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Volume of cylinder bounded by two dependent planes, ideas?

  1. Sep 17, 2016 #1
    1. The problem statement, all variables and given/known data

    Calculate the volume bounded by the plane/cylinder x^2+y^2=1 and the planes x+z=1 and y-z=-1.

    2. Relevant equations / The attempt at a solution

    It is pretty basic triple integral in cylindrical coordinates. For some reason, I can't get the right answer. I'm using bounds: 1-x ≤ z ≤ y + 1, -3π/4 ≤ θ ≤ π/4, 0 ≤ r ≤ 1.

    The shape is a cylindrical wedge of sorts but it is so twisted I'm not sure this is correct. Any ideas?
     
  2. jcsd
  3. Sep 17, 2016 #2

    LCKurtz

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    This is a tricky problem. The problem is that the two planes cross each other inside the cylinder. If you set the z values equal you get ##y = -x##. If you look at the projection in the ##xy## plane that line divides the circle into two regions, $$-\frac \pi 4 \le \theta \le \frac {3\pi} 4 \text{ and }\frac {3\pi} 4 \le \theta \le \frac{7\pi} 4$$I think you will find that in the first region ##z=1+y## is the upper surface and in the second region ##z=1-x## is the upper surface. Work out the integrals for those two cases and see if that helps.

    Edit, added: Here's a picture. It isn't oriented in the usual direction in order to get a nice view of it. But you can see the shape of it:
    picture.jpg
     
    Last edited: Sep 17, 2016
  4. Sep 18, 2016 #3
    Ok, so I get that the volume would then be Vtot = V1 + V2 = (2 * √(2)) / 3 + (2 * √(2)) / 3 = (4 * √(2)) / 3

    Is this correct, does someone get different answers?
     
  5. Sep 18, 2016 #4

    LCKurtz

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    Not quite what I got. Let's see your integrals.
    [Edit] Woops! Cancel that. I agree with your answers.
     
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