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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


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    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,
    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
    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


    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


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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

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


    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


    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


    Staff: Mentor

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