1. ## Area of oval

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

$\displaystyle {x}^{4}+{y}^{4}=-2\, \left( y+1 \right) \left( y-2 \right) {x}^{2}+{y}^{2} \left( 2\,y+3 \right)$

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).

2. 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:

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

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

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

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

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

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

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

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

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

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

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

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

3. I'm pretty sure they both have the same loci of points. That is, if $\displaystyle \mathbb{S}=\left\{(r,\theta):r=2+\sin(\theta)\righ t\}$ and $\displaystyle \mathbb{G}=\left\{(r,\theta):r=-2+\sin(\theta)\right\}$, then $\displaystyle \mathbb{S}=\mathbb{G}$. 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 $\displaystyle (|r|,\theta+\pi)$. So it's just one limicon and can integrate it accordingly.

4. 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: .$\displaystyle (r - \sin\theta - 2)(r - \sin\theta + 2) \;=\;0$

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

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 $\displaystyle 9\pi/2$.

5. Originally Posted by shawsend
I'm pretty sure they both have the same loci of points. That is, if $\displaystyle \mathbb{S}=\left\{(r,\theta):r=2+\sin(\theta)\righ t\}$ and $\displaystyle \mathbb{G}=\left\{(r,\theta):r=-2+\sin(\theta)\right\}$, then $\displaystyle \mathbb{S}=\mathbb{G}$. 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 $\displaystyle (|r|,\theta+\pi)$. 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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