Why the expectation of distance to center of disk is $r/3$ and not $r/2$?












1












$begingroup$


A person throw an arrow on a target of radius $r$. The position of the arrow on the target is uniformly distributed. Let $X$ the distance between the arrow and the center of the circle. The score obtained by a person is $r-X$. What is the average score ? The answer is $frac{r}{3}$, whereas I found $frac{r}{2}$ as follow



We have that $X$ is uniform on $[0,r]$. If $Y=r-X$, then $$mathbb E[Y]=int_0^r (r-x)f_X(x)dx=frac{1}{r}int_0^r (r-x)dx=frac{r}{2}.$$



Maybe there is a subtlety than I don't see ?






probability expected-value







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  • 2




    $begingroup$
    Why do you think $X$ is uniform on $[0,r]$?
    $endgroup$
    – 5xum
    Nov 30 '18 at 9:13










  • $begingroup$
    @5xum : I set $Z=(Rcos Theta, Rsin Theta)$ with $R$ uniform on $[0,r]$ and $Theta$ uniform on $[0,2pi)$. Then $mathbb P{Xleq x}=mathbb P{Rleq x, Theta in [0,2pi]}=frac{x}{r}.$ It doesn't work ?
    $endgroup$
    – idm
    Nov 30 '18 at 9:14












  • $begingroup$
    Uniform over the disc does not mean the distribution of the distance from the center is uniform. The probability density of points in a circle sharing a centre with the disc will be inversely proportional to the radius of the circle (if less than $r$), not a constant.$$dfrac{mathsf d ~~}{mathsf d~x}mathsf P(Xleqslant x)~propto~dfrac{1}{x}mathbf 1_{0< xleqslant r}$$
    $endgroup$
    – Graham Kemp
    Nov 30 '18 at 9:23










  • $begingroup$
    We just had a question like this half a day ago: Average distance from center of circle
    $endgroup$
    – Rahul
    Nov 30 '18 at 9:25






  • 1




    $begingroup$
    $mathbb P{Xleq x} = frac{pi x^2}{pi r^2}$ as you are equally likely to land at any point on the area of the board, not equally likely to land at any radial distance from the centre.
    $endgroup$
    – Paul
    Nov 30 '18 at 9:30
















1












$begingroup$


A person throw an arrow on a target of radius $r$. The position of the arrow on the target is uniformly distributed. Let $X$ the distance between the arrow and the center of the circle. The score obtained by a person is $r-X$. What is the average score ? The answer is $frac{r}{3}$, whereas I found $frac{r}{2}$ as follow



We have that $X$ is uniform on $[0,r]$. If $Y=r-X$, then $$mathbb E[Y]=int_0^r (r-x)f_X(x)dx=frac{1}{r}int_0^r (r-x)dx=frac{r}{2}.$$



Maybe there is a subtlety than I don't see ?






probability expected-value







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Why do you think $X$ is uniform on $[0,r]$?
    $endgroup$
    – 5xum
    Nov 30 '18 at 9:13










  • $begingroup$
    @5xum : I set $Z=(Rcos Theta, Rsin Theta)$ with $R$ uniform on $[0,r]$ and $Theta$ uniform on $[0,2pi)$. Then $mathbb P{Xleq x}=mathbb P{Rleq x, Theta in [0,2pi]}=frac{x}{r}.$ It doesn't work ?
    $endgroup$
    – idm
    Nov 30 '18 at 9:14












  • $begingroup$
    Uniform over the disc does not mean the distribution of the distance from the center is uniform. The probability density of points in a circle sharing a centre with the disc will be inversely proportional to the radius of the circle (if less than $r$), not a constant.$$dfrac{mathsf d ~~}{mathsf d~x}mathsf P(Xleqslant x)~propto~dfrac{1}{x}mathbf 1_{0< xleqslant r}$$
    $endgroup$
    – Graham Kemp
    Nov 30 '18 at 9:23










  • $begingroup$
    We just had a question like this half a day ago: Average distance from center of circle
    $endgroup$
    – Rahul
    Nov 30 '18 at 9:25






