# How to calculate the area of an ellipse

An ellipse is defined as a locus of all the points in a plane.

## What is the Area of an Ellipse Calculator?

‘Cuemath’s Area of an Ellipse Calculator’ is an online tool that helps to calculate the area of an ellipse. Cuemath’s online Area of an Ellipse Calculator calculator helps you to calculate the area of an ellipse within a few seconds.

NOTE: Enter the values up to four digits only.

## How to Use the Area of an Ellipse Calculator?

• Step 1: Enter the length of the semi-major axis and the semi-minor axis in the given input box.
• Step 2: Click on the “Calculate” button to find the area of an ellipse.
• Step 3: Click on the “Reset” button to clear the fields and enter the new values.

## How to Find the Area of an Ellipse?

The area of an ellipse is defined as the amount of space enclosed within the boundary of an ellipse. It is measured in square units. The formula to calculate the area of an ellipse is given by

Area of an ellipse = π a b square units

Where ‘a’ is the length of the semi-major axis and ‘b’ is the length of the semi-minor axis.

### Solved Example:

Find the area of an ellipse whose length of the semi-major axis is 5 units and the length of the semi-minor axis is 9 units

### Solution:

Given: Semi-major axis(a) = 5 units and semi-minor axis(b) = 9 units

= 141.37 square units

Therefore, the area of an ellipse is 141.37 square units.

Similarly, you can try the calculator to find the area of an ellipse for the following:

• Length of the semi-major axis(a) = 7 units and semi-minor axis(b) = 9 units
• Length of the semi-major axis(a) = 11 units and semi-minor axis(b) = 14 units

The area of an ellipse is defined as the amount of region present inside it. Alternatively, the area of an ellipse is the total number of unit squares that can be fit in it. You might have observed many ellipse-shaped shapes in our daily lives, for example, a cricket ground, a badminton racket, orbits of planets, etc. Let us explore a bit more about this shape while discussing its area, and the formula for the area of the ellipse with solved examples in the following sections.

 1 What Is the Area of Ellipse? 2 Area of Ellipse Formula 3 Proof of Formula of Area of Ellipse 4 Area of Ellipse Using Integration 5 How to Find the Area of Ellipse? 6 FAQs on Area of Ellipse

## What Is the Area of Ellipse?

The area or region covered by the ellipse in two dimensions is defined as the area of an ellipse. The area of an ellipse is expressed in square units like in 2 , cm 2 , m 2 , yd 2 , ft 2 , etc. Ellipse is a 2-D shape obtained by connecting all the points which are at a constant distance from the two fixed points on the plane. The fixed points are called foci. F1 and F 2 are the two foci. As an ellipse is not a perfect circle, the distance from the center of the ellipse to the points on the circumference is not constant. So, an ellipse has two radii. The longest chord in the ellipse is called the major axis of the ellipse. The minor axis is the chord that is the perpendicular bisector to the major axis.

Ellipse is the locus of all the points, whose sum of distances from two fixed points on a plane is constant. For example, in the images given below the points P1, P2 are located in such a way that the sum of the distances of point P1 from the fixed points F1 and F2 is equal to the sum of the distances of point P2 from the fixed points F1 and F2. That means if we join all such points P1, P2, P3 , etc; we will get a shape called an ellipse.

• $$F_1, F_2$$ = Two fixed points on the plane
• $$P_1, P_2, P_3$$ . = Points forming locus of ellipse

## Area of Ellipse Formula

We can calculate the area of an ellipse using a general formula, given the lengths of the major and minor axis. The formula to find the area of an ellipse is given by,

Area of ellipse = π a b
where,

• a = length of semi-major axis
• b = length of semi-minor axis

## Proof of Formula of Area of Ellipse

Let E be an ellipse, with major axis of length 2a and minor axis of length 2b, aligned in a cartesian plane in the reduced form. Then, from the equation of ellipse in reduced form, we get,

Consider a circle, C, of radius ‘a’ whose center is at the origin. From the equation of circle formula with origin at center, we have,

We observe from the formulas that each ordinate of the ellipse is $$b/a$$ times the ordinate of the circle. The same thing is true for the vertical chords. We can thus also relate the areas of the two shapes as,

Area of ellipse = (b/a) × Area of circle

We know, area of a circle with the given equation, $$x^2 + y^2 = a^2$$, is: A = πa 2 , where ‘a’ is the radius of the circle.

