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Proving Inequality

  1. Oct 28, 2013 #1
    1. The problem statement, all variables and given/known data
    Prove the inequality double integral (dA / (4+x^2+y^2)) is less than or equal to pi, where the double integral has a sub D where D is the disk x^2 + y^2 less than or equal to four

    2. Relevant equations

    3. The attempt at a solution
    I really have no idea, anyone want to give me a clue to help me get started?
  2. jcsd
  3. Oct 28, 2013 #2
    Are you sure this is the exact question?
    Last edited: Oct 28, 2013
  4. Oct 28, 2013 #3


    Staff: Mentor

    I believe it is.

    Here's the integral and inequality:
    $$\int_D \frac{dA}{4 + x^2 + y^2} \leq \pi$$
    where D is the disk x2 + y2 ≤ 4.

    The key here, I believe, is that ##\frac{1}{4 + x^2 + y^2} \leq \frac 1 4##.
  5. Oct 28, 2013 #4
    Are you really sure?
  6. Oct 28, 2013 #5


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    Staff Emeritus
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    Gold Member

    Obvously only Psychonaut can be sure but the problem statement is a true statement if that's what you're trying to get at.
  7. Oct 28, 2013 #6
    What do you mean by a "true statement"?

    [tex] \int_0^4 \frac{r}{4+r^2}\ \mbox{d}r = \ln(\sqrt5) = 0.8 [/tex]
  8. Oct 28, 2013 #7


    Staff: Mentor

    By "true statement" I think Office_Shredder means that the problem as described in the OP represents a problem that can be solved. In this case, the problem is fairly simple. If I'm missing something, please enlighten me.
  9. Oct 28, 2013 #8

    D H

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    You have the wrong integration limits here. They should be from 0 to 2, not from 0 to 4.
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