  • 1




    $begingroup$
    $mathbb P{Xleq x} = frac{pi x^2}{pi r^2}$ as you are equally likely to land at any point on the area of the board, not equally likely to land at any radial distance from the centre.
    $endgroup$
    – Paul
    Nov 30 '18 at 9:30














1












1








1


0



$begingroup$


A person throw an arrow on a target of radius $r$. The position of the arrow on the target is uniformly distributed. Let $X$ the distance between the arrow and the center of the circle. The score obtained by a person is $r-X$. What is the average score ? The answer is $frac{r}{3}$, whereas I found $frac{r}{2}$ as follow



We have that $X$ is uniform on $[0,r]$. If $Y=r-X$, then $$mathbb E[Y]=int_0^r (r-x)f_X(x)dx=frac{1}{r}int_0^r (r-x)dx=frac{r}{2}.$$



Maybe there is a subtlety than I don't see ?






probability expected-value







share|cite|improve this question











$endgroup$




A person throw an arrow on a target of radius $r$. The position of the arrow on the target is uniformly distributed. Let $X$ the distance between the arrow and the center of the circle. The score obtained by a person is $r-X$. What is the average score ? The answer is $frac{r}{3}$, whereas I found $frac{r}{2}$ as follow



We have that $X$ is uniform on $[0,r]$. If $Y=r-X$, then $$mathbb E[Y]=int_0^r (r-x)f_X(x)dx=frac{1}{r}int_0^r (r-x)dx=frac{r}{2}.$$



Maybe there is a subtlety than I don't see ?






probability expected-value




probability expected-value






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share|cite|improve this question













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edited Nov 30 '18 at 9:14









David G. Stork

10.2k21332




10.2k21332










asked Nov 30 '18 at 9:11









idmidm

8,56921345




8,56921345








  • 2




    $begingroup$
    Why do you think $X$ is uniform on $[0,r]$?
    $endgroup$
    – 5xum
    Nov 30 '18 at 9:13










  • $begingroup$
    @5xum : I set $Z=(Rcos Theta, Rsin Theta)$ with $R$ uniform on $[0,r]$ and $Theta$ uniform on $[0,2pi)$. Then $mathbb P{Xleq x}=mathbb P{Rleq x, Theta in [0,2pi]}=frac{x}{r}.$ It doesn't work ?
    $endgroup$
    – idm
    Nov 30 '18 at 9:14












  • $begingroup$
    Uniform over the disc does not mean the distribution of the distance from the center is uniform. The probability density of points in a circle sharing a centre with the disc will be inversely proportional to the radius of the circle (if less than $r$), not a constant.$$dfrac{mathsf d ~~}{mathsf d~x}mathsf P(Xleqslant x)~propto~dfrac{1}{x}mathbf 1_{0< xleqslant r}$$
    $endgroup$
    – Graham Kemp
    Nov 30 '18 at 9:23










  • $begingroup$
    We just had a question like this half a day ago: Average distance from center of circle
    $endgroup$
    – Rahul
    Nov 30 '18 at 9:25






  • 1




    $begingroup$
    $mathbb P{Xleq x} = frac{pi x^2}{pi r^2}$ as you are equally likely to land at any point on the area of the board, not equally likely to land at any radial distance from the centre.
    $endgroup$
    – Paul
    Nov 30 '18 at 9:30














  • 2




    $begingroup$
    Why do you think $X$ is uniform on $[0,r]$?
    $endgroup$
    – 5xum
    Nov 30 '18 at 9:13










  • $begingroup$
    @5xum : I set $Z=(Rcos Theta, Rsin Theta)$ with $R$ uniform on $[0,r]$ and $Theta$ uniform on $[0,2pi)$. Then $mathbb P{Xleq x}=mathbb P{Rleq x, Theta in [0,2pi]}=frac{x}{r}.$ It doesn't work ?
    $endgroup$
    – idm
    Nov 30 '18 at 9:14












