Finding surface area and volume of a sphere using only Pappus' Centroid Theorem












0












$begingroup$


I wonder if it is possible to derive surface area and volume of a sphere seperately using techniques involving pappus' theorem.



I did some calculation and found out the ratio of surface area and volume. Here is my work,




  1. My key observation is finding out that the centroid of a semidisk of raidus $r$ is also the centroid of a semicircle of raidus $2r/3$ when the centers coincide. (By "centers" I mean the centers of full circle and disc induced by half of them.)


  2. I sliced the semidisk identical pieces of triangles(infinitely many) and by locating each triangle's centroid I form a semicircle of radius $2r/3$ which must have the same centroid with semidisk.


  3. Then let's say our centroid is located $h$ distance above the center. We still don't know it.


  4. Using the theorem, the circular path taken by the centroid of the semidisk times the area of the centroid should give the volume of the sphere of radius $r$. And similarly, the circular path taken by the same centroid of the semicircle times the arc length of semicircle should give the surface area of a sphere of radius $2r/3$.


  5. We know that when the radius increases with a proportion, corresponding surface area will also increase with the square of that proportion. Thus, we need to multiply the surface area of sphere of radius $2r/3$ by the factor $9/4$ to get the surface area of the sphere of radius $r$.



Here is the calculations,



$$V = 2pi h * pi r^2/2=pi^2r^2h$$
$$S = 9/4*2pi h * pi (2r/3) = 3pi^2rh$$



Since we don't know $h$, I simply divide them to cancel it out and get,
$$V/S = r/3.$$






spheres centroid







share|cite|improve this question









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    0












    $begingroup$


    I wonder if it is possible to derive surface area and volume of a sphere seperately using techniques involving pappus' theorem.



    I did some calculation and found out the ratio of surface area and volume. Here is my work,




    1. My key observation is finding out that the centroid of a semidisk of raidus $r$ is also the centroid of a semicircle of raidus $2r/3$ when the centers coincide. (By "centers" I mean the centers of full circle and disc induced by half of them.)


    2. I sliced the semidisk identical pieces of triangles(infinitely many) and by locating each triangle's centroid I form a semicircle of radius $2r/3$ which must have the same centroid with semidisk.


    3. Then let's say our centroid is located $h$ distance above the center. We still don't know it.


    4. Using the theorem, the circular path taken by the centroid of the semidisk times the area of the centroid should give the volume of the sphere of radius $r$. And similarly, the circular path taken by the same centroid of the semicircle times the arc length of semicircle should give the surface area of a sphere of radius $2r/3$.


    5. We know that when the radius increases with a proportion, corresponding surface area will also increase with the square of that proportion. Thus, we need to multiply the surface area of sphere of radius $2r/3$ by the factor $9/4$ to get the surface area of the sphere of radius $r$.



    Here is the calculations,



    $$V = 2pi h * pi r^2/2=pi^2r^2h$$
    $$S = 9/4*2pi h * pi (2r/3) = 3pi^2rh$$



    Since we don't know $h$, I simply divide them to cancel it out and get,
    $$V/S = r/3.$$






    spheres centroid







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I wonder if it is possible to derive surface area and volume of a sphere seperately using techniques involving pappus' theorem.



      I did some calculation and found out the ratio of surface area and volume. Here is my work,




      1. My key observation is finding out that the centroid of a semidisk of raidus $r$ is also the centroid of a semicircle of raidus $2r/3$ when the centers coincide. (By "centers" I mean the centers of full circle and disc induced by half of them.)


      2. I sliced the semidisk identical pieces of triangles(infinitely many) and by locating each triangle's centroid I form a semicircle of radius $2r/3$ which must have the same centroid with semidisk.


      3. Then let's say our centroid is located $h$ distance above the center. We still don't know it.


      4. Using the theorem, the circular path taken by the centroid of the semidisk times the area of the centroid should give the volume of the sphere of radius $r$. And similarly, the circular path taken by the same centroid of the semicircle times the arc length of semicircle should give the surface area of a sphere of radius $2r/3$.


      5. We know that when the radius increases with a proportion, corresponding surface area will also increase with the square of that proportion. Thus, we need to multiply the surface area of sphere of radius $2r/3$ by the factor $9/4$ to get the surface area of the sphere of radius $r$.



