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Double integral problem

  • Thread starter 8614smith
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  • #1
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Homework Statement


[tex]I=\int^{a}_{0}dx\int^{\sqrt{a^{2}-x^{2}}_{0}[/tex][tex](x-y)dy[/tex]


Homework Equations



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


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.
 

Answers and Replies

  • #2
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that top integral bottom limit should be zero, and the equation is (x-y). I must have messed up the latex code
 
  • #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.
 
  • #4
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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?
 
  • #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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