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Evaluate the integral over R

  1. Feb 7, 2013 #1

    Zondrina

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

    Let R be the region in the first quadrant (x,y ≥ 0) bounded by the lines [itex]y = x - 1[/itex] and [itex]y = x + 1[/itex] and the circles [itex]x^2 + y^2 = 1[/itex] and [itex]x^2 + y^2 = 10[/itex]

    Evaluate the integral :

    [itex]\int \int_R (x^2 + y^2)^2(y-x)^2(y+x)dxdy[/itex]

    2. Relevant equations

    Polar co-ordinates, maybe substitution?

    3. The attempt at a solution

    So at first I considered x = rcosθ and y = rsinθ.

    So our lines turn into : r(cosθ - sinθ) = 1 and r(sinθ - cosθ) = 1

    Our circles become : r = 1 and r = √10 ( Some nice bounds for r ).

    My problem now is how to find my limits for theta. I'm having trouble seeing it since the lines don't exactly turn into something nice.
     
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  3. Feb 7, 2013 #2

    Simon Bridge

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    I'd rotate the coordinate system by 45 degrees so the two lines are parallel to the x axis.

    In polar coordinates, the limits for one of your variables will be a function of the other one.
     
  4. Feb 7, 2013 #3
    The question looks like it wants to be solved using substitution. (What I mean is that it's set up in a way that using substitution would be the easiest). I'll start you off:

    Let ##u = x^2 + 1##
    ##1 \le u \le 10##

    ##y=x−1## and ##y=x+1##
    ##y - x = -1## and ##y - x = 1##

    Let ##v = y - x##
    ##-1 \le v \le 1##
     
  5. Feb 7, 2013 #4
    To add to Simon's suggestion, you both rotate the coordinates by 45 degrees, but you have to calculate the value b (not equal to 1) so that the two lines are now y = b and y = - b. Maybe you think it silly of me to assume you wouldn't know that, but I just wanted to warn you anyway.
     
  6. Feb 7, 2013 #5

    Zondrina

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    I'm not quite sure how to rotate it unless you mean take θ = (θ - π/4)? At least i think that's what you mean.

    I know that since I solved for some nice values of r ill be holding it fixed for as long as possible and allowing θ to vary over the integration.

    r(cosθ - sinθ) = 1 and r(sinθ - cosθ) = 1 though... I don't know how to isolate for θ here?

    Also @ Carnage that makes sense, but I think your substitution for u is off a bit?
     
  7. Feb 7, 2013 #6
    You're right, a little typo there! It should be:

    ##u = x^2 + y^2##
     
  8. Feb 7, 2013 #7

    Zondrina

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    The only problem is what to do with y+x then?
     
  9. Feb 7, 2013 #8
    Using the fact that ##u = x^2 + y^2## and ##v = y -x## you can deduce that ##y + x = \sqrt{2u - v^2}## by taking only the solution which satisfies that ##x,y > 0##.

    Also, don't forget you also have to find the Jacobian since you're doing a change of variables.
    Note that the Jacobian is the determinant of ##\frac{\partial (x,y)}{\partial (u,v)}##. So your answer will be in the form:

    ##\displaystyle\int\limits_{-1}^1 \int\limits_{1}^{10} u^2v^2 \sqrt{2u - v^2} \left|\frac{\partial (x,y)}{\partial (u,v)}\right| \ du\ dv##
     
    Last edited: Feb 7, 2013
  10. Feb 7, 2013 #9

    Zondrina

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    Okay so.

    Let [itex]u = x^2 + y^2[/itex]. Then the circles give u = 1 and u = 10.

    Let [itex]v = y - x[/itex]. Then our lines give v = -1 and v = 1.

    EDIT : Whoops just realized that you posted the right method in the prior post.
     
    Last edited: Feb 7, 2013
  11. Feb 7, 2013 #10

    Dick

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    u=x+y, v=x-y sounds like a simpler substitution to me.
     
  12. Feb 7, 2013 #11

    Zondrina

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    Wait do you mean u = x+y and v = y-x?

    If v = y - x, then the lines give v = -1 and v = 1.

    The substitution for u yields no useful results though?
     
  13. Feb 7, 2013 #12

    Dick

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    v=y-x, if you wish. They both work. Sure it's useful. Your integral contains (x+y) so you can just replace it with u.
     
  14. Feb 7, 2013 #13

    Zondrina

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    Hmm it's the limits for u that I'm confused about if I were to make that substitution instead.

    u = x+y ...

    According to the lines x+y = 2x - 1 or x+y = 2x + 1. I can't see where to go with this one beyond that though.
     
  15. Feb 7, 2013 #14

    Dick

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    True, it is complicated. I think you do want to move to polar coordinates at some point. Did you draw a sketch of the region? The inner limit on r is always 1 but the outer limit won't always be sqrt(10). It will depend on the value of theta. It will be where the ray of angle theta hits one of the two lines.
     
  16. Feb 7, 2013 #15

    Zondrina

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    Yeah the region is two circles being sliced by both the lines which creates a small region in Q1.

