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Surface area bounded by 2 different planes

  1. Oct 24, 2016 #1
    1. The problem statement, all variables and given/known data
    Find the surface area of portion of plane x + y + z = 3 that lies above the disc (x^2) + (y^2) < 2 in the first octant ...




    2. Relevant equations


    3. The attempt at a solution
    Here's the solution provided by the author .....
    I think it's wrong .... I think it should be the green coloured area + the black area ....

    If it's only the black area , then the problem is find the surface area of portion of plane x + y + z = 3 that lies above the cylinder (x^2) + (y^2) < 2 in the first octant..
     
    Last edited by a moderator: Oct 24, 2016
  2. jcsd
  3. Oct 24, 2016 #2
    Image here
     

    Attached Files:

    • 397.jpg
      397.jpg
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  4. Oct 24, 2016 #3

    Mark44

    Staff: Mentor

    No, it's just the black area.
    That's not what they wrote. The disc they described consists of all the points in the x-y plane that lie inside the circle ##x^2 + y^2 = 2##.
     
  5. Oct 25, 2016 #4
    The disc here refers to the circle with radius 2 lie on the xy plane where z = 0???
     
  6. Oct 25, 2016 #5

    Mark44

    Staff: Mentor

    Almost -- the radius is ##\sqrt{2}##. And yes, the disc is in the x-y plane.
     
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