  • $begingroup$
    Uniform over the disc does not mean the distribution of the distance from the center is uniform. The probability density of points in a circle sharing a centre with the disc will be inversely proportional to the radius of the circle (if less than $r$), not a constant.$$dfrac{mathsf d ~~}{mathsf d~x}mathsf P(Xleqslant x)~propto~dfrac{1}{x}mathbf 1_{0< xleqslant r}$$
    $endgroup$
    – Graham Kemp
    Nov 30 '18 at 9:23










  • $begingroup$
    We just had a question like this half a day ago: Average distance from center of circle
    $endgroup$
    – Rahul
    Nov 30 '18 at 9:25






  • 1




    $begingroup$
    $mathbb P{Xleq x} = frac{pi x^2}{pi r^2}$ as you are equally likely to land at any point on the area of the board, not equally likely to land at any radial distance from the centre.
    $endgroup$
    – Paul
    Nov 30 '18 at 9:30








2




2




$begingroup$
Why do you think $X$ is uniform on $[0,r]$?
$endgroup$
– 5xum
Nov 30 '18 at 9:13




$begingroup$
Why do you think $X$ is uniform on $[0,r]$?
$endgroup$
– 5xum
Nov 30 '18 at 9:13












$begingroup$
@5xum : I set $Z=(Rcos Theta, Rsin Theta)$ with $R$ uniform on $[0,r]$ and $Theta$ uniform on $[0,2pi)$. Then $mathbb P{Xleq x}=mathbb P{Rleq x, Theta in [0,2pi]}=frac{x}{r}.$ It doesn't work ?
$endgroup$
– idm
Nov 30 '18 at 9:14






$begingroup$
@5xum : I set $Z=(Rcos Theta, Rsin Theta)$ with $R$ uniform on $[0,r]$ and $Theta$ uniform on $[0,2pi)$. Then $mathbb P{Xleq x}=mathbb P{Rleq x, Theta in [0,2pi]}=frac{x}{r}.$ It doesn't work ?
$endgroup$
– idm
Nov 30 '18 at 9:14














$begingroup$
Uniform over the disc does not mean the distribution of the distance from the center is uniform. The probability density of points in a circle sharing a centre with the disc will be inversely proportional to the radius of the circle (if less than $r$), not a constant.$$dfrac{mathsf d ~~}{mathsf d~x}mathsf P(Xleqslant x)~propto~dfrac{1}{x}mathbf 1_{0< xleqslant r}$$
$endgroup$
– Graham Kemp
Nov 30 '18 at 9:23




$begingroup$
Uniform over the disc does not mean the distribution of the distance from the center is uniform. The probability density of points in a circle sharing a centre with the disc will be inversely proportional to the radius of the circle (if less than $r$), not a constant.$$dfrac{mathsf d ~~}{mathsf d~x}mathsf P(Xleqslant x)~propto~dfrac{1}{x}mathbf 1_{0< xleqslant r}$$
$endgroup$
– Graham Kemp
Nov 30 '18 at 9:23












$begingroup$
We just had a question like this half a day ago: Average distance from center of circle
$endgroup$
– Rahul
Nov 30 '18 at 9:25




$begingroup$
We just had a question like this half a day ago: Average distance from center of circle
$endgroup$
– Rahul
Nov 30 '18 at 9:25




1




1




$begingroup$
$mathbb P{Xleq x} = frac{pi x^2}{pi r^2}$ as you are equally likely to land at any point on the area of the board, not equally likely to land at any radial distance from the centre.
$endgroup$
– Paul
Nov 30 '18 at 9:30




$begingroup$
$mathbb P{Xleq x} = frac{pi x^2}{pi r^2}$ as you are equally likely to land at any point on the area of the board, not equally likely to land at any radial distance from the centre.
$endgroup$
– Paul
Nov 30 '18 at 9:30










