Area of an Ellipse

Quote:

Originally Posted by sfgiants13

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