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Area of plane region

  1. Dec 24, 2011 #1
    1. The problem statement, all variables and given/known data

    Find the area of the plane region bounded by the curve
    $$
    (x^2+y^2)^3 = x^4+y^4
    $$

    2. Relevant equations
    The change of variables formula:
    $$
    \int\int_R F(x,y)dxdy = \int\int_S G(u,v)\left| \frac{∂(x,y)}{∂(u,v)}\right| dudv
    $$

    3. The attempt at a solution

    I recognize this as a change of variables problem, and in general I understand how to do change of variables, but for this one I cannot figure out what to use as the new variables u = u(x,y) and v = v(x,y). Previously I could tell what I needed by inspection, but I cannot tell with this problem. Is there some method for directly computing what u and v must be?

    I thought of using [itex]{u = x^2 + y^2}[/itex] and [itex]{v = x^4 + y^4}[/itex], but I can't figure out how to use this to solve for x and y, so I reached a dead end.

    Any help would be appreciated.
     
  2. jcsd
  3. Dec 24, 2011 #2

    SammyS

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    Try [itex]\displaystyle u=x^2+y^2\ \ \text{ and }\ \ v=2xy\,.[/itex]
     
  4. Dec 24, 2011 #3
    Thank you for your help! Using the above change of variables, I solved for x and y in terms of u and v (to compute the Jacobian [itex] \frac{∂(x,y)}{∂(u,v)} [/itex]):

    [itex] v = 2xy \Rightarrow y = \frac{v}{2x} [/itex]

    [itex] u = x^2+y^2=x^2 + \frac{v^2}{4x^2} \Rightarrow 4x^4-4ux^2+v^2=0[/itex]

    Let [itex] z=x^2 [/itex]. Then

    [itex] 4z^2-4uz+v^2=0 \Rightarrow z = \frac{1}{2}\left( u \pm \sqrt{u^2-v^2} \right) [/itex]

    [itex] \Rightarrow x = \sqrt{ \frac{1}{2}\left( u \pm \sqrt{u^2-v^2} \right) } [/itex]

    [itex] \Rightarrow y = \frac{v}{2\sqrt{ \frac{1}{2}\left( u \pm \sqrt{u^2-v^2} \right) }} [/itex]

    The curve becomes

    [itex] u^3 = x^4+y^4 = (x+y)^4 - 2xy(2y^2+3xy+2x^2) = (u+v)^2-4xy(x+y)^2+2x^2y^2 = (u+v)^2 - 2v(u+v)+\frac{1}{2}v^2 = u^2-\frac{1}{2}v^2[/itex]

    [itex] \Rightarrow u^3 = u^2-\frac{1}{2}v^2 [/itex]

    This seems overly complicated, like I made a mistake somewhere, but I cannot find it - does it look correct? Also, when I graphed the new curve, I got the below graph, so I think I didn't do something right. Isn't it supposed to be a nice curve?
     

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  5. Dec 24, 2011 #4

    SammyS

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    I think that's all right. I got the same graph.

    I transformed the expression by noticing that [itex]\displaystyle u^2=(x^2+y^2)^2=x^4+2x^2y^2+y^4\,.[/itex]

    So subtract 2v2 from that to get x4+y4 .

    Of course, (x2+y2)3 was obvious.

    I haven't looked at the resulting integral, but solving for v is fairly easy:
    [itex]v=\pm\sqrt{2(u^2-u^3)}=\pm|u|\sqrt{2}\sqrt{1-u}\,.[/itex]​

    Your graph is consistent with this from the point of view that the loop corresponds to 0 ≤ u ≤ 1. [itex]\sqrt{1-u}[/itex] is real there.

    Have you worked out the Jacobian yet?
     
  6. Dec 24, 2011 #5

    SammyS

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    After looking at the graph of u^3 < u^2-v^2, I also have my doubts about this transformation.
     
  7. Dec 25, 2011 #6

    SammyS

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    Before I go on about my previous suggestion for a translation, have you tried polar coordinates? ...

    After messing around with a few sets of ordered pairs and a few other details, like graphing the region in the xy-plane, I like this transformation again.

    Here's WolframAlpha's graph of the region in the xy-plane, along with the graphs of two circles, one with radius 1 the other with radius = 1/√(2) :
    attachment.php?attachmentid=42201&stc=1&d=1324788926.gif

    Here's WolframAlpha's graph of the region in the xy-plane, along with the graph of y = ±x .
    attachment.php?attachmentid=42202&stc=1&d=1324789056.gif

    Since x2 + y2 ≥ 0, u ≥ 0, for the region of interest.

    The point (x,y)=(a, ±a) → (u,v)=(2a2, ±2a2), where the signs are correlated.
    Specifically, the point (x,y)=(1/2, ±1/2) → (u,v)=(1/2, ±1/2) .​

    The point (x,y)=(±a,0) → (u,v)=(a2,0) also the point (x,y)=(0, ±a) → (u,v)=(a2,0) .
    Specifically, the points (x,y)=(±1,0) and (x,y)=(0, ±1) all go to (u,v)=(1,0)​
     

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  8. Dec 25, 2011 #7

    SammyS

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    Definitely, use polar coordinates.
     
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