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Double integral substitution

  1. Mar 4, 2014 #1
    Hey.

    1. The problem statement, all variables and given/known data

    ∫∫x^3 dxdy, with the area of integration: D={(x,y)∈R^2: 1<=x^2+9y^2<=9, x>=3y}

    3. The attempt at a solution

    Did the variable substitution u=x and v=3y so the area of integration became 1<=u^2 + v^2 <=9, u>=v. And the integral became ∫∫(1/3)u^3 dudv. Then I switched to polar coordinates and the integral ∫∫(1/3)r^4cos^3θ drdθ, with 1<=r<=9 and -3/4<=θ<=pi/4.

    Any errors so far?

    I really appreciate any help you can provide.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Mar 4, 2014
  2. jcsd
  3. Mar 4, 2014 #2

    HallsofIvy

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    Well, first, if you take u= x, v= 3y then 1<= x^2+ y^2<= 9 does NOT become 1<= u^2+ v^2<= 9, it becomes 1<= u^2+ v^2/9<= 9, the region between two ellipses. Since you then change to polar coordinates, I don't see any good reason to do that. Rather, change to polar coordinates initially. The line x= 3y, or y= x/3 has [itex]\theta= arctan(1/3)[/itex] so the integral is [tex]\int_{\theta= -\pi+ arctan(1/3)}^{arctan(1/3)}\int_{r= 1}^3 r^4 cos^3(\theta) drd\theta[/tex].
     
  4. Mar 4, 2014 #3
    My mistake, it should have said 1<= x^2+ 9y^2<= 9
     
  5. Mar 5, 2014 #4
    Can someone please confirm if my way of solving is correct? I made a typo in the first post, but I have corrected it now.
     
  6. Mar 5, 2014 #5

    vela

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    Your upper limit for ##r## is wrong.
     
  7. Mar 5, 2014 #6
    I had a feeling it was, because my answer was in the right form, just too large. Can you please explain why it is wrong?
     
  8. Mar 5, 2014 #7
    Just figured it out, the radius is not 9, it is √9=3. Is that the correct limit?
     
  9. Mar 5, 2014 #8

    vela

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    Yup!
     
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