3 Answers
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2












$begingroup$

$$F_X(x) = frac{pi x^2}{pi r^2}$$



$$f_X(x)=frac{2x}{r^2}$$



begin{align}
E[r-X]&=r-E[X] \
&=r - frac1{r^2}int_0^r 2x^2, dx\
&= r - frac1{r^2}frac{2r^3}3\
&= frac{r}{3}
end{align}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I don't get why $F_X(x)=frac{x^2}{r^2}$
    $endgroup$
    – idm
    Nov 30 '18 at 9:29










  • $begingroup$
    That is the meaning of uniform over an area, if you draw a circle of the same size on the target, it is equally likely to hit either of them.
    $endgroup$
    – Siong Thye Goh
    Nov 30 '18 at 9:30












  • $begingroup$
    Ok I see, thank you. But if $Z=(RcosTheta,RsinTheta)$ with $R$ uniform in $[0,r]$ and $Theta$ uniform on $[0,2pi]$, why $$mathbb P{|Z|leq x}=mathbb P{Rleq x,Thetain [0,2pi]}=mathbb P{Rleq x}=frac{x}{r}$$ is not true ? I really don't get this point
    $endgroup$
    – idm
    Nov 30 '18 at 9:41










  • $begingroup$
    the assumption that $R$ is uniform in $[0,r]$ is not valid.
    $endgroup$
    – Siong Thye Goh
    Nov 30 '18 at 9:45










  • $begingroup$
    ok, stange... thank you :)
    $endgroup$
    – idm
    Nov 30 '18 at 9:46



















1












$begingroup$

You may first construct the probability density as follows:




  • At distance $x$ from the center of the circle a corresponding annulus of "thickness" $dx$ has a probability weight of
    $$frac{1}{pi r^2}cdot 2pi cdot x cdot dx$$


So, you get
$$E(Y) = frac{1}{pi r^2} int_0^r (r-x)2pi cdot x; dx = cdots = frac{r}{3}$$






share|cite|improve this answer









$endgroup$





















    -1












    $begingroup$

    $$frac{int_0^r(r-x)x.dx}{int_0^r x.dx}$$



    This gives $r/3$.



    The problem with the integral you gave at first, is that the $(r-x)$ needs to be weighted by an $x.dx$, as this is the elemental area of the disc presented by a strip of thickness $dx$ at radius $x$ (or rather $2pi x.dx$ ... but then the $2pi$ appears on the top & bottom & obviously cancels ... or to put it another way, we could do the calculation for any sector of the disc & get the same result) - the catchment, if you like, at radius $x$.



    Basically you're calculating the mean value of $r-x$ over a sector of a disc, or over a whole disc - it makes no difference.






    share|cite|improve this answer











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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      $$F_X(x) = frac{pi x^2}{pi r^2}$$



      $$f_X(x)=frac{2x}{r^2}$$



      begin{align}
      E[r-X]&=r-E[X] \
      &=r - frac1{r^2}int_0^r 2x^2, dx\
      &= r - frac1{r^2}frac{2r^3}3\
      &= frac{r}{3}
      end{align}






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        I don't get why $F_X(x)=frac{x^2}{r^2}$
        $endgroup$
        – idm
        Nov 30 '18 at 9:29










      • $begingroup$
        That is the meaning of uniform over an area, if you draw a circle of the same size on the target, it is equally likely to hit either of them.
        $endgroup$
        – Siong Thye Goh
        Nov 30 '18 at 9:30












      • $begingroup$
        Ok I see, thank you. But if $Z=(RcosTheta,RsinTheta)$ with $R$ uniform in $[0,r]$ and $Theta$ uniform on $[0,2pi]$, why $$mathbb P{|Z|leq x}=mathbb P{Rleq x,Thetain [0,2pi]}=mathbb P{Rleq x}=frac{x}{r}$$ is not true ? I really don't get this point
        $endgroup$
        – idm
        Nov 30 '18 at 9:41










      • $begingroup$
        the assumption that $R$ is uniform in $[0,r]$ is not valid.
        $endgroup$
        – Siong Thye Goh
        Nov 30 '18 at 9:45










      • $begingroup$
        ok, stange... thank you :)
        $endgroup$
        – idm
        Nov 30 '18 at 9:46
















