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Intergation problem involving square root

  1. Sep 5, 2011 #1
    1. The problem statement, all variables and given/known data
    [tex]\int_{1}^{3}\sqrt{y^2+\frac{1}{2}+\frac{1}{16y^4}}dy[/tex]


    2. Relevant equations



    3. The attempt at a solution
    I don't know what to do with this integral, but i do feel like this can be simplified into something much simpler.
    I tried rewriting it as [itex]\frac{1}{4}\int_{1}^{3}\sqrt{\frac{16y^6+8y^4+1}{y^4}}dy[/itex]
    but I'm still not getting the same answer as in the back of the book
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 5, 2011 #2

    gb7nash

    User Avatar
    Homework Helper

    It can. Fill in the blanks.

    [itex]y^2+\frac{1}{2}+\frac{1}{16y^4}[/itex] = (_ + _)2

    edit: I take it back. I thought it was 1/(16y2)
     
    Last edited: Sep 5, 2011
  4. Sep 5, 2011 #3
    Is it [tex]\left(y^2+\frac{1}{4y^2}\right)^{2}[/tex]?
     
  5. Sep 5, 2011 #4

    Mark44

    Staff: Mentor

    No, given the original problem.

    Are you sure that you copied it correctly?

    Could the problem have been this integral?

    [tex]\int_{1}^{3}\sqrt{y^4+\frac{1}{2}+\frac{1}{16y^4}}dy[/tex]

    If that was the problem, the part inside the radical factors nicely.
     
  6. Sep 5, 2011 #5
    yeah that doesn't work.
    well its an arc length problem
    [tex]x=\frac{y^3}{3}+\frac{1}{4y}[/tex]
    [tex]\frac{dx}{dy}=y^2-\frac{1}{4y^2}[/tex]
    [tex]\int_{1}^{3}\sqrt{1+\left(y^2-\frac{1}{4y^2}\right)^{2}}dy[/tex]
    [tex]\int_{1}^{3}\sqrt{1+y^2-\frac{1}{2}+\frac{1}{16y^4}}dy[/tex]
    [tex]\int_{1}^{3}\sqrt{y^2+\frac{1}{2}+\frac{1}{16y^4}}dy[/tex]
     
  7. Sep 5, 2011 #6

    gb7nash

    User Avatar
    Homework Helper

    The second line isn't right. Look again.
     
  8. Sep 5, 2011 #7
    Yeah I see what I'm doing wrong.
    It's supposed to be [itex]y^4[/itex]
    thanks mark44 and gb7nash
     
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