⇒ Area of ellipse = (b/a)(πa 2 ) = πab

## Area of an Ellipse Using Integration

The formula of area of an ellipse can also be derived using integration. The general equation for an ellipse is given as, $$\dfrac>>+\dfrac>> = 1$$

We observe that the ellipse is divided into four quadrants. So, we can calculate the area of 1 quadrant and multiply by 4 to calculate the area of an ellipse.

Substituting, x = a sin t, we get, dx = a cos t . dt

⇒ x = 0 changes to t = 0 and x = a changes to t = π/2

• a = length of semi-major axis
• b = length of semi-minor axis

## How to Find the Area of Ellipse?

The steps that can be followed to calculate the area of an ellipse using the length of the major and minor axis are given as,

• Step 1: Find the distance from the farthest point on the ellipse from the center (‘a’ i.e., length of the semi-major axis).
• Step 2: Find the distance from the closest point on the ellipse from the center (‘b’ i.e., length of the semi-minor axis).
• Step 3: Multiply the product of a and b with π.
• Step 4: Mention the area in square units.

Important Notes

• Ellipse is the locus of all the points, whose sum of distances from two fixed points on a plane is constant.
• Area of ellipse = π a b
• Take the value of π as 3.14 or 22/7.
• Find the values of the semi-major axis (a) and the semi-minor axis (b) and then use the formula for the area of an ellipse.

## Solved Examples on Area of Ellipse

Example 1: The floor of a whispering gallery is constructed in the shape of an ellipse. Find its area, if the length of its major axis is 14 units and the length of its minor axis is 10 units.

Solution:

Given that,
Length of major axis = 14 units
⇒ length of semi-major axis is = (1/2) × 14 = 7 units.
Length of minor axis = 10 units
⇒ length of semi-minor axis is = (1/2) × 10 = 5 units.

Area of the ellipse is πab.

⇒ Area of whispering gallery = πab = π × 7 × 5 = (22/7) × 7 × 5 = 22 × 5 = 110 square units

Answer: Area of whispering gallery is 110 square units.

Example 2: Find the cross-sectional area of football with its semi-major axis as 5.5 inches and semi-minor axis as 3.5 inches using the area of an ellipse formula.

Solution:

American football is in the shape of an ellipse.

Length of semi-major axis = a = 5.5 inches

Length of semi-minor axis = b = 3.5 inches

The area of the ellipse is,

πab = π × 5.5 × 3.5 = (22/7) × 5.5 × 3.5 = 11 × 5.5 = 60.5 square inches

Answer: The cross-sectional area of football is 60.5 square inches.

In geometry, an is a two-dimensional flat elongated circle that is symmetrical along its shortest and longest diameters. An ellipse resembles an oval shape. In an ellipse, the longest diameter is known as the major axis, whereas the shortest diameter is known as the minor axis.

The distance of two points in the interior of an ellipse from a point on the ellipse is same as the distance of any other point on the ellipse from the same point. These points inside the ellipse are termed as foci. In this article, you will what an ellipse is, and how to find its area by using the area of an ellipse formula. But first see its few applications first.

Ellipses have multiple applications in the field of engineering, medicine, science, etc. For example, the planets revolve in their orbits which are elliptical in shape.

In an atom, it is believed that, electrons revolve around the nucleus in elliptical orbits.

The concept of ellipses is used in medicine for treatment of kidney stones (lithotripsy). Other real-world examples of elliptical shapes are the huge elliptical park in front of the White House in Washington DC and the St. Paul’s Cathedral building.

Up to this point, you have got an idea of what an ellipse is, let’s now proceed by looking at how to calculate area of an ellipse.

## How to Find the Area of an Ellipse?

To calculate the area of an ellipse, you need the measurements of both the major radius and minor radius.

