Is there a formula for getting the area of an ellipse?

Question

for example the formula for a circle is pi r squared

Answer 1

There's a way to derive the area of an ellipse via Calculus 2. Might as well work it out here.

The general form of an ellipse is

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

If we solve for y, we can get

(y^2)/(b^2) = 1 - (x^2)/(a^2)

y^2 = b^2 - [(b^2)/(a^2)]x^2

We're going to take only the *positive* square root, meaning

y = sqrt ( b^2 - [(b^2)/(a^2)]x^2 )

Let's make a single fraction inside the square root.

y = sqrt ( [a^2 b^2 - b^2 x^2] / (a^2) )

And we can factor out a b^2 and 1/a^2, inside the square root, which will come outside of the square root as b/a.

y = (b/a) * sqrt ( { [a^2 - x^2] } )

What we want to solve for is the area under this curve. This curve is half of an ellipse, which means we just double our result to obtain the area. We want to take the integral from -a to a, since "a" represents the distance from the center of the ellipse to its horizontal endpoints.

A = Integral (-a to a, (b/a) * sqrt ( { [a^2 - x^2] } ) dx)

But, by symmetry, why not just go from 0 to a, and double our result?

A = 2 * Integral (0 to a, (b/a) * sqrt ( { [a^2 - x^2] } ) dx)

Pulling out the constant (b/a), we obtain

A = (2b/a) * Integral (0 to a, sqrt ( { [a^2 - x^2] } ) dx )

We solve this using trigonometric substitution.

Let x = asin(t) {When x = 0, t = 0. When x = a, t = pi/2
dx = a cos(t) dt

A = (2b/a) * Integral (0 to pi/2, sqrt (a^2 - a^2sin^2(t)) a cos(t) dt )

Pulling out the constant a^2 comes out as an a, which will in turn cancel with the a in the denominator. Also, we have another a, wihich will merge with the 2b with make 2ab.

A = (2ab) * Integral (0 to pi/2, sqrt (1 - sin^2(t)) cos(t) dt )

A = (2ab) * Integral (0 to pi/2, sqrt ( cos^2(t) ) cos(t) dt )

The square root of cosine squared is itself, so we will end up with

A = (2ab) * Integral (0 to pi/2, cos^2(t) dt)

We solve this using our half angle identity.

A = (2ab) * Integral (0 to pi/2, (1/2) (1 + cos2t) dt)

Pulling out the (1/2) will merge with the 2ab as ab.

A = (ab) * Integral (0 to pi/2, (1 + cos2t) dt )

Now, we solve for the integral easily.

A = (ab) [t + (1/2)sin(2t)] {evaluated from 0 to pi/2}

A = (ab) { [pi/2 + (1/2)sin(pi)] - [0 + (1/2)sin(0)] }

A = (ab) { pi/2 }

A = (pi/2) ab

Note that this is only half of the ellipse, so we have to double this, and we obtain our final area.

A = pi (a)(b)

Answer 2

Yes!!!
The formula is 'pi x a x b'
Where 'a' is the major axis
and 'b' is the minor axis.

The Major axis is the distance from the centre of the ellipse (NOT either of the foci).to the furthest point of its circumference.
The Minor axis is the distance from the centre of the ellipse to the nearest point on the circumference.The Major and Minor axes are a right angles to each other.

Answer 3

Yes, R Wall like the person before me has said, for an ellipse which is doubly symmetric with a semimajor axis of 2a and a semiminor axis of 2b the area of the ellipse is pi*ab, its when either a-->b or b-->a that you get pi*a^2 or pi*b^2 which would be the area of a circle....

Answer 4

For example, the following is a standard equation for such an ellipse centered at the origin:


(x2/A2) + (y2/B2) = 1.
The area of such an ellipse is


Area = Pi * A * B ,

a very natural generalization of the formula for a circle!

Answer 5

pi*a*b where "2*a" is the width and "2*b" is the length of the ellipse.
A circle can be viewed as a special type of ellipse, where the width equals the length. In this case the area equals pi*a^2 and, of course, "a" is half the width of the circle or it's radius.

Answer 6

An ellipse is a much more complex form than a circle. The only way to find the area is by integration.

You can find a full explanation on Mathworld:

http://mathworld.wolfram.com/Ellipse.html

Answer 7

area of ellipse-semi-axes
are a and b

let the centre of the ellipse
be the origin,and let x be
measured along the semi-
axis whose length is a

then the equation of the
ellipse is;
x^2/a^2+y^2/b^2=1
giving y=b/a*sqrt{a^2-x^2}
therefore,area of first
quadrant
=b/a*int{sqrt(a^2-x^2)}dx
between x=0 and x=a
to integrate this expression,
let x=a sint
then dx=acostdt
also,when x=0,t=0;
and when x=a,sint=1,
that is, t=pi/2
therefore,
area of first quadrant
=b/a*int{a^2*(cost)^2}dt
for t between 0 and pi/2
=b/a*a^2/2[t+1/2sin2t]
for t between 0 and pi/2
when t=pi/2,sin2t=sin180=0;
therefore,
area of first quadrant
=b/a*a^2/2*pi/2=pi*ab/4;
therefore, the area of the
whole ellipse=pi*ab.

i hope that this helps

Answer 8

if the ellipse is in standard form equation
x^2/a^2+y^2/b^2=1 then

the area of the ellipse is = pi*a*b

Answer 9

yes

y^2/b^2=1-x^2/a^2

u can do integration of this function limits 2a to 0 and

the result can b multiplied by 4(as it would b area of one part)

and u will get pie*a*b

as u can c from this equation circle is special case of ellipse
when a=b

Answer 10

Area
The area enclosed by an ellipse is πab, where 'a' and 'b' are the semimajor and semiminor axes and π is Archimedes's constant. In the case of a circle where a = b, the expression reduces to the familiar πa2.