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Intersecting Cylinders

  1. Aug 8, 2007 #1
    1. The problem statement, all variables and given/known data

    Find the surface area of the region common to the intersecting cylinders
    x^2 + y^2 = 1 and x^2 + z^2 = 1.

    2. Relevant equations



    3. The attempt at a solution

    I know that the answer is 16 but why? How can we parametrize this surfaces?





    3. The attempt at a solution
     
  2. jcsd
  3. Aug 8, 2007 #2

    berkeman

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    Staff: Mentor

    Welcome to the PF, agent_cooper. You need to show us your work in order for us to offer tutorial help. What integrals are the relevant equations for calculating the surface area of shapes with boundaries? How do those equations apply here?
     
  4. Aug 8, 2007 #3

    Dick

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    Have you tried actually thinking about the algebraic conditions that define the intersection of those two cylinders? As in your last post you may want to think about how understanding the geometry can help you to avoid an explicit integral.
     
  5. Aug 8, 2007 #4
    We can solve these equations together for z and x(we need two parameters):
    z = y or z = -y & x = sqrt(1 - y^2) or x = -sqrt(1 - y^2) . The surface area can be formulized as (integral) z ds.
    Here ds = sqrt (1 + (dx/dy)^2) dy. Since (dx/dy)^2 = (y^2) / (1 - y^2), we get ds = 1 / sqrt(1 - y^2). We have z = y , and thus we get
    S = (integral from 0 to r) [y / (1 - y^2)] dy . We can solve it by using an appropriate substitution(and the value of this integral is 1 actually).
    My question is why we multiply this integral by 16? Maybe it's easy but i can't see it for now.
     
  6. Aug 8, 2007 #5

    Dick

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    I like z=+/-y. The cylinders are cut by planes. That's the geometric insight. I really don't get the rest of your post, but the it's late here and I'm tired. Cylinders are fundamentally flat. You can unwrap the surface sections onto a plane and solve them there.
     
  7. Aug 9, 2007 #6
    Anyway, thanks. It's such a good site and later i want to contribute,too.
     
  8. Aug 9, 2007 #7

    Dick

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    It is a good site, innit it? Thank the moderators for keeping it sane.
     
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