How to calculate area of an ellipse based on its formula?












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How can I determine the area of a half-ellipse if all that is given is $y = sqrt{1-n^2x^2}$? I have tried both geometry and calculus, but without convincing results…



Thank you






calculus geometry area







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  • 1




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    This equation corresponds to half of an ellipse if I am not wrong ...
    $endgroup$
    – Damien
    Dec 11 '18 at 11:41






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    The area of an ellipse is equal to $pi ab$, $a,b$ being the lengths of the semi-major and semi-minor axes of the ellipse.
    $endgroup$
    – Shubham Johri
    Dec 11 '18 at 11:42










  • $begingroup$
    @damien you are right: I have edited accordingly...
    $endgroup$
    – Pregunto
    Dec 11 '18 at 11:43










  • $begingroup$
    Try to manipulate the formula of the ellipse to make apparent its two lengths, i.e. try to find its canonic form
    $endgroup$
    – Damien
    Dec 11 '18 at 11:49
















1












$begingroup$


How can I determine the area of a half-ellipse if all that is given is $y = sqrt{1-n^2x^2}$? I have tried both geometry and calculus, but without convincing results…



Thank you






calculus geometry area







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    This equation corresponds to half of an ellipse if I am not wrong ...
    $endgroup$
    – Damien
    Dec 11 '18 at 11:41






  • 1




    $begingroup$
    The area of an ellipse is equal to $pi ab$, $a,b$ being the lengths of the semi-major and semi-minor axes of the ellipse.
    $endgroup$
    – Shubham Johri
    Dec 11 '18 at 11:42










  • $begingroup$
    @damien you are right: I have edited accordingly...
    $endgroup$
    – Pregunto
    Dec 11 '18 at 11:43










  • $begingroup$
    Try to manipulate the formula of the ellipse to make apparent its two lengths, i.e. try to find its canonic form
    $endgroup$
    – Damien
    Dec 11 '18 at 11:49














1












1








1





$begingroup$


How can I determine the area of a half-ellipse if all that is given is $y = sqrt{1-n^2x^2}$? I have tried both geometry and calculus, but without convincing results…



Thank you






calculus geometry area







share|cite|improve this question











$endgroup$




How can I determine the area of a half-ellipse if all that is given is $y = sqrt{1-n^2x^2}$? I have tried both geometry and calculus, but without convincing results…



Thank you






calculus geometry area




calculus geometry area






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 11 '18 at 11:43







Pregunto

















asked Dec 11 '18 at 11:31









PreguntoPregunto

1476




1476








  • 1




    $begingroup$
    This equation corresponds to half of an ellipse if I am not wrong ...
    $endgroup$
    – Damien
    Dec 11 '18 at 11:41






  • 1




    $begingroup$
    The area of an ellipse is equal to $pi ab$, $a,b$ being the lengths of the semi-major and semi-minor axes of the ellipse.
    $endgroup$
    – Shubham Johri
    Dec 11 '18 at 11:42










  • $begingroup$
    @damien you are right: I have edited accordingly...
    $endgroup$
    – Pregunto
    Dec 11 '18 at 11:43










  • $begingroup$
    Try to manipulate the formula of the ellipse to make apparent its two lengths, i.e. try to find its canonic form
    $endgroup$
    – Damien
    Dec 11 '18 at 11:49














  • 1




    $begingroup$
    This equation corresponds to half of an ellipse if I am not wrong ...
    $endgroup$
    – Damien
    Dec 11 '18 at 11:41






  • 1




    $begingroup$
    The area of an ellipse is equal to $pi ab$, $a,b$ being the lengths of the semi-major and semi-minor axes of the ellipse.
    $endgroup$
    – Shubham Johri
    Dec 11 '18 at 11:42










  • $begingroup$
    @damien you are right: I have edited accordingly...
    $endgroup$
    – Pregunto
    Dec 11 '18 at 11:43










  • $begingroup$
    Try to manipulate the formula of the ellipse to make apparent its two lengths, i.e. try to find its canonic form
    $endgroup$
    – Damien
    Dec 11 '18 at 11:49








1




1




$begingroup$
This equation corresponds to half of an ellipse if I am not wrong ...
$endgroup$
– Damien
Dec 11 '18 at 11:41




$begingroup$
This equation corresponds to half of an ellipse if I am not wrong ...
$endgroup$
– Damien
Dec 11 '18 at 11:41




1




1




$begingroup$
The area of an ellipse is equal to $pi ab$, $a,b$ being the lengths of the semi-major and semi-minor axes of the ellipse.
$endgroup$
– Shubham Johri
Dec 11 '18 at 11:42




$begingroup$
The area of an ellipse is equal to $pi ab$, $a,b$ being the lengths of the semi-major and semi-minor axes of the ellipse.
$endgroup$
– Shubham Johri
Dec 11 '18 at 11:42












$begingroup$
@damien you are right: I have edited accordingly...
$endgroup$
– Pregunto
Dec 11 '18 at 11:43




$begingroup$
@damien you are right: I have edited accordingly...
$endgroup$
– Pregunto
Dec 11 '18 at 11:43












$begingroup$
Try to manipulate the formula of the ellipse to make apparent its two lengths, i.e. try to find its canonic form
$endgroup$
– Damien
Dec 11 '18 at 11:49




$begingroup$
Try to manipulate the formula of the ellipse to make apparent its two lengths, i.e. try to find its canonic form
$endgroup$
– Damien
Dec 11 '18 at 11:49










2 Answers
2






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1












$begingroup$

Figure



The area of the half-ellipse is given by the integral $displaystyleint_{-1/n}^{1/n} ydx=2int_0^{1/n}sqrt{1-n^2x^2} dx$



