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Just wondering if I set this up oK

  1. Sep 16, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.

    2. Relevant equations

    y = x^2 +1 , y = 9-x^2: about y = -1

    3. The attempt at a solution

    I used disks. I said that r = 1 + ((9-x^2)-(x^2 +1 )) = 9 - 2x^2
    So my integral all together is

    V = ∏ ∫ (9 - 2x^2)^2 dx from [-2,2]

    Does this set up look correct.???
     
  2. jcsd
  3. Sep 16, 2013 #2
    Also I have another one I would like to confirm.
    This one your supposed to set up the integral not evaluate for the volume of a solid.
    y = tanx, y = x, x = pi/3 : about the y axis.
    I have
    V = 2∏∫ x(tanx - x) dx between 0 and ∏/3
    I think it is good but just to be safe.
     
  4. Sep 16, 2013 #3

    Mark44

    Staff: Mentor

    Did you draw a sketch? Actually, two sketches would be useful - one of the region defined by the two curves, and another that shows the solid of revolution.

    Disks won't work, but washers will. A washer is a disk with a hole in its middle. For a typical volume element, the outer radius goes from the line y = -1 to the curve y = 9 - x2. The inner radius goes from the line y = -1 to the curve y = x2 + 1.
     
  5. Sep 16, 2013 #4

    Mark44

    Staff: Mentor

    Looks OK to me. It would have been helpful to indicate that your using shells.
     
  6. Sep 16, 2013 #5
    Yeah I always draw it. Sorry for the confusion with the other problem not indicating I was using shells.

    So what about this. r = inner radius, R = outer radius
    r = 1 + ( x^2 + 1) R = 1+ (9-x^2)
    So V = ∏∫( x^2 +2)^2 - (10-x^2)^2 dx I think that is it but I have a little trouble with the radius sometimes.
     
    Last edited: Sep 16, 2013
  7. Sep 16, 2013 #6

    Mark44

    Staff: Mentor

    Your radii are correct but the integral isn't. The volume of a washer is ##\pi(R^2 - r^2)(\text{thickness})##.
     
  8. Sep 16, 2013 #7
    I just mean minus. Check it out I edited.
     
  9. Sep 16, 2013 #8
    V = ∏∫(10-x^2)^2 -( x^2 +2)^2 dx GRR.
     
  10. Sep 16, 2013 #9

    Mark44

    Staff: Mentor

    That's better.
     
  11. Sep 16, 2013 #10
    It is like I'm thinking one thing and typing another. Thanks for the confirmation.
     
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