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

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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- August 17th 2008, 07:43 AMjbuddenhArea of oval
The following 4th degree polynomial in two variables encloses an oval shape in the x-y plane:

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). - August 17th 2008, 09:30 AMSoroban
Hello, jbuddenh!

I have a start on this problem ... but I'm puzzled.

. . Have I made an error ...*several*errors?

Quote:

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

. Find the area of the oval.

We have: .

. . .

. . . . .

. . . . . . . . Convert to polar coordinates

. . . . . . . .

. . . . . . . . . .

Divide by

. . . . . . . . . .

. . which factors: .

. . and we have: .

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

- August 17th 2008, 01:33 PMshawsend
I'm pretty sure they both have the same loci of points. That is, if and , then . 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 . So it's just one limicon and can integrate it accordingly.

- August 18th 2008, 01:47 PMjbuddenh
No errors, good job converting to polar coordinates.

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:

http://i36.tinypic.com/iwt0l1.gif

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 . - August 18th 2008, 01:52 PMjbuddenh