Results 1 to 2 of 2

Math Help - Area of an Ellipse

  1. #1
    Junior Member
    Joined
    May 2008
    Posts
    56

    Area of an Ellipse

    x^2/16 + y^2/0=1

    Find the function y=f(x) that gives the curve bounding the top of the ellipse.

    Use deltax=1 and midpoints to approximate the area of the part of the ellipse lying in the 1st quadrant

    Approximate the total area (which I assume is just c x 4)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Santa Cruz, CA
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by sfgiants13 View Post
    x^2/16 + y^2/9=1

    Find the function y=f(x) that gives the curve bounding the top of the ellipse.

    Use deltax=1 and midpoints to approximate the area of the part of the ellipse lying in the 1st quadrant

    Approximate the total area (which I assume is just c x 4)
    I think you meant 9 for the denominator in the y^2 term...for you were originally dividing by zero...

    \frac{x^2}{16}+\frac{y^2}{9}=1

    First, solve for y:

    \frac{y^2}{9}=1-\frac{x^2}{16}\implies y^2=\tfrac{9}{16}(16-x^2)\implies y=\pm\tfrac{3}{4}\sqrt{16-x^2}

    Since we are told to find the area under the ellipse in the first quadrant, we take y=f(x)=\tfrac{3}{4}\sqrt{16-x^2}

    Now, we are to use a midpoint Riemann Sum:

    We are told that \Delta x=1, and the ellipse's major axis is along the x-axis and has a length of 4 units. Thus, we are applying the Riemann sum over the interval [0,4].

    What is the midpoint of each section?

    Our midpoint x values are \tfrac{1}{2},~\tfrac{3}{2},~\tfrac{5}{2},\text{ and }\tfrac{7}{2}

    So our Riemann sum can be expressed as \sum_{k=1}^{4}f\left[\tfrac{1}{2}(2k-1)\right]\Delta x. Evaluating, we see that we have:

    A=\tfrac{3}{4}\left[\sqrt{16-\left(\tfrac{1}{2}\right)^2}+\sqrt{16-\left(\tfrac{3}{2}\right)^2}+\sqrt{16-\left(\tfrac{5}{2}\right)^2}+\sqrt{16-\left(\tfrac{7}{2}\right)^2}\right] \approx \color{red}\boxed{9.55}

    Thus, the total area would be 4 times this amount.

    Does this make sense?

    --Chris
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. ellipse area
    Posted in the Geometry Forum
    Replies: 5
    Last Post: August 2nd 2011, 12:36 AM
  2. Area of an ellipse
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 26th 2010, 10:49 AM
  3. ellipse area
    Posted in the Geometry Forum
    Replies: 6
    Last Post: November 17th 2010, 05:59 AM
  4. Area of a rectangle within an ellipse
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 5th 2010, 10:47 AM
  5. area of ellipse
    Posted in the Geometry Forum
    Replies: 2
    Last Post: April 1st 2009, 09:22 PM

Search Tags


/mathhelpforum @mathhelpforum