      2












      $begingroup$

      $$F_X(x) = frac{pi x^2}{pi r^2}$$



      $$f_X(x)=frac{2x}{r^2}$$



      begin{align}
      E[r-X]&=r-E[X] \
      &=r - frac1{r^2}int_0^r 2x^2, dx\
      &= r - frac1{r^2}frac{2r^3}3\
      &= frac{r}{3}
      end{align}






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        I don't get why $F_X(x)=frac{x^2}{r^2}$
        $endgroup$
        – idm
        Nov 30 '18 at 9:29










      • $begingroup$
        That is the meaning of uniform over an area, if you draw a circle of the same size on the target, it is equally likely to hit either of them.
        $endgroup$
        – Siong Thye Goh
        Nov 30 '18 at 9:30












      • $begingroup$
        Ok I see, thank you. But if $Z=(RcosTheta,RsinTheta)$ with $R$ uniform in $[0,r]$ and $Theta$ uniform on $[0,2pi]$, why $$mathbb P{|Z|leq x}=mathbb P{Rleq x,Thetain [0,2pi]}=mathbb P{Rleq x}=frac{x}{r}$$ is not true ? I really don't get this point
        $endgroup$
        – idm
        Nov 30 '18 at 9:41










      • $begingroup$
        the assumption that $R$ is uniform in $[0,r]$ is not valid.
        $endgroup$
        – Siong Thye Goh
        Nov 30 '18 at 9:45










      • $begingroup$
        ok, stange... thank you :)
        $endgroup$
        – idm
        Nov 30 '18 at 9:46














      2












      2








      2





      $begingroup$

      $$F_X(x) = frac{pi x^2}{pi r^2}$$



      $$f_X(x)=frac{2x}{r^2}$$



      begin{align}
      E[r-X]&=r-E[X] \
      &=r - frac1{r^2}int_0^r 2x^2, dx\
      &= r - frac1{r^2}frac{2r^3}3\
      &= frac{r}{3}
      end{align}






      share|cite|improve this answer









      $endgroup$



      $$F_X(x) = frac{pi x^2}{pi r^2}$$



      $$f_X(x)=frac{2x}{r^2}$$



      begin{align}
      E[r-X]&=r-E[X] \
      &=r - frac1{r^2}int_0^r 2x^2, dx\
      &= r - frac1{r^2}frac{2r^3}3\
      &= frac{r}{3}
      end{align}







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Nov 30 '18 at 9:25









      Siong Thye GohSiong Thye Goh

      100k1465117




      100k1465117












      • $begingroup$
        I don't get why $F_X(x)=frac{x^2}{r^2}$
        $endgroup$
        – idm
        Nov 30 '18 at 9:29










      • $begingroup$
        That is the meaning of uniform over an area, if you draw a circle of the same size on the target, it is equally likely to hit either of them.
        $endgroup$
        – Siong Thye Goh
        Nov 30 '18 at 9:30












      • $begingroup$
        Ok I see, thank you. But if $Z=(RcosTheta,RsinTheta)$ with $R$ uniform in $[0,r]$ and $Theta$ uniform on $[0,2pi]$, why $$mathbb P{|Z|leq x}=mathbb P{Rleq x,Thetain [0,2pi]}=mathbb P{Rleq x}=frac{x}{r}$$ is not true ? I really don't get this point
        $endgroup$
        – idm
        Nov 30 '18 at 9:41










      • $begingroup$
        the assumption that $R$ is uniform in $[0,r]$ is not valid.
        $endgroup$
        – Siong Thye Goh
        Nov 30 '18 at 9:45










      • $begingroup$
        ok, stange... thank you :)
        $endgroup$
        – idm
        Nov 30 '18 at 9:46


















      • $begingroup$
        I don't get why $F_X(x)=frac{x^2}{r^2}$
        $endgroup$
        – idm
        Nov 30 '18 at 9:29










      • $begingroup$
        That is the meaning of uniform over an area, if you draw a circle of the same size on the target, it is equally likely to hit either of them.
        $endgroup$
        – Siong Thye Goh
        Nov 30 '18 at 9:30












