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).
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:
. 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.
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.
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:
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 .