Proof of this integration shortcut: $int_a^b frac{dx}{sqrt{(x-a)(b-x)}}=pi$












9












$begingroup$


I came across this as one of the shortcuts in my textbook without any proof.

When $bgt a$,




$$intlimits_a^b dfrac{dx}{sqrt{(x-a)(b-x)}}=pi$$





My attempt :



I notice that the the denominator is $0$ at both the bounds. I thought of substituting $x=a+(b-a)t$ so that the integral becomes
$$intlimits_0^1 dfrac{dt}{sqrt{t(1-t)}}$$



This doesn't look simple, but I'm wondering if the answer can be seen using symmetry/geometry ?






calculus integration definite-integrals







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    It's better to reduce to the $(a,b)=(-1,1)$ case rather than the $(0,1)$ case.
    $endgroup$
    – Lord Shark the Unknown
    Jul 16 '18 at 17:59










  • $begingroup$
    @InterstellarProbe here is an example
    $endgroup$
    – rsadhvika
    Jul 16 '18 at 18:04












  • $begingroup$
    @InterstellarProbe So much the worse for Wolfie!
    $endgroup$
    – Lord Shark the Unknown
    Jul 16 '18 at 18:06






  • 1




    $begingroup$
    $intlimits_0^1 dfrac{dt}{sqrt{t(1-t)}}=beta(dfrac14,dfrac14)=pi$
    $endgroup$
    – Nosrati
    Jul 16 '18 at 18:16






  • 1




    $begingroup$
    maybe try the substitution $t=cos theta$ or $t=sintheta$? I haven't tried it but the denominator would simplify
    $endgroup$
    – usr0192
    Jul 16 '18 at 18:18
















9












$begingroup$


I came across this as one of the shortcuts in my textbook without any proof.

When $bgt a$,




$$intlimits_a^b dfrac{dx}{sqrt{(x-a)(b-x)}}=pi$$





My attempt :



I notice that the the denominator is $0$ at both the bounds. I thought of substituting $x=a+(b-a)t$ so that the integral becomes
$$intlimits_0^1 dfrac{dt}{sqrt{t(1-t)}}$$



This doesn't look simple, but I'm wondering if the answer can be seen using symmetry/geometry ?






calculus integration definite-integrals







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    It's better to reduce to the $(a,b)=(-1,1)$ case rather than the $(0,1)$ case.
    $endgroup$
    – Lord Shark the Unknown
    Jul 16 '18 at 17:59










  • $begingroup$
    @InterstellarProbe here is an example
    $endgroup$
    – rsadhvika
    Jul 16 '18 at 18:04












  • $begingroup$
    @InterstellarProbe So much the worse for Wolfie!
    $endgroup$
    – Lord Shark the Unknown
    Jul 16 '18 at 18:06






  • 1




    $begingroup$
    $intlimits_0^1 dfrac{dt}{sqrt{t(1-t)}}=beta(dfrac14,dfrac14)=pi$
    $endgroup$
    – Nosrati
    Jul 16 '18 at 18:16






  • 1




    $begingroup$
    maybe try the substitution $t=cos theta$ or $t=sintheta$? I haven't tried it but the denominator would simplify
    $endgroup$
    – usr0192
    Jul 16 '18 at 18:18














9












9








9


0



$begingroup$


I came across this as one of the shortcuts in my textbook without any proof.

When $bgt a$,




$$intlimits_a^b dfrac{dx}{sqrt{(x-a)(b-x)}}=pi$$





My attempt :



I notice that the the denominator is $0$ at both the bounds. I thought of substituting $x=a+(b-a)t$ so that the integral becomes
$$intlimits_0^1 dfrac{dt}{sqrt{t(1-t)}}$$



This doesn't look simple, but I'm wondering if the answer can be seen using symmetry/geometry ?






calculus integration definite-integrals







share|cite|improve this question











$endgroup$




I came across this as one of the shortcuts in my textbook without any proof.

