Question

Given the parametric equations `x=acos theta` and `y=b sin theta`. What is the bounded area?

I got `-abpi` which is negative and clearly incorrect. What did I do wrong?

Answer 1

The sign is wrong as the parametric equations trace out in an anti-clockwise direction which will give a negative area,

Explanation:

The area in parametric coordinates is given by:

`A = int_alpha^beta \ y \ dx`

As we trace out in clockwise direction.

Now, we have parametric equations:

`{ (x=acos theta), (y=b sin theta) :} => dx/(d theta) = -asin theta`

And these equations trace out in an anti-clockwise direction and so we must take account that; So the area of the ellipse is given by:

`A = -int_0^(2pi) \ (b sin theta)(-asin theta) \ d theta`
`\ \ \ = ab \ int_0^(2pi) \ sin^2 theta \ d theta`

Using the identity:

`cos2A -= 1-2sin^2A => sin^2A -= 1/2(1-cos2A)`

Give us:

`A = ab \ int_0^(2pi) \ 1/2(1-cos2 theta) \ d theta`
`\ \ \ = (ab)/2 \ int_0^(2pi) \ 1-cos2 theta \ d theta`
`\ \ \ = (ab)/2 \ [theta-1/2sin2 theta ]_0^(2pi)`
`\ \ \ = (ab)/2 \ {(2pi-1/2sin4pi) - (0-1/2sin0)}`
`\ \ \ = (ab)/2 * 2pi`
`\ \ \ = abpi`