### Area of an ellipse formula

The formula for area of an ellipse is given as:

Area of an ellipse = πr1r2

Where, π = 3.14, r1 and r2 are the minor and the major radii respectively.

Note: Minor radius = semi -minor axis (minor axis/2) and the major radius = Semi- major axis (major axis/2)

Let’s test our understanding of the area of an ellipse formula by solving a few example problems.

Example 1

What is the area of an ellipse whose minor and major radii are, 12 cm and 7 cm, respectively?

Area of an ellipse = πr1r2

Example 2

The major axis and minor axis of an ellipse are, 14 m and 12 m, respectively. What is the area of the ellipse?

Major axis = 14m ⇒ major radius, r2 =14/2 = 7 m

Minor axis = 12 m ⇒ minor radius, r1 = 12/2 = 6 m.

Area of an ellipse = πr1r2

Example 3

The area of an ellipse is 50.24 square yards. If the major radius of the ellipse is 6 yards more than the minor radius. Find the minor and major radii of the ellipse.

Area = 50.24 square yards

Let the minor radius = x

The major radius = x + 6

But, area of an ellipse = πr1r2

⇒50.24 = 3.14 * x *(x + 6)

⇒50.24 = 3.14x (x + 6)

By applying the distributive property of multiplication on the RHS, we get,

⇒50.24 = 3.14x 2 + 18.84x

Divide both sides by 3.14

⇒x 2 + 8x – 2x – 16 = 0

⇒ x (x + 8) – 2 (x + 8) = 0

Substitute x = 2 for the two equations of radii

The major radius = x + 6 ⇒ 8 yards

The minor radius = x = 2 yards

So, the major radius of the ellipse is 8 yards and the minor radius is 2 yards.

Example 4

Find the area of an ellipse whose radii area 50 ft and 30 ft respectively.

r1 = 30 ft and r2 = 50 ft

Area of an ellipse = πr1r2

Hence, the area of the ellipse is 4,710 ft 2 .

Example 5

Calculate the area of the ellipse shown below.

Area of an ellipse = πr1r2

Area of a semi – ellipse (h2)

A semi ellipse is a half an ellipse. Since we know the area of an ellipse as πr1r2, therefore, the area of a semi ellipse is half the area of an ellipse.

Area of a semi ellipse = ½ πr1r2

Example 6

Find the area of a semi – ellipse of radii 8 cm and 5 cm.

Area of a semi ellipse = ½ πr1r2

To calculate the area of ellipse you need to know the radius 1 and radius 2 of the ellipse.

1. r1 : (Datatype int): to represent the radius 1 of the ellipse.

2. r2 : (Datatype int): to represent the radius 2 of the ellipse.

3. PI : (Datatype float): static variable to represent constant PI value up to 3 decimal places i.e 3.142

5. area_of_ellipse : (Datatype long): to store area of ellipse calculated.

Volume of Ellipse : ПЂ x r1 x r2

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Given an ellipse with a semi-major axis of length a and semi-minor axis of length b. The task is to find the area of an ellipse.
In mathematics, an ellipse is a curve in a plane surrounding by two focal points such that the sum of the distances to the two focal points is constant for every point on the curve or we can say that it is a generalization of the circle.

Important points related to Ellipse:

• Center: A point inside the ellipse which is the midpoint of the line segment which links the two foci. In other words, it is the intersection of minor and major axes.
• Major Axis: The longest diameter of an ellipse is termed as the major axis.
• Minor Axis: The shortest diameter of an ellipse is termed as minor axis.
• Chord: A line segment that links any two points on an ellipse.
• Focus: These are the two fixed points that define an ellipse.
• Latus Rectum: The line segments which passes through the focus of an ellipse and perpendicular to the major axis of an ellipse, is called as the latus rectum of an ellipse.

Area of an ellipse: The formula to find the area of an ellipse is given below:

where a and b are the semi-major axis and semi-minor axis respectively and 3.142 is the value of π.
Examples:

An oval is a geometric shape that looks like an elongated circle. Another name for an oval is an ‘ellipse‘. You might be asked to recognize an oval shape in primary school, secondary school and at university. The area of an oval is the amount of space contained within it. You can calculate the area of an oval by following a simple mathematical equation.