When $bgt a$,




$$intlimits_a^b dfrac{dx}{sqrt{(x-a)(b-x)}}=pi$$





My attempt :



I notice that the the denominator is $0$ at both the bounds. I thought of substituting $x=a+(b-a)t$ so that the integral becomes
$$intlimits_0^1 dfrac{dt}{sqrt{t(1-t)}}$$



This doesn't look simple, but I'm wondering if the answer can be seen using symmetry/geometry ?






calculus integration definite-integrals




calculus integration definite-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jul 16 '18 at 18:31









Nosrati

26.5k62354




26.5k62354










asked Jul 16 '18 at 17:57









rsadhvikarsadhvika

1,6681228




1,6681228








  • 2




    $begingroup$
    It's better to reduce to the $(a,b)=(-1,1)$ case rather than the $(0,1)$ case.
    $endgroup$
    – Lord Shark the Unknown
    Jul 16 '18 at 17:59










  • $begingroup$
    @InterstellarProbe here is an example
    $endgroup$
    – rsadhvika
    Jul 16 '18 at 18:04












  • $begingroup$
    @InterstellarProbe So much the worse for Wolfie!
    $endgroup$
    – Lord Shark the Unknown
    Jul 16 '18 at 18:06






  • 1




    $begingroup$
    $intlimits_0^1 dfrac{dt}{sqrt{t(1-t)}}=beta(dfrac14,dfrac14)=pi$
    $endgroup$
    – Nosrati
    Jul 16 '18 at 18:16






  • 1




    $begingroup$
    maybe try the substitution $t=cos theta$ or $t=sintheta$? I haven't tried it but the denominator would simplify
    $endgroup$
    – usr0192
    Jul 16 '18 at 18:18














  • 2




    $begingroup$
    It's better to reduce to the $(a,b)=(-1,1)$ case rather than the $(0,1)$ case.
    $endgroup$
    – Lord Shark the Unknown
    Jul 16 '18 at 17:59










  • $begingroup$
    @InterstellarProbe here is an example
    $endgroup$
    – rsadhvika
    Jul 16 '18 at 18:04












  • $begingroup$
    @InterstellarProbe So much the worse for Wolfie!
    $endgroup$
    – Lord Shark the Unknown
    Jul 16 '18 at 18:06






  • 1




    $begingroup$
    $intlimits_0^1 dfrac{dt}{sqrt{t(1-t)}}=beta(dfrac14,dfrac14)=pi$
    $endgroup$
    – Nosrati
    Jul 16 '18 at 18:16






  • 1




    $begingroup$
    maybe try the substitution $t=cos theta$ or $t=sintheta$? I haven't tried it but the denominator would simplify
    $endgroup$
    – usr0192
    Jul 16 '18 at 18:18








2




2




$begingroup$
It's better to reduce to the $(a,b)=(-1,1)$ case rather than the $(0,1)$ case.
$endgroup$
– Lord Shark the Unknown
Jul 16 '18 at 17:59




$begingroup$
It's better to reduce to the $(a,b)=(-1,1)$ case rather than the $(0,1)$ case.
$endgroup$
– Lord Shark the Unknown
Jul 16 '18 at 17:59












$begingroup$
@InterstellarProbe here is an example
$endgroup$
– rsadhvika
Jul 16 '18 at 18:04






$begingroup$
@InterstellarProbe here is an example
$endgroup$
– rsadhvika
Jul 16 '18 at 18:04














$begingroup$
@InterstellarProbe So much the worse for Wolfie!
$endgroup$
– Lord Shark the Unknown
Jul 16 '18 at 18:06




$begingroup$
@InterstellarProbe So much the worse for Wolfie!
$endgroup$
– Lord Shark the Unknown
Jul 16 '18 at 18:06




1




1




$begingroup$
$intlimits_0^1 dfrac{dt}{sqrt{t(1-t)}}=beta(dfrac14,dfrac14)=pi$
$endgroup$
– Nosrati
Jul 16 '18 at 18:16




