# How do you determine the area enclosed by an ellipse x^2/5 + y^2/ 3 using the trapezoidal rule?

Oct 26, 2015

Assuming that the ellipse is ${x}^{2} / 5 + {y}^{2} / 3 = 1$, see the explanation section below.

#### Explanation:

The center of this ellipse is on the $x$ axis, so the area of the ellipse is twice the area of the upper half of the ellipse.

Solve ${x}^{2} / 5 + {y}^{2} / 3 = 1$ for $y$ to get the upper half of the ellipse.

The upper half of the ellipse is described by the function

$y = f \left(x\right) = \sqrt{\frac{15 - 3 {x}^{2}}{5}}$.

Note that the values of $x$ vary from $- \sqrt{5}$ to $\sqrt{5}$.

To find the area using the trapezoidal rule we need an positive integer $n$. Since none is given we'll just go through the general method of solution.

$\left\{a , b\right] = \left[- \sqrt{5} , \sqrt{5}\right]$, and
.

So we get
$\Delta x = \frac{b - a}{n} = \frac{\sqrt{5} - \left(- \sqrt{5}\right)}{n} = \frac{2 \sqrt{5}}{n}$

The endpoints of the subintervals are found by beginning at $a = - \sqrt{5}$ and successively adding $\Delta x = \frac{2 \sqrt{5}}{n}$ to find the points until we get to ${x}_{n} = b = \pi$.

${x}_{0} = - \sqrt{5}$, ${x}_{1} = - \sqrt{5} + \frac{2 \sqrt{5}}{n}$, ${x}_{2} = - \sqrt{5} + 2 \frac{2 \sqrt{5}}{n}$, . . . , ${x}_{i} = - \sqrt{5} + i \frac{2 \sqrt{5}}{n}$, . . . , and ${x}_{n} = - \sqrt{5} + n \frac{2 \sqrt{5}}{n} = \sqrt{5} = b$

Now apply the formula (do the arithmetic) for $f \left(x\right) = \sqrt{\frac{15 - 3 {x}^{2}}{5}}$..

${T}_{n} = \frac{\Delta x}{2} \left[f \left({x}_{0}\right) + 2 f \left({x}_{1}\right) + 2 f \left({x}_{2}\right) + \cdot \cdot \cdot + 2 f \left({x}_{n - 1}\right) + f \left({x}_{n}\right)\right]$

The total area of the ellipse is $\approx 2 {T}_{n}$