Question

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

Answer 1

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(x) = sqrt((15-3x^2)/5)`.

Note that the values of `x` vary from `-sqrt5` to `sqrt5`.

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.

`{a,b] = [-sqrt5, sqrt5]`, and
.

So we get
`Delta x = (b-a)/n = (sqrt5-(-sqrt5))/n = (2sqrt5)/n`

The endpoints of the subintervals are found by beginning at `a=-sqrt5` and successively adding `Delta x = (2sqrt5)/n` to find the points until we get to `x_n = b = pi`.

`x_0 = -sqrt5`, `x_1 = -sqrt5 + (2sqrt5)/n`, `x_2 = -sqrt5 + 2(2sqrt5)/n`, . . . , `x_i = -sqrt5 + i(2sqrt5)/n`, . . . , and `x_n = -sqrt5 + n(2sqrt5)/n= sqrt5 = b`

Now apply the formula (do the arithmetic) for `f(x) = sqrt((15-3x^2)/5)`..

`T_n=(Deltax)/2 [f(x_0)+2f(x_1)+2f(x_2)+ * * * +2f(x_(n-1))+f(x_n)]`

The total area of the ellipse is `~~2T_n`