# Thread: Area of ellipse in terms of Integral

1. ## Area of ellipse in terms of Integral

Hello,
If 0<a<b, we have

$\frac12 \displaystyle\int_0^{2\pi}\frac{d\theta}{(b+a*cos( \theta))^2}d\theta=\frac{\pi b}{(b^2-a^2)^{\frac32}}$

The LHS of this equation is the area of ellipse written as $\frac12\displaystyle\int_0^{2\pi} \rho(\theta)^2d\theta$

My question is what are $\rho$ and $\theta?$ I know to compute an area of ellipse using calculus with trigonometric substitutions. But I don't know this integral used as area of ellipse.

2. ## Re: Area of ellipse in terms of Integral

This is the equation of an ellipse with center (-C,0) and foci at (-2C,0) and (0,0)
so that its equation in Cartesian coordinates would be

$$\sqrt{x^2+y^2}+\sqrt{(x+2C)^2+y^2}=2A$$

where $0<C<A$

and in polar coordinates

$$(x+2C)^2+y^2=(2A-\rho )^2$$

$$\rho ^2+4C x+4C^2=4A^2-4A \rho +\rho ^2$$

$$A \rho + C x=A^2-C^2$$

$$A \rho +C \rho \cos \theta =A^2-C^2$$

3. ## Re: Area of ellipse in terms of Integral

Hello,
It seems that the vertices of the ellipse in your reply are (x,0)and (-x-2C,0) because center of the ellipse is (-C,0). The major axis of the ellipse is parallel to x-axis because y-co-ordinates of the vertices and the foci are same.

The standard of such ellipse not centered at origin is $\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$ with center (h,k). The equation which you have given is length of major axis which is 2A. C is the length of semi-minor axis.

Is this correct? If yes, I shall ask my further querries in regard to polar co-ordinates of this ellipse.

4. ## Re: Area of ellipse in terms of Integral

Is $\rho=\sqrt{(x^2+y^2)}?$

5. ## Re: Area of ellipse in terms of Integral

Originally Posted by WMDhamnekar
Hello,
It seems that the vertices of the ellipse in your reply are (x,0)and (-x-2C,0) because center of the ellipse is (-C,0). The major axis of the ellipse is parallel to x-axis because y-co-ordinates of the vertices and the foci are same.

The standard form of equation of such ellipse not centered at origin is $\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$ with center (h,k). The equation which you have given is length of major axis which is 2A. C is the length of semi-minor axis.

Is this correct? If yes, I shall ask my further querries in regard to polar co-ordinates of this ellipse.
Hello,
I know to compute an area of ellipse using double integrals (Change of variables and Jacobian) also.

6. ## Re: Area of ellipse in terms of Integral

Originally Posted by Idea
This is the equation of an ellipse with center (-C,0) and foci at (-2C,0) and (0,0)
so that its equation in Cartesian coordinates would be

$$\sqrt{x^2+y^2}+\sqrt{(x+2C)^2+y^2}=2A$$

where $0<C<A$

and in polar coordinates

$$(x+2C)^2+y^2=(2A-\rho )^2$$

$$\rho ^2+4C x+4C^2=4A^2-4A \rho +\rho ^2$$

$$A \rho + C x=A^2-C^2$$

$$A \rho +C \rho \cos \theta =A^2-C^2$$
Hello,

How does your last step is = $\frac12\displaystyle\int_0^{2\pi}\rho(\theta)^2 d\theta?$

7. ## Re: Area of ellipse in terms of Integral

Vertices of the ellipse are at $(-A-C,0)$ and $(A-C,0)$

Now, the equation in polar form from my previous post

$$\rho =\frac{A^2-C^2}{A+C \cos \theta }=\frac{1}{\frac{A}{A^2-C^2}+\frac{C}{A^2-C^2}\cos \theta }=\frac{1}{b+ a \cos \theta }$$

where

$b=\frac{A}{A^2-C^2}$ and $a=\frac{C}{A^2-C^2}$

semi-major axis $= A$

semi-minor axis $=\sqrt{A^2-C^2}$

area $=\pi A\sqrt{A^2-C^2}$

A simple calculation shows that

$$b^2-a^2=\frac{1}{A^2-C^2}$$

therefore the area equals

$\pi b\left(A^2-C^2\right)\sqrt{A^2-C^2}=\pi b \left(A^2-C^2\right)^{3/2}=\frac{\pi b}{\left(b^2-a^2\right)^{3/2}}$

