Question

Given an ellipse with major and minor axis of `a` and `b` find the area of the ellipse? Show your work. the perimeter of the ellipse is rather difficult to derive in closed form, discuss how you would approach it?

Answer 1

`1/4 piab`

Explanation:

Let us consider an ellipse having center at origin `(0,0)` with semi-major axis and semi-minor axis of lengths `a and b` along `x`-axis and `y`-axis respectively as shown in the figure above.

The equation of this ellipse can be written in the standard form

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

As the ellipse is symmetrical with respect to `x and y` axes, the total area `A` is `4times` the area in one quadrant.

Consider first quadrant only. Solving equation (1) for `y` we get

`y = b sqrt(1 - x^2 / a^2 )`
retaining only `+ve` root for the quadrant of interest.

Area `dA` of an infinitesimal vertical strip of ellipse of width `dx` at distance `x` is given by

`dA=b sqrt(( 1 - x^2 / a^2 ))cdot dx`

Area `A/4` is found by integrating this with respect to `x` over the limits `x=0 " to " x=a`

`A/4=int_0^ab sqrt(( 1 - x^2 / a^2 ))cdot dx` ......(2)

Substituting in (2) `x/a=sin t` and `dx = a cos t cdot dt`, with limits are from `t= 0" to "pi/2`, we get

`A/4=int_0^(pi/2) b sqrt(( 1-sin^2t ))cdotacostcdot dt`

Using `sqrt(( 1 - sin^2 t )) = cos t`

`=>A/4=abint_0^(pi/2) cos^2tcdot dt`

Rewriting using the identity `cos^2 t = ( cos 2t + 1 ) / 2`

`=>A/4=abint_0^(pi/2) ( cos 2t + 1 ) / 2cdot dt`

Evaluating the integral

`A/4 = 1/2 ab [ (sin 2t)/2 + t ]_0^( pi/2)`
`=>A/4 = 1/2 ab [ (sin pi)/2 + pi/2 ]`
`=>A/4 = (ab pi)/4`

`:.`Area of ellipse `A= pi a b` # =pixx"length of semi-major axis"xx"length of semi-minor axis"#
`color (white)( WWWWWWWWWWWWWWWWWWWWWWW)` ......(3)

In the given problem we have an ellipse with major axis and minor axis of lengths `a` and `b` respectively as shown below.

Inserting given values in equation (3) we get

`A=pixxa/2xxb/2`
`=>A=1/4 piab`

Perimeter part follows.

Answer 2

Part 2. Perimeter of an ellipse.

Explanation:

As for area part of the question let us consider an ellipse having center at origin `(0,0)` with semi-major axis and semi-minor axis of lengths `a and b` along `x`-axis and `y`-axis respectively as shown in the figure above.

The equation of this ellipse is written in the standard form

`x^2 / a^2 + y^2 / b^2 = 1` .......(1)
where it is assumed that `0 " < " b " < "a`

Now the ellipse can be represented by the parametric equations

`x=acostheta andy=bsintheta`
`0≤θ≤2π`

The ellipse is symmetrical with respect to `x and y` axes, therefore, the total perimeter `p` is `4times` the perimeter of one quadrant. Considering first quadrant.

Length `dp` of an infinitesimal element of ellipse `dx` at an angle `theta` from the `x`-axis is given by

`dp=sqrt(((dx)/(d theta))^2+((dy)/(d theta))^2)`
`:.p= 4int_0^(π/2)sqrt(((dx)/(d theta))^2+((dy)/(d theta))^2)cdot d theta`
`=>p= 4int_0^(π/2)sqrt(a^2sin^2theta+b^2cos^2theta)cdot d theta`
`=>p= 4int_0^(π/2)sqrt(a^2(1-cos^2theta)+b^2cos^2theta)cdot d theta`
`=>p= 4int_0^(π/2)sqrt(a^2-(a^2-b^2)cos^2theta)cdot d theta`

Multiplying and dividing with `a` we get

`=>p= 4int_0^(π/2)asqrt(1-(1-b^2/a^2)cos^2theta)cdot d theta`

Let us define eccentricity of ellipse `epsilon-=sqrt(1-b^2/a^2)`. Substituting in expression for `p` above we get

`p= 4aint_0^(π/2)sqrt(1-epsilon^2cos^2theta)cdot d theta`

This is called an elliptic integral and unfortunately can’t be evaluated using standard functions.

We can make use of Binomial theorem to rewrite the integrand as a sum of infinite series. Note that this series converges for all values of `theta` as `0<=epsilon^2cos^2theta" < "0`. Integrating term-wise the series we can find the perimeter of the ellipse.

Explicitly we get the following expression
`p=2pia[1-(1/2)^2epsilon^2/1-((1cdot3)/(2cdot4))^2epsilon^4/3-((1cdot3cdot5)/(2cdot4cdot6))^2epsilon^6/5....]`
.-.-.-.-.-.-.-.-.-.-.-

We may also note that for a circle of radius `a`, eccentricity `epsilon=0`. The expression for perimeter (circumference) reduces to

`p=2pia`