$begingroup$
$intlimits_0^1 dfrac{dt}{sqrt{t(1-t)}}=beta(dfrac14,dfrac14)=pi$
$endgroup$
– Nosrati
Jul 16 '18 at 18:16




1




1




$begingroup$
maybe try the substitution $t=cos theta$ or $t=sintheta$? I haven't tried it but the denominator would simplify
$endgroup$
– usr0192
Jul 16 '18 at 18:18




$begingroup$
maybe try the substitution $t=cos theta$ or $t=sintheta$? I haven't tried it but the denominator would simplify
$endgroup$
– usr0192
Jul 16 '18 at 18:18










5 Answers
5






active

oldest

votes


















20












$begingroup$

Other way is substitution $t=sin^2theta$ so
$$intlimits_0^1 dfrac{dt}{sqrt{t(1-t)}}=intlimits_0^frac{pi}{2} 2dt=pi$$






share|cite|improve this answer









$endgroup$





















    4












    $begingroup$

    It's called an Abel Integral ( at least in my language ). You can write that
    $$
    frac{1}{sqrt{left(x-aright)left(b-xright)}}=frac{2}{a-b}frac{1}{sqrt{1-left(frac{2}{a-b}left(x-frac{b+a}{2}right)right)^2}}$$



    that goes into arcsinus




    $$int_{a}^{b}frac{text{d}x}{sqrt{left(x-aright)left(b-xright)}}=text{arcsin}left(frac{2}{b-a}frac{b-a}{2}right)+text{arcsin}left(frac{2}{a-b}frac{a-b}{2}right)=2text{arcsin}left(1right)=pi$$







    share|cite|improve this answer









    $endgroup$





















      4












      $begingroup$

      begin{align}
      tan^2 theta &= frac{x-a}{b-x} \
      2tan theta sec^2 theta , dtheta &=
      frac{b-a}{(b-x)^2} , dx \
      2sqrt{frac{x-a}{b-x}} times frac{(x-a)+(b-x)}{b-x} , dtheta &=
      frac{b-a}{(b-x)^2} , dx \
      2, dtheta &= frac{dx}{sqrt{(x-a)(b-x)}} \
      int frac{dx}{sqrt{(x-a)(b-x)}} &=
      2tan^{-1} sqrt{frac{x-a}{b-x}}
      end{align}



      The singularity in Wolfram Alpha comes from the upper limit $b$.




      Geometrical interpretation



      Considering circular arc $(x,y)=(sqrt{b-u},sqrt{u-a})$



      begin{align}
      ds &= frac{sqrt{b-a} , du}{2sqrt{(u-a)(b-u)}} \
      tan theta &= sqrt{frac{u-a}{b-u}} \
      begin{pmatrix}
      x \ y
      end{pmatrix} &=
      begin{pmatrix}
      sqrt{b-a} cos theta \
      sqrt{b-a} sin theta
      end{pmatrix} \
      ds &= sqrt{b-a} , dtheta
      end{align}




      See also another integral here.






      share|cite|improve this answer











      $endgroup$





















        4












        $begingroup$

        Let $m = frac{b+a}{2}$ and $r = frac{b-a}{2}$. Consider the circle



        $$ (x - m)^2 + y^2 = r^2. $$



        Part of this locus with $y geq 0$ is given by $y = sqrt{r^2 - (x-m)^2} = sqrt{(x-a)(b-x)}$ for $a leq x leq b$. By the implicit differentiation, this function satisfies $ 2(x - m) dx + 2ydy = 0 $ and hence



        $$ frac{dy}{dx} = -frac{x-m}{y}. $$



        So the length of the upper-circular arc is



        $$ pi r = int_{a}^{b} sqrt{1+left(frac{dy}{dx}right)^2} , dx = int_{a}^{b} sqrt{frac{(x-m)^2 + y^2}{y^2}} , dx = int_{a}^{b} frac{r}{sqrt{(x-a)(b-x)}} , dx. $$



        Dividing both sides by $r$ gives the desired answer.