      Here is the calculations,



      $$V = 2pi h * pi r^2/2=pi^2r^2h$$
      $$S = 9/4*2pi h * pi (2r/3) = 3pi^2rh$$



      Since we don't know $h$, I simply divide them to cancel it out and get,
      $$V/S = r/3.$$






      spheres centroid







      share|cite|improve this question









      $endgroup$




      I wonder if it is possible to derive surface area and volume of a sphere seperately using techniques involving pappus' theorem.



      I did some calculation and found out the ratio of surface area and volume. Here is my work,




      1. My key observation is finding out that the centroid of a semidisk of raidus $r$ is also the centroid of a semicircle of raidus $2r/3$ when the centers coincide. (By "centers" I mean the centers of full circle and disc induced by half of them.)


      2. I sliced the semidisk identical pieces of triangles(infinitely many) and by locating each triangle's centroid I form a semicircle of radius $2r/3$ which must have the same centroid with semidisk.


      3. Then let's say our centroid is located $h$ distance above the center. We still don't know it.


      4. Using the theorem, the circular path taken by the centroid of the semidisk times the area of the centroid should give the volume of the sphere of radius $r$. And similarly, the circular path taken by the same centroid of the semicircle times the arc length of semicircle should give the surface area of a sphere of radius $2r/3$.


      5. We know that when the radius increases with a proportion, corresponding surface area will also increase with the square of that proportion. Thus, we need to multiply the surface area of sphere of radius $2r/3$ by the factor $9/4$ to get the surface area of the sphere of radius $r$.



      Here is the calculations,



      $$V = 2pi h * pi r^2/2=pi^2r^2h$$
      $$S = 9/4*2pi h * pi (2r/3) = 3pi^2rh$$



      Since we don't know $h$, I simply divide them to cancel it out and get,
      $$V/S = r/3.$$






      spheres centroid




      spheres centroid






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      share|cite|improve this question











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      asked Jan 18 '17 at 8:52









      Ahmed BilâlAhmed Bilâl

      6510




      6510






















          1 Answer
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          0












          $begingroup$

          The centroid of semi-circle should be $dfrac{4r}{3pi}$ from the origin.



          So the volume is



          $$V=2pi times frac{4r}{3pi}times frac{pi r^2}{2}=frac{4}{3}pi r^3$$



          The centroid of semi-circular arc should be $dfrac{2r}{pi}$ from the origin.



          So the surface area is



          $$S=2pi times frac{2r}{pi}times pi r=4pi r^2$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Yeah but how to get those centroids mathematically without using calculus or hanging them or something?
            $endgroup$
            – Ahmed Bilâl
            Jan 19 '17 at 5:58










          • $begingroup$
            I didn't claim that the centroids popped out from stone but just point out your mistakes. The key point is how to apply the Pappus' theorem. Lazy guys love wikipedia so that no needs to do everything from scratch.
            $endgroup$
            – Ng Chung Tak
            Jan 19 '17 at 11:09












          • $begingroup$
            Where are the mistakes? You just put the value of $h$ and simplified the expressions.
            $endgroup$
            – Ahmed Bilâl
            Jan 19 '17 at 11:14










          • $begingroup$
            You say $dfrac{2r}{3}$
            $endgroup$
            – Ng Chung Tak
            Jan 19 '17 at 11:17










          • $begingroup$
            Ok. I said the centroid of a semicircle of radius 2r/3 is the same point that is the centroid of a semidisk of radius r when their centers coincide. Put another way, if they have same radius, disk's centroid should be at a distance above center 2/3 times of the circle's. My english suffers a lot so sorry for that.
            $endgroup$
            – Ahmed Bilâl
            Jan 19 '17 at 11:34











          Your Answer





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          1 Answer
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          active

          oldest

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          1 Answer
          1






          active

          oldest

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          active

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          active

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          0












          $begingroup$

          The centroid of semi-circle should be $dfrac{4r}{3pi}$ from the origin.



          So the volume is



          $$V=2pi times frac{4r}{3pi}times frac{pi r^2}{2}=frac{4}{3}pi r^3$$



          The centroid of semi-circular arc should be $dfrac{2r}{pi}$ from the origin.



