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Math Help - Area of oval

  1. #1
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    Area of oval

    The following 4th degree polynomial in two variables encloses an oval shape in the x-y plane:

    {x}^{4}+{y}^{4}=-2\, \left( y+1 \right)  \left( y-2 \right) {x}^{2}+{y}^{2} \left( 2\,y+3 \right) <br />

    Find the area of the oval.

    (Note: just thought this might be a nice problem -- I am not a student and not asking for help).
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  2. #2
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    Hello, jbuddenh!

    I have a start on this problem ... but I'm puzzled.
    . . Have I made an error ... several errors?


    The following 4th degree polynomial in two variables encloses an oval shape in the x-y plane:

    x^4 + y^4 \;=\; -2x^2(y+1)(y-2) + y^2(2y+3) . Find the area of the oval.

    We have: . x^4 + y^4 \;=\;-2x^2y^2 + 2x^2y + 4x^2 + 2y^3 + 3y^2

    . . . x^4 + 2x^2y^2 + y^4 \;=\;4x^2 + 3y^2 + 2x^2y + 2y^3

    . . . . . (x^2+y^2)^2 \;=\;\overbrace{x^2 + 3x^2} +3y^2 + 2x^2y + 2y^3

    . . . . . (x^2+y^2)^2 \;=\;x^2 + 3(x^2+y^2) + 2y(x^2+y^2) . . . Convert to polar coordinates

    . . . . . . . . (r^2)^2 \;=\;(r\cos\theta)^2 + 3r^2 + 2(r\sin\theta)r^2

    . . . . . . . . . . r^4 \;=\;r^2\cos^2\!\theta + 3r^2 + 2r^3\sin\theta

    Divide by r^2\!:\;\;r^2 \;=\;\cos^2\!\theta + 3 + 2r\sin\theta

    . . . . . . . . . . r^2 \;=\;1-\sin^2\!\theta + 3 + 2r\sin\theta \quad\Rightarrow\quad r^2 - 2\sin\theta\!\cdot r + \sin^2\theta - 4 \:=\:0

    . . which factors: . (r - \sin\theta - 2)(r - \sin\theta + 2) \;=\;0

    . . and we have: . r \;=\;\sin\theta \pm2


    But this is not an oval . . . We have two limacons.

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  3. #3
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    I'm pretty sure they both have the same loci of points. That is, if \mathbb{S}=\left\{(r,\theta):r=2+\sin(\theta)\righ  t\} and \mathbb{G}=\left\{(r,\theta):r=-2+\sin(\theta)\right\}, then \mathbb{S}=\mathbb{G}. I'm just a little unaccustomed to dealing with negative radii: when the radius is negative, the point in the polar plane is actually located at (|r|,\theta+\pi). So it's just one limicon and can integrate it accordingly.
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  4. #4
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    Quote Originally Posted by Soroban View Post
    Hello, jbuddenh!

    I have a start on this problem ... but I'm puzzled.
    . . Have I made an error ... several errors?
    No errors, good job converting to polar coordinates.

    Quote Originally Posted by Soroban View Post
    [...snip...]

    . . which factors: . (r - \sin\theta - 2)(r - \sin\theta + 2) \;=\;0

    . . and we have: . r \;=\;\sin\theta \pm2


    But this is not an oval . . . We have two limacons.

    I called it an 'oval' because it is a simple closed curve, convex, and with a line of symmetry. I don't think 'oval' has a precise mathematical definition, but that seemed oval to me. Here is a picture of it, created with maple:

    I lied a little because the curve also has an isolated point namely the origin (0,0) which must have dropped out when you divided by r^2. Yes it is a limacon, but as another person pointed out, not two of them. For the area I got 9\pi/2.
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  5. #5
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    Quote Originally Posted by shawsend View Post
    I'm pretty sure they both have the same loci of points. That is, if \mathbb{S}=\left\{(r,\theta):r=2+\sin(\theta)\righ  t\} and \mathbb{G}=\left\{(r,\theta):r=-2+\sin(\theta)\right\}, then \mathbb{S}=\mathbb{G}. I'm just a little unaccustomed to dealing with negative radii: when the radius is negative, the point in the polar plane is actually located at (|r|,\theta+\pi). So it's just one limicon and can integrate it accordingly.
    Yes, I think you are right. Thanks for clarifying the limacon situation. See also the graph of the original x-y equation created with maple, and part of my reply to Soroban in this thread.
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