Area of oval

**jbuddenh** · August 17th 2008, 07:43 AM

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).

**Soroban** · August 17th 2008, 09:30 AM

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.

**shawsend** · August 17th 2008, 01:33 PM

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.

**jbuddenh** · August 18th 2008, 01:47 PM

Originally Posted by Soroban

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.

Originally Posted by Soroban

[...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 $9\pi/2$ .

**jbuddenh** · August 18th 2008, 01:52 PM

Originally Posted by shawsend

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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