Question

Use the parametric equations of an ellipse, x=acosθ, y=bsinθ, 0≤θ≤2π , to find the area that it encloses.

Answer 1

Area of ellipse=
`\pi ab`

Explanation:

We are given that

`x=acos\theta`

`y=bsin\theta`

`0\leq\theta\leq 2\pi`

We have to find the area enclose by it.

`x/a=cos\theta, y/b=sin\theta`

`sin^2\theta+cos^2\theta=x^2/a^2+y^2/b^2`

Using the formula

`sin^2x+cos^2x=1`

`(x^2)/(a^2)+(y^2)/(b^2)=1`

This is the equation of ellipse.

Area of ellipse

=
`4\int_(0)^(a)(b)/(a)√(a^2-x^2)dx`

When x=0,
`\theta=\pi/2`

When x=a,
`\theta=0`

Using the formula

Area of ellipse

=
`(4b)/(a)\int_(\pi/2)^(0)√(a^2-a^2cos^2\theta)(-asin\theta)d\theta`

Area of ellipse=
`-4ba\int_(\pi/2)^(0)√(1-cos^2\theta)(sin\theta)d\theta`

Area of ellipse=
`-4ba\int_(\pi/2)^(0) sin^2\theta d\theta`

Area of ellipse=
`-2ba\int_(\pi/2)^(0)(2sin^2\theta)d\theta`

Area of ellipse=
`-2ba\int_(\pi/2)^(0)(1-cos2\theta)d\theta`

Using the formula

`1-cos2\theta=2sin^2\theta`

Area of ellipse=
`-2ba[\theta-1/2sin(2\theta)]^(0)_(\pi/2)`

Area of ellipse
`=-2ba(-\pi/2-0)`

Area of ellipse=
`\pi ab`