There is a formula for solving integrals of the form $intsqrt{a^2-x^2} dx$, but we will not use it directly. Instead, substitute $nx=sinthetaimplies dx=frac{costheta}ndtheta$



$implies A=frac2nint_0^{pi/2}cos^2theta dtheta=frac1nint_0^{pi/2}(cos2theta+1) dtheta=fracpi{2n}$



Can you now show similarly that the area enclosed by the ellipse $frac{x^2}{a^2}+frac{y^2}{b^2}=1$ is equal to $pi ab$?






share|cite|improve this answer









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    0












    $begingroup$

    With the use of generalized polar coordinates
    begin{aligned}x&=arcos t\
    y&=brsin tend{aligned}
    where $a=frac 1n,; b=1,; t in [0,pi]; text{and}; rin [0,1]$ in the given case. The Jacobian is ${rover n}$ and the area
    $$cal{A}=int_0^{pi} int_0^1 1cdot {rover n};dr ;dt=frac{pi}{2n}$$






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      Figure



      The area of the half-ellipse is given by the integral $displaystyleint_{-1/n}^{1/n} ydx=2int_0^{1/n}sqrt{1-n^2x^2} dx$



      There is a formula for solving integrals of the form $intsqrt{a^2-x^2} dx$, but we will not use it directly. Instead, substitute $nx=sinthetaimplies dx=frac{costheta}ndtheta$



      $implies A=frac2nint_0^{pi/2}cos^2theta dtheta=frac1nint_0^{pi/2}(cos2theta+1) dtheta=fracpi{2n}$



      Can you now show similarly that the area enclosed by the ellipse $frac{x^2}{a^2}+frac{y^2}{b^2}=1$ is equal to $pi ab$?






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Figure



        The area of the half-ellipse is given by the integral $displaystyleint_{-1/n}^{1/n} ydx=2int_0^{1/n}sqrt{1-n^2x^2} dx$



        There is a formula for solving integrals of the form $intsqrt{a^2-x^2} dx$, but we will not use it directly. Instead, substitute $nx=sinthetaimplies dx=frac{costheta}ndtheta$



        $implies A=frac2nint_0^{pi/2}cos^2theta dtheta=frac1nint_0^{pi/2}(cos2theta+1) dtheta=fracpi{2n}$



        Can you now show similarly that the area enclosed by the ellipse $frac{x^2}{a^2}+frac{y^2}{b^2}=1$ is equal to $pi ab$?






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Figure



          The area of the half-ellipse is given by the integral $displaystyleint_{-1/n}^{1/n} ydx=2int_0^{1/n}sqrt{1-n^2x^2} dx$



          There is a formula for solving integrals of the form $intsqrt{a^2-x^2} dx$, but we will not use it directly. Instead, substitute $nx=sinthetaimplies dx=frac{costheta}ndtheta$



          $implies A=frac2nint_0^{pi/2}cos^2theta dtheta=frac1nint_0^{pi/2}(cos2theta+1) dtheta=fracpi{2n}$



          Can you now show similarly that the area enclosed by the ellipse $frac{x^2}{a^2}+frac{y^2}{b^2}=1$ is equal to $pi ab$?






          share|cite|improve this answer









          $endgroup$



          Figure



          The area of the half-ellipse is given by the integral $displaystyleint_{-1/n}^{1/n} ydx=2int_0^{1/n}sqrt{1-n^2x^2} dx$



          There is a formula for solving integrals of the form $intsqrt{a^2-x^2} dx$, but we will not use it directly. Instead, substitute $nx=sinthetaimplies dx=frac{costheta}ndtheta$



          $implies A=frac2nint_0^{pi/2}cos^2theta dtheta=frac1nint_0^{pi/2}(cos2theta+1) dtheta=fracpi{2n}$



          Can you now show similarly that the area enclosed by the ellipse $frac{x^2}{a^2}+frac{y^2}{b^2}=1$ is equal to $pi ab$?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 11 '18 at 11:55









          Shubham JohriShubham Johri

          5,477818




          5,477818























              0












              $begingroup$

              With the use of generalized polar coordinates
              begin{aligned}x&=arcos t\
              y&=brsin tend{aligned}
              where $a=frac 1n,; b=1,; t in [0,pi]; text{and}; rin [0,1]$ in the given case. The Jacobian is ${rover n}$ and the area
              $$cal{A}=int_0^{pi} int_0^1 1cdot {rover n};dr ;dt=frac{pi}{2n}$$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                With the use of generalized polar coordinates
                begin{aligned}x&=arcos t\
                y&=brsin tend{aligned}
                where $a=frac 1n,; b=1,; t in [0,pi]; text{and}; rin [0,1]$ in the given case. The Jacobian is ${rover n}$ and the area
                $$cal{A}=int_0^{pi} int_0^1 1cdot {rover n};dr ;dt=frac{pi}{2n}$$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  With the use of generalized polar coordinates
                  begin{aligned}x&=arcos t\
                  y&=brsin tend{aligned}
                  where $a=frac 1n,; b=1,; t in [0,pi]; text{and}; rin [0,1]$ in the given case. The Jacobian is ${rover n}$ and the area
                  $$cal{A}=int_0^{pi} int_0^1 1cdot {rover n};dr ;dt=frac{pi}{2n}$$






                  share|cite|improve this answer









                  $endgroup$



                  With the use of generalized polar coordinates
                  begin{aligned}x&=arcos t\
                  y&=brsin tend{aligned}
                  where $a=frac 1n,; b=1,; t in [0,pi]; text{and}; rin [0,1]$ in the given case. The Jacobian is ${rover n}$ and the area
                  $$cal{A}=int_0^{pi} int_0^1 1cdot {rover n};dr ;dt=frac{pi}{2n}$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 11 '18 at 12:29









                  user376343user376343

                  3,9584829




                  3,9584829






























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