8. ## Re: Area of ellipse in terms of Integral

Originally Posted by Idea
Vertices of the ellipse are at $(-A-C,0)$ and $(A-C,0)$

Now, the equation in polar form from my previous post

$$\rho =\frac{A^2-C^2}{A+C \cos \theta }=\frac{1}{\frac{A}{A^2-C^2}+\frac{C}{A^2-C^2}\cos \theta }=\frac{1}{b+ a \cos \theta }$$

where

$b=\frac{A}{A^2-C^2}$ and $a=\frac{C}{A^2-C^2}$

semi-major axis $= A$

semi-minor axis $=\sqrt{A^2-C^2}$

area $=\pi A\sqrt{A^2-C^2}$

A simple calculation shows that

$$b^2-a^2=\frac{1}{A^2-C^2}$$

therefore the area equals

$\pi b\left(A^2-C^2\right)\sqrt{A^2-C^2}=\pi b \left(A^2-C^2\right)^{3/2}=\frac{\pi b}{\left(b^2-a^2\right)^{3/2}}$
Hello,
Your answer is correct. But I didn't understand how did you compute the length of semi-minor axis? I understood the answer to this question. (Length of semi-major axis-length of foci to center of ellipse)^2=(Length of semi-minor axis)^2

9. ## Re: Area of ellipse in terms of Integral

in the drawing there is a right triangle with hypotenuse $A=5$

$C=4$ is the focal distance from the center of the ellipse to the focus at $(0,0)$

the semi-minor axis has length $\sqrt {A^2-C^2}$

10. ## Re: Area of ellipse in terms of Integral

Originally Posted by Vinod
Is $\rho=\sqrt{(x^2+y^2)}?$
yes, the polar $\rho$ expressed in terms of the Cartesian coordinates $x,y$

11. ## Re: Area of ellipse in terms of Integral

Originally Posted by WMDhamnekar
Hello,
Your answer is correct. But I didn't understand how did you compute the length of semi-minor axis? I understood the answer to this question. (Length of semi-major axis-length of foci to center of ellipse)^2=(Length of semi-minor axis)^2
Please read ' Length of semi-major axis^2 - Length of foci to center of ellipse^2=Length of semi-minor axis^2

12. ## Re: Area of ellipse in terms of Integral

Originally Posted by Idea
Vertices of the ellipse are at $(-A-C,0)$ and $(A-C,0)$

Now, the equation in polar form from my previous post

$$\rho =\frac{A^2-C^2}{A+C \cos \theta }=\frac{1}{\frac{A}{A^2-C^2}+\frac{C}{A^2-C^2}\cos \theta }=\frac{1}{b+ a \cos \theta }$$

where

$b=\frac{A}{A^2-C^2}$ and $a=\frac{C}{A^2-C^2}$

semi-major axis $= A$

semi-minor axis $=\sqrt{A^2-C^2}$

area $=\pi A\sqrt{A^2-C^2}$

A simple calculation shows that

$$b^2-a^2=\frac{1}{A^2-C^2}$$

therefore the area equals

$\pi b\left(A^2-C^2\right)\sqrt{A^2-C^2}=\pi b \left(A^2-C^2\right)^{3/2}=\frac{\pi b}{\left(b^2-a^2\right)^{3/2}}$
Hello,

If $\rho(\theta)$ is written as $\frac{1}{b+a*cos(\theta)}$, How to write $\rho(r*e^{i*\theta})$, where $z=a+b*i$ or $z=r*e^{(i*\theta)}$

13. ## Re: Area of ellipse in terms of Integral

Originally Posted by WMDhamnekar
Hello,

If $\rho(\theta)$ is written as $\frac{1}{b+a*cos(\theta)}$, How to write $\rho(r*e^{i*\theta})$, where $z=a+b*i$ or $z=r*e^{(i*\theta)}$
Fisrt, why would you want to?

Second, $\displaystyle \rho = \dfrac{1}{a + b cos( \theta ) } \to \rho e^{ i \theta }$ only works for complex numbers. Nothing in this problem deals with complex numbers so why did you bring this up? Did you give us the full problem?

-Dan