        share|cite|improve this answer









        $endgroup$





















          3












          $begingroup$

          I was taught to use the substitution $x=a sin^2 theta+b cos^2 theta$






          share|cite|improve this answer









          $endgroup$













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            5 Answers
            5






            active

            oldest

            votes








            5 Answers
            5






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            20












            $begingroup$

            Other way is substitution $t=sin^2theta$ so
            $$intlimits_0^1 dfrac{dt}{sqrt{t(1-t)}}=intlimits_0^frac{pi}{2} 2dt=pi$$






            share|cite|improve this answer









            $endgroup$


















              20












              $begingroup$

              Other way is substitution $t=sin^2theta$ so
              $$intlimits_0^1 dfrac{dt}{sqrt{t(1-t)}}=intlimits_0^frac{pi}{2} 2dt=pi$$






              share|cite|improve this answer









              $endgroup$
















                20












                20








                20





                $begingroup$

                Other way is substitution $t=sin^2theta$ so
                $$intlimits_0^1 dfrac{dt}{sqrt{t(1-t)}}=intlimits_0^frac{pi}{2} 2dt=pi$$






                share|cite|improve this answer









                $endgroup$



                Other way is substitution $t=sin^2theta$ so
                $$intlimits_0^1 dfrac{dt}{sqrt{t(1-t)}}=intlimits_0^frac{pi}{2} 2dt=pi$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jul 16 '18 at 18:19









                NosratiNosrati

                26.5k62354




                26.5k62354























                    4












                    $begingroup$

                    It's called an Abel Integral ( at least in my language ). You can write that
                    $$
                    frac{1}{sqrt{left(x-aright)left(b-xright)}}=frac{2}{a-b}frac{1}{sqrt{1-left(frac{2}{a-b}left(x-frac{b+a}{2}right)right)^2}}$$



                    that goes into arcsinus




                    $$int_{a}^{b}frac{text{d}x}{sqrt{left(x-aright)left(b-xright)}}=text{arcsin}left(frac{2}{b-a}frac{b-a}{2}right)+text{arcsin}left(frac{2}{a-b}frac{a-b}{2}right)=2text{arcsin}left(1right)=pi$$







                    share|cite|improve this answer









                    $endgroup$


















                      4












                      $begingroup$

                      It's called an Abel Integral ( at least in my language ). You can write that
                      $$
                      frac{1}{sqrt{left(x-aright)left(b-xright)}}=frac{2}{a-b}frac{1}{sqrt{1-left(frac{2}{a-b}left(x-frac{b+a}{2}right)right)^2}}$$



                      that goes into arcsinus




                      $$int_{a}^{b}frac{text{d}x}{sqrt{left(x-aright)left(b-xright)}}=text{arcsin}left(frac{2}{b-a}frac{b-a}{2}right)+text{arcsin}left(frac{2}{a-b}frac{a-b}{2}right)=2text{arcsin}left(1right)=pi$$







                      share|cite|improve this answer









                      $endgroup$
















                        4












                        4








                        4





                        $begingroup$

                        It's called an Abel Integral ( at least in my language ). You can write that
                        $$
                        frac{1}{sqrt{left(x-aright)left(b-xright)}}=frac{2}{a-b}frac{1}{sqrt{1-left(frac{2}{a-b}left(x-frac{b+a}{2}right)right)^2}}$$



                        that goes into arcsinus




                        $$int_{a}^{b}frac{text{d}x}{sqrt{left(x-aright)left(b-xright)}}=text{arcsin}left(frac{2}{b-a}frac{b-a}{2}right)+text{arcsin}left(frac{2}{a-b}frac{a-b}{2}right)=2text{arcsin}left(1right)=pi$$







                        share|cite|improve this answer









                        $endgroup$



                        It's called an Abel Integral ( at least in my language ). You can write that
                        $$
                        frac{1}{sqrt{left(x-aright)left(b-xright)}}=frac{2}{a-b}frac{1}{sqrt{1-left(frac{2}{a-b}left(x-frac{b+a}{2}right)right)^2}}$$



                        that goes into arcsinus




                        $$int_{a}^{b}frac{text{d}x}{sqrt{left(x-aright)left(b-xright)}}=text{arcsin}left(frac{2}{b-a}frac{b-a}{2}right)+text{arcsin}left(frac{2}{a-b}frac{a-b}{2}right)=2text{arcsin}left(1right)=pi$$








                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered Jul 16 '18 at 18:22









                        AtmosAtmos

                        4,737419




                        4,737419























                            4












                            $begingroup$

                            begin{align}
                            tan^2 theta &= frac{x-a}{b-x} \
                            2tan theta sec^2 theta , dtheta &=
                            frac{b-a}{(b-x)^2} , dx \
                            2sqrt{frac{x-a}{b-x}} times frac{(x-a)+(b-x)}{b-x} , dtheta &=
                            frac{b-a}{(b-x)^2} , dx \
                            2, dtheta &= frac{dx}{sqrt{(x-a)(b-x)}} \
                            int frac{dx}{sqrt{(x-a)(b-x)}} &=
                            2tan^{-1} sqrt{frac{x-a}{b-x}}
                            end{align}



                            The singularity in Wolfram Alpha comes from the upper limit $b$.




                            Geometrical interpretation



                            Considering circular arc $(x,y)=(sqrt{b-u},sqrt{u-a})$



                            begin{align}
                            ds &= frac{sqrt{b-a} , du}{2sqrt{(u-a)(b-u)}} \
                            tan theta &= sqrt{frac{u-a}{b-u}} \
                            begin{pmatrix}
                            x \ y
                            end{pmatrix} &=
                            begin{pmatrix}
                            sqrt{b-a} cos theta \
                            sqrt{b-a} sin theta
                            end{pmatrix} \
                            ds &= sqrt{b-a} , dtheta
                            end{align}




                            See also another integral here.






                            share|cite|improve this answer











                            $endgroup$


















                              4












                              $begingroup$

                              begin{align}
                              tan^2 theta &= frac{x-a}{b-x} \
                              2tan theta sec^2 theta , dtheta &=
                              frac{b-a}{(b-x)^2} , dx \
                              2sqrt{frac{x-a}{b-x}} times frac{(x-a)+(b-x)}{b-x} , dtheta &=
                              frac{b-a}{(b-x)^2} , dx \
                              2, dtheta &= frac{dx}{sqrt{(x-a)(b-x)}} \
                              int frac{dx}{sqrt{(x-a)(b-x)}} &=
                              2tan^{-1} sqrt{frac{x-a}{b-x}}
                              end{align}



                              The singularity in Wolfram Alpha comes from the upper limit $b$.




                              Geometrical interpretation



                              Considering circular arc $(x,y)=(sqrt{b-u},sqrt{u-a})$



                              begin{align}
                              ds &= frac{sqrt{b-a} , du}{2sqrt{(u-a)(b-u)}} \
                              tan theta &= sqrt{frac{u-a}{b-u}} \
                              begin{pmatrix}
                              x \ y
                              end{pmatrix} &=
                              begin{pmatrix}
                              sqrt{b-a} cos theta \
                              sqrt{b-a} sin theta
                              end{pmatrix} \
                              ds &= sqrt{b-a} , dtheta
                              end{align}




                              See also another integral here.






                              share|cite|improve this answer











                              $endgroup$
















                                4












                                4








                                4





                                $begingroup$

                                begin{align}
                                tan^2 theta &= frac{x-a}{b-x} \
                                2tan theta sec^2 theta , dtheta &=
                                frac{b-a}{(b-x)^2} , dx \
                                2sqrt{frac{x-a}{b-x}} times frac{(x-a)+(b-x)}{b-x} , dtheta &=
                                frac{b-a}{(b-x)^2} , dx \
                                2, dtheta &= frac{dx}{sqrt{(x-a)(b-x)}} \
                                int frac{dx}{sqrt{(x-a)(b-x)}} &=
                                2tan^{-1} sqrt{frac{x-a}{b-x}}
                                end{align}



                                The singularity in Wolfram Alpha comes from the upper limit $b$.




                                Geometrical interpretation



                                Considering circular arc $(x,y)=(sqrt{b-u},sqrt{u-a})$



                                begin{align}
                                ds &= frac{sqrt{b-a} , du}{2sqrt{(u-a)(b-u)}} \
                                tan theta &= sqrt{frac{u-a}{b-u}} \
                                begin{pmatrix}
                                x \ y
                                end{pmatrix} &=
                                begin{pmatrix}
                                sqrt{b-a} cos theta \
                                sqrt{b-a} sin theta
                                end{pmatrix} \
                                ds &= sqrt{b-a} , dtheta
                                end{align}




                                See also another integral here.






                                share|cite|improve this answer











                                $endgroup$



                                begin{align}
                                tan^2 theta &= frac{x-a}{b-x} \
                                2tan theta sec^2 theta , dtheta &=
                                frac{b-a}{(b-x)^2} , dx \
                                2sqrt{frac{x-a}{b-x}} times frac{(x-a)+(b-x)}{b-x} , dtheta &=
                                frac{b-a}{(b-x)^2} , dx \
                                2, dtheta &= frac{dx}{sqrt{(x-a)(b-x)}} \
                                int frac{dx}{sqrt{(x-a)(b-x)}} &=
                                2tan^{-1} sqrt{frac{x-a}{b-x}}
                                end{align}



                                The singularity in Wolfram Alpha comes from the upper limit $b$.




                                Geometrical interpretation



                                Considering circular arc $(x,y)=(sqrt{b-u},sqrt{u-a})$



                                begin{align}
                                ds &= frac{sqrt{b-a} , du}{2sqrt{(u-a)(b-u)}} \
                                tan theta &= sqrt{frac{u-a}{b-u}} \
                                begin{pmatrix}
                                x \ y
                                end{pmatrix} &=
                                begin{pmatrix}
                                sqrt{b-a} cos theta \
                                sqrt{b-a} sin theta
                                end{pmatrix} \
                                ds &= sqrt{b-a} , dtheta
                                end{align}




                                See also another integral here.







                                share|cite|improve this answer














                                share|cite|improve this answer



                                share|cite|improve this answer








                                edited Jul 17 '18 at 3:37

























                                answered Jul 16 '18 at 18:19









                                Ng Chung TakNg Chung Tak

                                14.3k31334




                                14.3k31334























                                    4












                                    $begingroup$

                                    Let $m = frac{b+a}{2}$ and $r = frac{b-a}{2}$. Consider the circle



                                    $$ (x - m)^2 + y^2 = r^2. $$



                                    Part of this locus with $y geq 0$ is given by $y = sqrt{r^2 - (x-m)^2} = sqrt{(x-a)(b-x)}$ for $a leq x leq b$. By the implicit differentiation, this function satisfies $ 2(x - m) dx + 2ydy = 0 $ and hence



                                    $$ frac{dy}{dx} = -frac{x-m}{y}. $$



                                    So the length of the upper-circular arc is



                                    $$ pi r = int_{a}^{b} sqrt{1+left(frac{dy}{dx}right)^2} , dx = int_{a}^{b} sqrt{frac{(x-m)^2 + y^2}{y^2}} , dx = int_{a}^{b} frac{r}{sqrt{(x-a)(b-x)}} , dx. $$



                                    Dividing both sides by $r$ gives the desired answer.






                                    share|cite|improve this answer









                                    $endgroup$


















                                      4












                                      $begingroup$

                                      Let $m = frac{b+a}{2}$ and $r = frac{b-a}{2}$. Consider the circle



                                      $$ (x - m)^2 + y^2 = r^2. $$



                                      Part of this locus with $y geq 0$ is given by $y = sqrt{r^2 - (x-m)^2} = sqrt{(x-a)(b-x)}$ for $a leq x leq b$. By the implicit differentiation, this function satisfies $ 2(x - m) dx + 2ydy = 0 $ and hence



                                      $$ frac{dy}{dx} = -frac{x-m}{y}. $$



                                      So the length of the upper-circular arc is



                                      $$ pi r = int_{a}^{b} sqrt{1+left(frac{dy}{dx}right)^2} , dx = int_{a}^{b} sqrt{frac{(x-m)^2 + y^2}{y^2}} , dx = int_{a}^{b} frac{r}{sqrt{(x-a)(b-x)}} , dx. $$



                                      Dividing both sides by $r$ gives the desired answer.






                                      share|cite|improve this answer









                                      $endgroup$
















                                        4












                                        4








                                        4





                                        $begingroup$

                                        Let $m = frac{b+a}{2}$ and $r = frac{b-a}{2}$. Consider the circle



                                        $$ (x - m)^2 + y^2 = r^2. $$



                                        Part of this locus with $y geq 0$ is given by $y = sqrt{r^2 - (x-m)^2} = sqrt{(x-a)(b-x)}$ for $a leq x leq b$. By the implicit differentiation, this function satisfies $ 2(x - m) dx + 2ydy = 0 $ and hence



                                        $$ frac{dy}{dx} = -frac{x-m}{y}. $$



                                        So the length of the upper-circular arc is



                                        $$ pi r = int_{a}^{b} sqrt{1+left(frac{dy}{dx}right)^2} , dx = int_{a}^{b} sqrt{frac{(x-m)^2 + y^2}{y^2}} , dx = int_{a}^{b} frac{r}{sqrt{(x-a)(b-x)}} , dx. $$



                                        Dividing both sides by $r$ gives the desired answer.






                                        share|cite|improve this answer









                                        $endgroup$



                                        Let $m = frac{b+a}{2}$ and $r = frac{b-a}{2}$. Consider the circle



                                        $$ (x - m)^2 + y^2 = r^2. $$



                                        Part of this locus with $y geq 0$ is given by $y = sqrt{r^2 - (x-m)^2} = sqrt{(x-a)(b-x)}$ for $a leq x leq b$. By the implicit differentiation, this function satisfies $ 2(x - m) dx + 2ydy = 0 $ and hence



                                        $$ frac{dy}{dx} = -frac{x-m}{y}. $$



                                        So the length of the upper-circular arc is



                                        $$ pi r = int_{a}^{b} sqrt{1+left(frac{dy}{dx}right)^2} , dx = int_{a}^{b} sqrt{frac{(x-m)^2 + y^2}{y^2}} , dx = int_{a}^{b} frac{r}{sqrt{(x-a)(b-x)}} , dx. $$



                                        Dividing both sides by $r$ gives the desired answer.







                                        share|cite|improve this answer












                                        share|cite|improve this answer



                                        share|cite|improve this answer










                                        answered Jul 17 '18 at 7:01









                                        Sangchul LeeSangchul Lee

                                        91.6k12164265




                                        91.6k12164265























                                            3












                                            $begingroup$

                                            I was taught to use the substitution $x=a sin^2 theta+b cos^2 theta$






                                            share|cite|improve this answer









                                            $endgroup$


















                                              3












                                              $begingroup$

                                              I was taught to use the substitution $x=a sin^2 theta+b cos^2 theta$






                                              share|cite|improve this answer









                                              $endgroup$
















                                                3












                                                3








                                                3





                                                $begingroup$

                                                I was taught to use the substitution $x=a sin^2 theta+b cos^2 theta$






                                                share|cite|improve this answer









                                                $endgroup$



                                                I was taught to use the substitution $x=a sin^2 theta+b cos^2 theta$







                                                share|cite|improve this answer












                                                share|cite|improve this answer



                                                share|cite|improve this answer










                                                answered Jul 17 '18 at 6:41









                                                Hari ShankarHari Shankar

                                                2,181139




                                                2,181139






























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