Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Area of an oval shape

  1. Jun 5, 2005 #1
    i want to know if there is a formula for the area of an oval shape because i have been working on it and want to know if it has been discovered
  2. jcsd
  3. Jun 5, 2005 #2


    User Avatar
    Science Advisor
    Homework Helper

    What's the mathematically correct term for "oval"...?

  4. Jun 5, 2005 #3
    hint: integrate the formula for an ellipse, using the boundaries carefully, and making sure that you sum any "negative" portions of the graph seperately.

    or you could just look it up in a book, i suppose...but that would be no fun.
  5. Jun 5, 2005 #4


    User Avatar
    Homework Helper

    If you mean an ellipse, then, yes, I would think the formula has been known for centuries. It's actually trivial to derive it yourself.

    The ellipse is basically a stretched version of the circle. It's said to have a shorter minor axis, [itex]a[/itex] and a longer major axis, [itex]b[/itex]. These axes are the analogues of the (uniform) radius of a circle. Observe that if you took a circle of radius [itex]a[/itex] and stretched it by a factor of [itex]\frac{b}{a}[/itex] in one direction, you'd have the exact shape of the ellipse.

    So we'll use that fact. Since the ellipse is stretched in only one dimension with respect to the circle, the ratio of the areas of the ellipse to the circle is going to be simply [itex]\frac{b}{a}[/itex]. We know that area of the circle is [itex]\pi a^2[/itex]. (I'm not going to derive this here, I'm assuming you've seen proofs of that before).


    [tex]A_{ellipse} = A_{circle}(\frac{b}{a}) = \pi a^2 (\frac{b}{a}) = \pi ab[/tex]

    So the area of the ellipse is simply [itex]\pi ab[/itex], Pi times the minor axis times the major axis.
  6. Jun 5, 2005 #5


    User Avatar
    Homework Helper

    But, if by oval, you mean the classical "egg" shape, the cross section of which has only one axis of symmetry, then you need to have a well defined formula (Cartesian or polar) for the bounding curves, then use integration to find the area bounded by the curves.
  7. Jun 6, 2005 #6


    User Avatar
    Science Advisor
    Homework Helper

    Many years ago my sister gave me a Christmas present that was a lump of brass in the shape of a solid of revolution with a central cross section given by the equation

    [tex]\left[\frac{|x|}{a}\right]^n + \left[\frac{|y|}{b}\right]^n = 1 [/tex]

    I don't recall the value for n that was used, but it was chosen to make the thing have stable equilibrium while standing on end, but just barely. On a flat surface it would roll nicely end to end, bobbing up and down, and had some nice characteristics for spinning more or less about the long axis. I seem to recall the shape had been used by some architects to design a sports arena.

    It is an interesting generalization of a symmetric oval, of which the ellipse is a special case. In general, n need not be an integer (I think it was not for my little novelty; it was probably about 2.5). With n = 1, it is a diamond. With n < 1 it has a concave cross section. With n > 1 you can generate oblong shapes comparable to oblong balls like those used in football and rugby. In fact, you can have different exponents on the two terms. With different exponents you can fine tune the "pointiness" of the ends
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook