Results 1 to 13 of 13
Like Tree4Thanks
  • 1 Post By Idea
  • 1 Post By Idea
  • 1 Post By Idea
  • 1 Post By Idea

Thread: Area of ellipse in terms of Integral

  1. #1
    Newbie
    Joined
    May 2019
    From
    Mumbai (Bombay),Maharashtra,India
    Posts
    15

    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.

    If any member knows the correct answer, he/she may reply with correct answer.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Jun 2013
    From
    Lebanon
    Posts
    1,039
    Thanks
    530

    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

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

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

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

    $$A \rho +C \rho \cos \theta =A^2-C^2$$
    Thanks from WMDhamnekar
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    May 2019
    From
    Mumbai (Bombay),Maharashtra,India
    Posts
    15

    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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member Vinod's Avatar
    Joined
    Sep 2011
    From
    Mumbai (Bombay),Maharashtra,India
    Posts
    392
    Thanks
    7

    Re: Area of ellipse in terms of Integral

    Is $\rho=\sqrt{(x^2+y^2)}?$
    Last edited by Vinod; Sep 8th 2019 at 04:01 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    May 2019
    From
    Mumbai (Bombay),Maharashtra,India
    Posts
    15

    Re: Area of ellipse in terms of Integral

    Quote Originally Posted by WMDhamnekar View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    May 2019
    From
    Mumbai (Bombay),Maharashtra,India
    Posts
    15

    Re: Area of ellipse in terms of Integral

    Quote Originally Posted by Idea View Post
    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 = $\frac12\displaystyle\int_0^{2\pi}\rho(\theta)^2 d\theta?$
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Jun 2013
    From
    Lebanon
    Posts
    1,039
    Thanks
    530

    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

    $\pi b\left(A^2-C^2\right)\sqrt{A^2-C^2}=\pi b \left(A^2-C^2\right)^{3/2}=\frac{\pi b}{\left(b^2-a^2\right)^{3/2}}$
    Last edited by Idea; Sep 8th 2019 at 06:20 AM.
    Thanks from WMDhamnekar
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    May 2019
    From
    Mumbai (Bombay),Maharashtra,India
    Posts
    15

    Re: Area of ellipse in terms of Integral

    Quote Originally Posted by Idea View Post
    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
    Last edited by WMDhamnekar; Sep 8th 2019 at 07:24 AM.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    Joined
    Jun 2013
    From
    Lebanon
    Posts
    1,039
    Thanks
    530

    Re: Area of ellipse in terms of Integral

    Area of ellipse in terms of Integral-ellipse.gif

    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 $\sqrt {A^2-C^2}$
    Thanks from WMDhamnekar
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    Joined
    Jun 2013
    From
    Lebanon
    Posts
    1,039
    Thanks
    530

    Re: Area of ellipse in terms of Integral

    Quote Originally Posted by Vinod View Post
    Is $\rho=\sqrt{(x^2+y^2)}?$
    yes, the polar $\rho$ expressed in terms of the Cartesian coordinates $x,y$
    Thanks from Vinod
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Newbie
    Joined
    May 2019
    From
    Mumbai (Bombay),Maharashtra,India
    Posts
    15

    Re: Area of ellipse in terms of Integral

    Quote Originally Posted by WMDhamnekar View Post
    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
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Newbie
    Joined
    May 2019
    From
    Mumbai (Bombay),Maharashtra,India
    Posts
    15

    Re: Area of ellipse in terms of Integral

    Quote Originally Posted by Idea View Post
    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 $z=r*e^{(i*\theta)}$
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    11,499
    Thanks
    901
    Awards
    1

    Re: Area of ellipse in terms of Integral

    Quote Originally Posted by WMDhamnekar View Post
    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
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Area of ellipse in terms of Integral
    Posted in the Calculus Forum
    Replies: 0
    Last Post: Sep 6th 2019, 12:15 AM
  2. Replies: 10
    Last Post: Jan 1st 2012, 12:59 AM
  3. Area of an ellipse
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Nov 26th 2010, 10:49 AM
  4. express the area in terms of an integral
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Aug 9th 2010, 01:16 PM
  5. Area of an Ellipse
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: Oct 1st 2008, 10:36 PM

/mathhelpforum @mathhelpforum