
    1. The problem statement, all variables and given/known data
    I am looking for the line of intersection of two cylinders (x^2+y^=1 and x^2+z^2=1) to find the curve for a line integral. However, I am getting z=y, which is a patently false answer. Mathematically, it seems to work out though.

    2. Relevant equations
    x^2+y^=1
    x^2+z^2=1

    3. The attempt at a solution
    image_123923953.JPG
     
  2. jcsd
  3. Dec 15, 2017 #2

    Charles Link

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    I have not previously worked such a problem, but suggestion is let ##x=t ##. Then ##y=\sqrt{1-t^2} ## and ## z=\sqrt{1-t^2} ##, (omitting ## \pm 's##) will parameterize the curve.
     
    Last edited: Dec 15, 2017
  4. Dec 15, 2017 #3

    SammyS

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    Hello @OpheliaM . Welcome to Physics Forums !!!

    Actually you get y2 = z2, which is somewhat different than y = z .
     
  5. Dec 15, 2017 #4
    @SammyS Could you explain further?
     
  6. Dec 15, 2017 #5

    Charles Link

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    There is a constraint on what ## x ## is doing that you are disregarding in simply saying that ## y^2=z^2 ##.
     
  7. Dec 15, 2017 #6

    SammyS

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    Graph the following function, or tabulate some values.

    ## f(x)=\sqrt{x^2\,} ##
     
  8. Dec 15, 2017 #7
    Specifically, I am looking for the “boundary of the region common to the cylinders x^2+y^2<1 and x^2+z^2<1 in the first octant”. Can you explain the restriction further?
     
  9. Dec 15, 2017 #8

    SammyS

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    Please use the "Reply" link, so we know who you and what you addressing.
     
  10. Dec 15, 2017 #9

    Charles Link

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    The boundary region is a curved line in space. Something like ## y^2=z^2 ## gives you basically a couple of planes in three dimensions.
     
  11. Dec 15, 2017 #10

    Charles Link

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    Generally, curved lines in 3 dimensions are described by ## x=f(t) ##, ## y=g(t) ##, and ## z=h(t) ##. They are parameterized with the dummy variable ## t ##.
     
  12. Dec 15, 2017 #11
    That makes sense. What is the specific restrixrion I ignored with y=z?
     
  13. Dec 15, 2017 #12

    Charles Link

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    You are ignoring the original equations that contain ## x ##. See my post 2 again.
     
  14. Dec 15, 2017 #13
    Thank you for your help. However, I am having trouble understanding specificiallg why am I unable to equate those two equationsin your post. Why can i not disregard x? Does it have to do with the +- signs? If so, specifically where were they disregarded?
     
  15. Dec 15, 2017 #14

    Charles Link

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    It is correct that ## y= \pm z ## along the curve. In getting this result though, you can't throw away both of the original equations. This is kind of a peculiar system of equations, and I don't have a really good answer for why what you did doesn't work, but it just doesn't...
     
  16. Dec 15, 2017 #15

    Charles Link

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    To look at something that is somewhat related to the type of result you need for this problem, examine what a straight line looks like in 3 dimensions: You will typically see something like ##\frac{x-5}{3}=\frac{y-2}{5}=\frac{z-6}{4}=t ## and this line passes through ## (5,2,6 ) ## and it is in the direction of the vector ## \vec{v}=3 \hat{i}+5 \hat{j}+ 4 \hat{k} ##. You can even normalize this vector ## \hat{v} ## to a unit vector in that direction. ## \\ ## Basically the equation of this line is the parameterization ## x=3t+5 ##, ## y=5t+2 ##, and ## z=4t+6 ##. It might appear clumsy, but we can't do a simple line in 3 dimensions like we do a line in two dimensions. e.g. the equation ## 3x+5y +2z+4=0 ## represents a plane, rather than a line.
     
  17. Dec 15, 2017 #16
    Thanks!
     
  18. Dec 15, 2017 #17

    LCKurtz

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    @OpheliaM: Your original result that ##z=y## (in the first octant) is correct and tells you that the required curve lies in that plane. Here are a couple of hints:
    1. Do you know how to parameterize the ##xy## cylinder in terms of ##\theta## and ##z##, where ##\theta## is the polar angle? If so, then do that and
    2. Use the fact that ##z = y##.
    That should give you a parameterization in the form ##\vec R(\theta) = \langle x(\theta),y(\theta),z(\theta) \rangle
    ##, which is the parametric form of a curve.
     
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