• Tape measurer or ruler
• Calculator

Measure the length of the ellipse from ‘A’ to ‘B’ and divide it by 2. This is known as ‘semi-major axis’ or ‘radius 1’.

Measure the width of the oval across its centremost point. The centre point is the middle point between ‘A’ and ‘B’. Once you measure the width of the oval, divide this value by 2. This is known as the ‘semi-minor axis’ or ‘radius 2’.

Multiply half of the width by half of the length. For example, if the length of the oval is 10 cm and the width of 6 cm, you would multiply 5 by 3 to get 15 cm.

Multiply this figure in Step 3 by pi, which is 3.14. Using the previous example, you would multiply 15 by 3.14 to get 47.1 cm², which is the area of the ellipse.

If you want to read similar articles to How To Calculate The Area Of An Oval, we recommend you visit our Learning category.

### Using Green’s theorem to find area

#### Suggested background

• The idea behind Green’s theorem
• Double integrals as area

Typically we use Green’s theorem as an alternative way to calculate a line integral $\dlint$. If, for example, we are in two dimension, $\dlc$ is a simple closed curve, and $\dlvf(x,y)$ is defined everywhere inside $\dlc$, we can use Green’s theorem to convert the line integral into to double integral. Instead of calculating line integral $\dlint$ directly, we calculate the double integral \begin \iint_\dlr \left(\pdiff<\dlvfc_2> – \pdiff<\dlvfc_1>\right) dA \end

Can we use Green’s theorem to go the other direction? If we are given a double integral, can we use Green’s theorem to convert the double integral into a line integral and calculate the line integral? If we are given the double integral \begin \iint_\dlr f(x,y) dA, \end we can use Green’s theorem only if there happens to be a vector field $\dlvf(x,y)$ so that \begin f(x,y) = \pdiff<\dlvfc_2> – \pdiff<\dlvfc_1>. \end However, we haven’t learned any method to find such a vector field $\dlvf$. So, we aren’t likely to use Green’s theorem in this direction very often.

There is one important exception to this rule, however, and that is when we are using a double integral to calculate the area of a region $\dlr$. The area of a region $\dlr$ is equal to the double integral of $f(x,y)=1$ over $\dlr$: \begin \text

= \iint_\dlr dA = \iint_\dlr 1\, dA. \end If $f(x,y) =1$, it is easy to find a vector field $\dlvf$ so that \begin \pdiff<\dlvfc_2> – \pdiff<\dlvfc_1> = f(x,y) = 1. \end There are many such vector fields $\dlvf$, but we’ll pick the vector field $\dlvf(x,y) = (-y/2, x/2)$. You can confirm that indeed $\displaystyle \pdiff<\dlvfc_2> – \pdiff<\dlvfc_1> =1$.

In summary, if $\dlc$ is a counterclockwise oriented simple closed curve that bounds a region where you can apply Green’s theorem, the area of the region $\dlr$ bounded by $\dlc = \partial \dlr$ is \begin \text

= \dlint = \frac<1> <2>\int_\dlc x\,dy – y\,dx, \end where $\dlvf(x,y) = (-y/2,x/2)$.

#### Example 1

Use Green’s Theorem to calculate the area of the disk $\dlr$ of radius $r$ defined by $x^2+y^2 \le r^2$.

Solution: Since we know the area of the disk of radius $r$ is $\pi r^2$, we better get $\pi r^2$ for our answer.

The boundary of $\dlr$ is the circle of radius $r$. We can parametrized it in a counterclockwise orientation using \begin \dllp(t) = (r\cos t, r\sin t), \quad 0 \le t \le 2\pi. \end Then \begin \dllp'(t) = (-r\sin t, r\cos t), \end and, by Green’s theorem, \begin \text

\dlr &= \iint dA\\ &= \frac<1> <2>\int_\dlc x\,dy – y\,dx\\ &= \frac<1> <2>\int_0^ <2\pi>[(r\cos t)(r\cos t) – (r\sin t)(-r\sin t)]dt \\ &= \frac<1> <2>\int_0^ <2\pi>r^2 (\sin^2t+\cos^2t) dt = \frac<2>\int_0^ <2\pi>dt = \pi r^2. \end Thankfully, our answer agrees with what we know it should be.

