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Double integral showing a range of values?

  1. Apr 9, 2009 #1
    Hi all,

    I am faced with this question. I am asked to show that

    [tex]
    2(\sqrt{5}-2)\pi\leq\iint_{R}\frac{1}{\sqrt{4+sin^2x+sin^2y}}dA\leq\frac{\pi}{2}
    [/tex]

    Noting that the double integral is to be performed on region R which is bounded by the circle [tex]
    x^2+y^2=1
    [/tex]

    From what I know, the double integral is a definite integral, then how come will it range from
    [tex]
    2(\sqrt{5}-2)\pi\leq
    [/tex] to [tex]\frac{\pi}{2}[/tex]?


    What I manage to do is by letting [tex]u=sin x[/tex] and [tex]v=sin y[/tex].

    Therefore it follows that:
    [tex]
    \iint_{R}\frac{1}{\sqrt{4+u^2+v^2}}dA
    [/tex]

    And converting it into polar coodinates by letting [tex]u=rcos\theta[/tex] and [tex]v=rsin\theta[/tex] and [tex]dA=rdrd\theta[/tex], I have:
    [tex]
    \int_{\theta=0}^{\theta=2\pi} \int_{r=0}^{r=1} \frac{r}{\sqrt{4+r^2}}drd\theta
    [/tex]

    And then solving the above will give me [tex]2(\sqrt{5}-2)\pi[/tex]. However even then, this is a definite integral. How will I get a range of values? And even for that matter, how am I going to prove that
    [tex]
    2(\sqrt{5}-2)\pi\leq\iint_{R}\frac{1}{\sqrt{4+sin^2x+sin^2y}}dA
    [/tex]
    when all I have now is:
    [tex]
    2(\sqrt{5}-2)\pi=\iint_{R}\frac{1}{\sqrt{4+sin^2x+sin^2y}}dA
    [/tex]
     
    Last edited: Apr 9, 2009
  2. jcsd
  3. Apr 10, 2009 #2
    I'm very sorry I still don't know how to use Latex, so I'll just give you a hint, it is wrong just let u=sinx, v=siny simply, if you do so you must change the coordinate using Jacobian,
    indeed, you just need a tiny correction, notice that x>sinx, y>siny, then replace sinx, siny by x and y, you get an inequality.
    for pi/2,just find the upper bond of the integrand, it's quite straightforward.
     
  4. Apr 10, 2009 #3
    well, yeah, another correction x^2>sin^2(x),this would work.And besides upper bound we still need the lower bound ,right?
    And mostly I was explaining this double integral does not equal to 2(sqrt5-2)pi,but larger than that
     
  5. Apr 10, 2009 #4
    uuuhhh, is that a typo? I just took your approach as correct. Yikes! Anyway x^2 \geq \sin^2x, etc. Sorry to previous poster....
     
  6. Apr 11, 2009 #5
    Hi,

    I still don't quite get you. Basically you are letting x^2 > sin^2(x) and y^2 > sin^2(y). Why do you need to set it to greater and not let them be equal? And how can that lead to inequality? Is there some websites or book I can read about this?

    Thanks in advance.



     
  7. Apr 11, 2009 #6
    Hi,

    So instead of letting x^2=sin^2(x), we can let x^2>sin^2(x)? can you explain to me in more details?

    Thanks!

     
  8. Apr 11, 2009 #7
    well...it's not hard but a bit complicated to explain:
    first, it is a definite integral an it has a certain value, but probably you cannot work it out, so it asks you for a range;
    second, when you use u=sinx, v=siny, that is fine, but the following steps are wrong, you change the variables, so you need also change the integral element, for example, when you use polar coordinate, x=r*cosa, y=r*sina, you also change dxdy into rdrda,not just drda,but you just used dudv to replace dxdy at first, indeed this requires a Jacobian, you can search it on Wikipedia.
    but using x^2>sin^2(x), you are not changing variables, but just construct an inequality in the original coordinate, so you won't need to change the integral element dxdy.
    third, using x^2>sin^2(x) to construct the inequality, you'll find a new integrand always smaller than the original integrand, then you do the integral, you'll find the lower bond of the integral
     
  9. Apr 11, 2009 #8
    Thanks kof9595995.

    How will using x^2>=sin^2(x), I will get a new integrand smaller than the original one? I'm lost on this.

    Secondly, even if I manage to find the lower bound, how do I approach to find the upper bound?

    Finally if there is any website or book that discussed about integral inequality, please let me know.

    Thanks.

     
  10. Apr 11, 2009 #9
    sqrt[4+x^2+y^2]>sqrt[4+sin^2(x)+sin^2(y)], and it's the denominator ,so the integrand is smaller
    the upper bound is rather easy, note that integrand is always smaller than 1/2, and the integrating area is pi*R^2=pi,so the integral must be smaller than pi/2
     
  11. Apr 11, 2009 #10
    Oh dear I think my calculus is really lousy. :( I'm so confused now.

    How is the integrand be smaller than 1/2 and how will that lead to upper bound being pi/2?

    I'm sorry I am really lost on this.



     
  12. Apr 11, 2009 #11
    because sqrt[4+sin^2(x)+sin^2(y)]>sqrt4=2, and if you think geometrically, the integral means a volume smaller than a cylinder with height 1/2 radius 1, algebraically, you can replace the integrand by 1/2, it'll also give the same result, but thinking geometrically is a shortcut.
    Don't be upset,, it's still not abstract at this stage, just think more and practice more.
     
  13. Apr 11, 2009 #12
    Erm..... please let me know if I am on the right track.

    the biggest value of sin^2(x) is 1. Hence, the biggest value of sqrt(4+sin^2(x)+sin^2(y)) is sqrt(6).

    But why did you compare it with sqrt(4)?

    Also I know that region A is defined by x^2+y^2=1, hence it's radius is 1. But why did you say that it has a height of 1/2 and not anything else?

     
  14. Apr 11, 2009 #13
    because you want one upper bond of the integrand, so you need 4+sin^2(x)+sin^2(y)>=4+0+0=4, so integrand<=1/sqrt4=1/2, that why I say the height is 1/2
     
  15. Apr 12, 2009 #14
    Hi kof9595995,

    Thanks for your help. I more or less what is going on. Basically to find the upper bound, you let sin^2(x) and sin^2(y) be 0. But why do you not let sin^2(x) and sin^2(y) be 1 in order to find the lower bound?

    Why do we proceed to convert the integrand to polar coordinates for lower bound, yet we let let sin^2(x) and sin^2(y) be 0 for upper bound?

     
  16. Apr 12, 2009 #15
    huh,it doesn't really matter but then you can't get the result, which the problem asks for
     
  17. Apr 23, 2009 #16
    Sorry for the late reply.

    Thank you very much for your help!

    Regards,
    Casey
     
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