      • $begingroup$
        Ok I see, thank you. But if $Z=(RcosTheta,RsinTheta)$ with $R$ uniform in $[0,r]$ and $Theta$ uniform on $[0,2pi]$, why $$mathbb P{|Z|leq x}=mathbb P{Rleq x,Thetain [0,2pi]}=mathbb P{Rleq x}=frac{x}{r}$$ is not true ? I really don't get this point
        $endgroup$
        – idm
        Nov 30 '18 at 9:41










      • $begingroup$
        the assumption that $R$ is uniform in $[0,r]$ is not valid.
        $endgroup$
        – Siong Thye Goh
        Nov 30 '18 at 9:45










      • $begingroup$
        ok, stange... thank you :)
        $endgroup$
        – idm
        Nov 30 '18 at 9:46
















      $begingroup$
      I don't get why $F_X(x)=frac{x^2}{r^2}$
      $endgroup$
      – idm
      Nov 30 '18 at 9:29




      $begingroup$
      I don't get why $F_X(x)=frac{x^2}{r^2}$
      $endgroup$
      – idm
      Nov 30 '18 at 9:29












      $begingroup$
      That is the meaning of uniform over an area, if you draw a circle of the same size on the target, it is equally likely to hit either of them.
      $endgroup$
      – Siong Thye Goh
      Nov 30 '18 at 9:30






      $begingroup$
      That is the meaning of uniform over an area, if you draw a circle of the same size on the target, it is equally likely to hit either of them.
      $endgroup$
      – Siong Thye Goh
      Nov 30 '18 at 9:30














      $begingroup$
      Ok I see, thank you. But if $Z=(RcosTheta,RsinTheta)$ with $R$ uniform in $[0,r]$ and $Theta$ uniform on $[0,2pi]$, why $$mathbb P{|Z|leq x}=mathbb P{Rleq x,Thetain [0,2pi]}=mathbb P{Rleq x}=frac{x}{r}$$ is not true ? I really don't get this point
      $endgroup$
      – idm
      Nov 30 '18 at 9:41




      $begingroup$
      Ok I see, thank you. But if $Z=(RcosTheta,RsinTheta)$ with $R$ uniform in $[0,r]$ and $Theta$ uniform on $[0,2pi]$, why $$mathbb P{|Z|leq x}=mathbb P{Rleq x,Thetain [0,2pi]}=mathbb P{Rleq x}=frac{x}{r}$$ is not true ? I really don't get this point
      $endgroup$
      – idm
      Nov 30 '18 at 9:41












      $begingroup$
      the assumption that $R$ is uniform in $[0,r]$ is not valid.
      $endgroup$
      – Siong Thye Goh
      Nov 30 '18 at 9:45




      $begingroup$
      the assumption that $R$ is uniform in $[0,r]$ is not valid.
      $endgroup$
      – Siong Thye Goh
      Nov 30 '18 at 9:45












      $begingroup$
      ok, stange... thank you :)
      $endgroup$
      – idm
      Nov 30 '18 at 9:46




      $begingroup$
      ok, stange... thank you :)
      $endgroup$
      – idm
      Nov 30 '18 at 9:46











      1












      $begingroup$

      You may first construct the probability density as follows:




      • At distance $x$ from the center of the circle a corresponding annulus of "thickness" $dx$ has a probability weight of
        $$frac{1}{pi r^2}cdot 2pi cdot x cdot dx$$


      So, you get
      $$E(Y) = frac{1}{pi r^2} int_0^r (r-x)2pi cdot x; dx = cdots = frac{r}{3}$$






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        You may first construct the probability density as follows:




        • At distance $x$ from the center of the circle a corresponding annulus of "thickness" $dx$ has a probability weight of
          $$frac{1}{pi r^2}cdot 2pi cdot x cdot dx$$


        So, you get
        $$E(Y) = frac{1}{pi r^2} int_0^r (r-x)2pi cdot x; dx = cdots = frac{r}{3}$$