#### Example 2

Calculate the area of the region $D$ bounded by the curve $\dlc$ parametrized by $\dllp(t)=\sin 2t\,\vc +\sin t\,\vc$ for $0 \le t \le \pi$. The region and curve are illustrated by the below applet.

Area inside sinusoidal curve. The curve $\dlc$ parameterized by $\dllp(t)=(\sin 2t,\sin t)$ for $0 \le t \le \pi$ is the counterclockwise oriented boundary of a region $D$, shown shaded in blue. As you specify $t$ by dragging the green point on the slider, the red point traces out the curve $\dllp(t)$. Alternatively, you can drag the red point around the curve, and the green point on the slider indicates the corresponding value of $t$. One can calculate the area of $D$ using Green’s theorem and the vector field $\dlvf(x,y)=(-y,x)/2$.

Solution: We’ll use Green’s theorem to calculate the area bounded by the curve. Since $\dlc$ is a counterclockwise oriented boundary of $D$, the area is just the line integral of the vector field $\dlvf(x,y) = \frac<1><2>(-y,x)$ around the curve $\dlc$ parametrized by $\dllp(t)$.

To integrate around $\dlc$, we need to calculate the derivative of the parametrization $\dllp'(t)=2\cos 2t \,\vc+\cos t\,\vc$. Then, by Green’s theorem, we turn the double integral defining the area into line integral and calculate the area. \begin \text

\dlr &= \iint dA\\ &= \dlint = \plint<0><\pi><\dlvf><\dllp>\\ &= \frac<1> <2>\int_0^ <\pi>( -\sin t, \sin 2t) \cdot ( 2\cos 2t,\cos t)dt\\ &= \frac<1> <2>\int_0^ <\pi>( -\sin t, 2 \sin t \cos t ) \cdot (2\cos^2 t – 2\sin^2 t,\cos t)dt\\ &= \int_0^ <\pi>\sin^3t dt =\int_0^ <\pi>(\sin t-\sin t\,\cos^2t)\, dt\\ &= \Bigl[-\cos t -\frac<1><3>\cos^3 t\Bigr]_0^ <\pi>= 1+\frac<1><3>=\frac<4> <3>\end

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Find the area of an ellipse using integrals and calculus.

Problem : Find the area of an ellipse with half axes a and b.

Solution to the problem:
The equation of the ellipse shown above may be written in the form
x 2 / a 2 + y 2 / b 2 = 1
Since the ellipse is symmetric with respect to the x and y axes, we can find the area of one quarter and multiply by 4 in order to obtain the total area.
Solve the above equation for y
y =

b √ [ 1 – x 2 / a 2 ]
The upper part of the ellipse (y positive) is given by
y = b √ [ 1 – x 2 / a 2 ]
We now use integrals to find the area of the upper right quarter of the ellipse as follows
(1 / 4) Area of ellipse = 0 a b √ [ 1 – x 2 / a 2 ] dx
We now make the substitution sin t = x / a so that dx = a cos t dt and the area is given by
(1 / 4) Area of ellipse = 0 π/2 a b ( √ [ 1 – sin 2 t ] ) cos t dt
√ [ 1 – sin 2 t ] = cos t since t varies from 0 to π/2, hence
(1 / 4) Area of ellipse = 0 π/2 a b cos 2 t dt
Use the trigonometric identity cos 2 t = ( cos 2t + 1 ) / 2 to linearize the integrand;
(1 / 4) Area of ellipse = 0 π/2 a b ( cos 2t + 1 ) / 2 dt
Evaluate the integral
(1 / 4) Area of ellipse = (1/2) b a [ (1/2) sin 2t + t ] 0 π/2
= (1/4) π a b
Obtain the total area of the ellipse by multiplying by 4
Area of ellipse = 4 * (1/4) π a b = π a b More references on integrals and their applications in calculus.