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Thread: Area of an Ellipse

  1. #1
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    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)
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    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 $\displaystyle y^2$ term...for you were originally dividing by zero...

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

    First, solve for y:

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

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

    Now, we are to use a midpoint Riemann Sum:

    We are told that $\displaystyle \Delta x=1$, 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 $\displaystyle [0,4]$.

    What is the midpoint of each section?

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

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

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

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

    Does this make sense?

    --Chris
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