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          You may first construct the probability density as follows:




          • At distance $x$ from the center of the circle a corresponding annulus of "thickness" $dx$ has a probability weight of
            $$frac{1}{pi r^2}cdot 2pi cdot x cdot dx$$


          So, you get
          $$E(Y) = frac{1}{pi r^2} int_0^r (r-x)2pi cdot x; dx = cdots = frac{r}{3}$$






          share|cite|improve this answer









          $endgroup$



          You may first construct the probability density as follows:




          • At distance $x$ from the center of the circle a corresponding annulus of "thickness" $dx$ has a probability weight of
            $$frac{1}{pi r^2}cdot 2pi cdot x cdot dx$$


          So, you get
          $$E(Y) = frac{1}{pi r^2} int_0^r (r-x)2pi cdot x; dx = cdots = frac{r}{3}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 30 '18 at 9:43









          trancelocationtrancelocation

          9,6801722




          9,6801722























              -1












              $begingroup$

              $$frac{int_0^r(r-x)x.dx}{int_0^r x.dx}$$



              This gives $r/3$.



              The problem with the integral you gave at first, is that the $(r-x)$ needs to be weighted by an $x.dx$, as this is the elemental area of the disc presented by a strip of thickness $dx$ at radius $x$ (or rather $2pi x.dx$ ... but then the $2pi$ appears on the top & bottom & obviously cancels ... or to put it another way, we could do the calculation for any sector of the disc & get the same result) - the catchment, if you like, at radius $x$.



              Basically you're calculating the mean value of $r-x$ over a sector of a disc, or over a whole disc - it makes no difference.






              share|cite|improve this answer











              $endgroup$


















                -1












                $begingroup$

                $$frac{int_0^r(r-x)x.dx}{int_0^r x.dx}$$



                This gives $r/3$.



                The problem with the integral you gave at first, is that the $(r-x)$ needs to be weighted by an $x.dx$, as this is the elemental area of the disc presented by a strip of thickness $dx$ at radius $x$ (or rather $2pi x.dx$ ... but then the $2pi$ appears on the top & bottom & obviously cancels ... or to put it another way, we could do the calculation for any sector of the disc & get the same result) - the catchment, if you like, at radius $x$.



                Basically you're calculating the mean value of $r-x$ over a sector of a disc, or over a whole disc - it makes no difference.






                share|cite|improve this answer











                $endgroup$
















                  -1












                  -1








                  -1





                  $begingroup$

                  $$frac{int_0^r(r-x)x.dx}{int_0^r x.dx}$$



                  This gives $r/3$.



                  The problem with the integral you gave at first, is that the $(r-x)$ needs to be weighted by an $x.dx$, as this is the elemental area of the disc presented by a strip of thickness $dx$ at radius $x$ (or rather $2pi x.dx$ ... but then the $2pi$ appears on the top & bottom & obviously cancels ... or to put it another way, we could do the calculation for any sector of the disc & get the same result) - the catchment, if you like, at radius $x$.



                  Basically you're calculating the mean value of $r-x$ over a sector of a disc, or over a whole disc - it makes no difference.






                  share|cite|improve this answer











                  $endgroup$



                  $$frac{int_0^r(r-x)x.dx}{int_0^r x.dx}$$



                  This gives $r/3$.



                  The problem with the integral you gave at first, is that the $(r-x)$ needs to be weighted by an $x.dx$, as this is the elemental area of the disc presented by a strip of thickness $dx$ at radius $x$ (or rather $2pi x.dx$ ... but then the $2pi$ appears on the top & bottom & obviously cancels ... or to put it another way, we could do the calculation for any sector of the disc & get the same result) - the catchment, if you like, at radius $x$.



                  Basically you're calculating the mean value of $r-x$ over a sector of a disc, or over a whole disc - it makes no difference.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 30 '18 at 9:37

























                  answered Nov 30 '18 at 9:23









                  AmbretteOrriseyAmbretteOrrisey

                  55410




                  55410






























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