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Double Integral

  1. Nov 21, 2009 #1
    1. The problem statement, all variables and given/known data

    [tex]\int\int[/tex] y [tex]\sqrt{x^2+y^2}[/tex]dx dy

    2. Relevant equations

    x[tex]\geq[/tex] 0, y[tex]\geq[/tex] 0, x^2+y^2 [tex]\leq[/tex] 4

    3. The attempt at a solution

    first of all, what are the limits of integration

    rearranging x^2+y^2 [tex]\leq[/tex] 4 you get x = 2 - y
    this would be my limit of integration for the inner integral yes?

    the limits for the outer integral cannot be a function so this would go between 2 and 0?

    when the first integration is performed can the square root and y be multiplied out to give

    yx + y^2 or is it a case of integrating by parts/substitution?
     
  2. jcsd
  3. Nov 21, 2009 #2

    Nabeshin

    User Avatar
    Science Advisor

    Ok first off, I think you should understand what the region is a little better. So,
    [tex]\left[x^2+y^2\leq4\right] \neq \left[x=2-y\right][/tex] that's just a simple algebra error (I assume when you took the square root you just applied it individually to 4 and y, but you can't do that!).

    So, then, what does that region describe? Start off by thinking about the similar bound,
    [tex]x^2+y^2=4[/tex]
    any idea what that is? And if you do know what it is, what kind of coordinates are very convenient for describing that kind of geometry? :wink:
     
  4. Nov 21, 2009 #3
    but if you use polar co-ordinates you don't have a value for theta

    r = x^2 + y^2
    =4
    also how can the y part be obtained?

    theta = arc tan (y/x)
    this bit confused me

    edit: theta can be obtained from what we know that x and y > 0. therefore theta = pi/2
     
    Last edited: Nov 21, 2009
  5. Nov 21, 2009 #4

    Mark44

    Staff: Mentor

    Your region is not described by r = 2; it is described by 0 <= r <= 2. And theta certainly does play a role.
     
    Last edited: Nov 21, 2009
  6. Nov 21, 2009 #5

    Nabeshin

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    Science Advisor

    Probably a semantics issue, but you shouldn't say theta equals anything. Theta has a range, just like Mark44 notes that r has a range, and is not simply one value.

    Although, r goes from 0 to 2, not 0 to 4.
     
  7. Nov 21, 2009 #6
    yes it is a semantic issue.
    what i have written on paper is different to my typing. just a bad habit to always press equals on a computer.
    what is my next step?

    i think once i get the limits sorted i will be fine

    the limits are 0<r<2
    0<=theta<=pi/2

    can the y factor be expressed as r sin theta?
     
  8. Nov 21, 2009 #7

    Mark44

    Staff: Mentor

    This is only 1/4 of the way around.
    Yes, but what about the rest of the integrand, and what does dx dy become?
     
  9. Nov 21, 2009 #8
    x and y are both > or = 0. which to me implies only a quarter way around

    dx and dy will become dr and d(theta)

    [tex]\int[/tex] theta 0 [tex]\int[/tex]2 0 2r sin theta r dr d(theta)
     
  10. Nov 22, 2009 #9

    Mark44

    Staff: Mentor

    You're right that theta ranges between 0 and pi/2. You put your inequalities x>=0 and y>=0 in the relevant equations rather than in the problem statement, and I missed them.

    Your integrand should not have 2r; the original integrand of y*sqrt(x^2 + y^2) becomes r*rsin(theta) = r^2*sin(theta). There is also the factor of r from r dr d(theta), which you correctly show.

    Your iterated integral should look like this:
    [tex]\int_{\theta = 0}^{\pi/2}~\int_{r = 0}^2 r^3~sin(\theta)~dr~d\theta[/tex]
     
  11. Nov 23, 2009 #10
    so i should get an answer of 4?
     
  12. Nov 23, 2009 #11

    Mark44

    Staff: Mentor

    Yep.
     
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