    Would the substitution given by Karn not suffice though? Dealing with u = x + y seems like more work than it should be.
     
  17. Feb 7, 2013 #16

    Simon Bridge

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    Rotate in cartesian coordinates if you want - as it stands the main difficulty is working the limits.

    From the sketch, can you see that the region becomes simple?

    In polar coordinates - it may help to divide the region into bits - there are constant r bits and bits where r depends on the angle.
     
  18. Feb 8, 2013 #17
    If you're still interested in solving using substitution, here is how to compute the Jacobian. First, find x and y in terms of u and v. There are two solutions because of the squares:

    ##x=\frac{1}{2} \left(-\sqrt{2 u-v^2}-v\right), y=\frac{1}{2} \left(v-\sqrt{2 u-v^2}\right)##

    or

    ##x=\frac{1}{2} \left(\sqrt{2 u-v^2}-v\right), y=\frac{1}{2} \left(\sqrt{2 u-v^2}+v\right)##

    It doesn't matter which we choose because we take the absolute value of the Jacobian during integration.

    ##\begin{align*}
    \displaystyle \left| \frac{\partial (x,y)}{\partial (u,v)} \right| & =
    \begin{vmatrix}
    \frac{1}{2\sqrt{2u - v^2}} & \frac{1}{2} \left(-\frac{v}{\sqrt{2 u-v^2}}-1\right)\\
    \frac{1}{2 \sqrt{2 u-v^2}} & \frac{1}{2} \left(1-\frac{v}{\sqrt{2 u-v^2}}\right)
    \end{vmatrix}\\\\
    & = \frac{1}{2 \sqrt{2 u-v^2}}\frac{1}{2} \left(1-\frac{v}{\sqrt{2 u-v^2}}\right) - \frac{1}{2} \left(-\frac{v}{\sqrt{2 u-v^2}}-1\right)\frac{1}{2 \sqrt{2 u-v^2}}\\\\
    & = \frac{1}{2 \sqrt{2 u-v^2}}
    \end{align*}##

    So we have,
    ##\begin{align*}
    \displaystyle\int\limits_{-1}^1 \int\limits_{1}^{10} u^2v^2 \sqrt{2u - v^2} \left|\frac{\partial (x,y)}{\partial (u,v)}\right| \ du\ dv &= \int\limits_{-1}^1 \int\limits_{1}^{10} u^2v^2 \sqrt{2u - v^2} \left|\frac{1}{2 \sqrt{2 u-v^2}}\right|\ du\ dv \\ &= \frac{1}{2} \int\limits_{-1}^1 \int\limits_{1}^{10} u^2v^2 \ du\ dv \\ &= 111
    \end{align*}##

    Can someone confirm this answer using a different method? The Jacobian and ##(y+x)## cancel nicely so that possibly suggests there is an easier method to do this, though I haven't thought of it.

    EDIT: I failed to consider we're working in the 1st quadrant. Does anyone know how to change the limits to restrict this to only in the 1st quadrant?
     
    Last edited: Feb 8, 2013
  19. Feb 8, 2013 #18

    Zondrina

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    I'll attempt going straight to polars again and see what happens.

    Let x = rcosθ and y = rsinθ.

    So our lines turn into : (1) r(cosθ - sinθ) = 1 and (2) r(sinθ - cosθ) = 1

    (1) [itex]r(cosθ - sinθ) = 1 \Rightarrow θ = \frac{\pi}{4} - arcsin(\frac{1}{r \sqrt{2}})[/itex]

    (2) [itex]r(sinθ - cosθ) = 1 \Rightarrow θ = \frac{\pi}{4} - arcsin(- \frac{1}{r \sqrt{2}})[/itex]

    Our circles become : r = 1 and r = √10 ( Some nice bounds for r ).

    The Jacobian of the polars is just r, so the integral over R becomes the integral over R' such that :

    [itex]\int_{1}^{\sqrt{10}} \int_{\frac{\pi}{4} - arcsin(- \frac{1}{r \sqrt{2}})}^{\frac{\pi}{4} - arcsin(\frac{1}{r \sqrt{2}})} r^7(1-sin(2θ))(sinθ+cosθ) dθdr[/itex]

    I'm not 100% certain about this though.

    EDIT : The integral evaluates to [itex]\frac{2}{15} - \frac{40 \sqrt{10}}{3}[/itex] which is roughly -42.

    This of course makes no sense at all since we are in a positive quadrant. No clue how -42 was produced.
     
    Last edited: Feb 8, 2013
  20. Feb 8, 2013 #19

    Dick

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    Yes, I plowed through the polar form after the transform u=x+y, v=x-y and after doing it several times I think I've got it right. I also get 111. I think if you go back and check the signs on everything you'll find you don't get any further restrictions from the 1st quadrant requirement.
     
  21. Feb 8, 2013 #20

    Zondrina

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    I'm assuming my take on the polars was off then? I tried to rotate it as Simon suggested, but if you're both getting 111 then chances are I've missed something.

    If u=x+y and v=x-y I'm not certain how you managed to transform the integral. I'm guessing you went from R → R' → R'' where R'' is in polars though.
     
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