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Double integral problem

  1. Jun 26, 2010 #1
    1. The problem statement, all variables and given/known data
    [tex]I=\int^{a}_{0}dx\int^{\sqrt{a^{2}-x^{2}}_{0}[/tex][tex](x-y)dy[/tex]


    2. Relevant equations

    [tex]r^{2}=x^{2}+y^{2}[/tex]


    3. The attempt at a solution

    Im thinking that the question is asking to integrate the first quarter of the circle of radius a between 0 and pi/2. In that case ive changed the limits to :-
    [tex]\int^{\frac{\pi}{2}}_{0}\int^{a}_{0}r.dr.d\theta[/tex] but now im not sure what to do with the (x-y) given in the question. I cant seem to be able to rearrange this - [tex]r^{2}=x^{2}+y^{2}[/tex] to give me a value i can use in polar coordinates.

    This integral apparantly is supposed to give an answer of 0. is this correct? i don't see how it can be zero if there is a limit of a.
     
  2. jcsd
  3. Jun 26, 2010 #2
    that top integral bottom limit should be zero, and the equation is (x-y). I must have messed up the latex code
     
  4. Jun 26, 2010 #3

    Hurkyl

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    x and y are, well, x and y. However, x and y are algebraically related to r and theta, and you may find it useful to do a conversion.


    While it would be good practice to do so, a naive brute force calculation is not the best way to solve this problem -- the solution is very, very simple if you make the right observation about what this integral is actually calculating. You've already made half of the observation, or nearly so -- you've identified geometrically the region of integration.
     
  5. Jun 27, 2010 #4
    i'm sure its really obvious, but i can't see the relation between (x-y) and r or theta, i don't see how an r can be substituted in if [tex]r^2=x^2+y^2[/tex] i can't make it fit. Are you also suggesting that i substitute the a for something else?
     
  6. Jun 27, 2010 #5

    Hurkyl

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    You knew one relationship between (x,y) and (r,theta): r²=x²+y². Where did that relationship come from?
     
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