          So the surface area is



          $$S=2pi times frac{2r}{pi}times pi r=4pi r^2$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Yeah but how to get those centroids mathematically without using calculus or hanging them or something?
            $endgroup$
            – Ahmed Bilâl
            Jan 19 '17 at 5:58










          • $begingroup$
            I didn't claim that the centroids popped out from stone but just point out your mistakes. The key point is how to apply the Pappus' theorem. Lazy guys love wikipedia so that no needs to do everything from scratch.
            $endgroup$
            – Ng Chung Tak
            Jan 19 '17 at 11:09












          • $begingroup$
            Where are the mistakes? You just put the value of $h$ and simplified the expressions.
            $endgroup$
            – Ahmed Bilâl
            Jan 19 '17 at 11:14










          • $begingroup$
            You say $dfrac{2r}{3}$
            $endgroup$
            – Ng Chung Tak
            Jan 19 '17 at 11:17










          • $begingroup$
            Ok. I said the centroid of a semicircle of radius 2r/3 is the same point that is the centroid of a semidisk of radius r when their centers coincide. Put another way, if they have same radius, disk's centroid should be at a distance above center 2/3 times of the circle's. My english suffers a lot so sorry for that.
            $endgroup$
            – Ahmed Bilâl
            Jan 19 '17 at 11:34
















          0












          $begingroup$

          The centroid of semi-circle should be $dfrac{4r}{3pi}$ from the origin.



          So the volume is



          $$V=2pi times frac{4r}{3pi}times frac{pi r^2}{2}=frac{4}{3}pi r^3$$



          The centroid of semi-circular arc should be $dfrac{2r}{pi}$ from the origin.



          So the surface area is



          $$S=2pi times frac{2r}{pi}times pi r=4pi r^2$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Yeah but how to get those centroids mathematically without using calculus or hanging them or something?
            $endgroup$
            – Ahmed Bilâl
            Jan 19 '17 at 5:58










          • $begingroup$
            I didn't claim that the centroids popped out from stone but just point out your mistakes. The key point is how to apply the Pappus' theorem. Lazy guys love wikipedia so that no needs to do everything from scratch.
            $endgroup$
            – Ng Chung Tak
            Jan 19 '17 at 11:09












          • $begingroup$
            Where are the mistakes? You just put the value of $h$ and simplified the expressions.
            $endgroup$
            – Ahmed Bilâl
            Jan 19 '17 at 11:14










          • $begingroup$
            You say $dfrac{2r}{3}$
            $endgroup$
            – Ng Chung Tak
            Jan 19 '17 at 11:17










          • $begingroup$
            Ok. I said the centroid of a semicircle of radius 2r/3 is the same point that is the centroid of a semidisk of radius r when their centers coincide. Put another way, if they have same radius, disk's centroid should be at a distance above center 2/3 times of the circle's. My english suffers a lot so sorry for that.
            $endgroup$
            – Ahmed Bilâl
            Jan 19 '17 at 11:34














          0












          0








          0





          $begingroup$

          The centroid of semi-circle should be $dfrac{4r}{3pi}$ from the origin.



          So the volume is



          $$V=2pi times frac{4r}{3pi}times frac{pi r^2}{2}=frac{4}{3}pi r^3$$



          The centroid of semi-circular arc should be $dfrac{2r}{pi}$ from the origin.



          So the surface area is



          $$S=2pi times frac{2r}{pi}times pi r=4pi r^2$$






          share|cite|improve this answer









          $endgroup$



          The centroid of semi-circle should be $dfrac{4r}{3pi}$ from the origin.



          So the volume is



          $$V=2pi times frac{4r}{3pi}times frac{pi r^2}{2}=frac{4}{3}pi r^3$$



          The centroid of semi-circular arc should be $dfrac{2r}{pi}$ from the origin.



          So the surface area is



          $$S=2pi times frac{2r}{pi}times pi r=4pi r^2$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 18 '17 at 23:03









          Ng Chung TakNg Chung Tak

          14.6k31334




          14.6k31334












          • $begingroup$
            Yeah but how to get those centroids mathematically without using calculus or hanging them or something?
            $endgroup$
            – Ahmed Bilâl
            Jan 19 '17 at 5:58










          • $begingroup$
            I didn't claim that the centroids popped out from stone but just point out your mistakes. The key point is how to apply the Pappus' theorem. Lazy guys love wikipedia so that no needs to do everything from scratch.
            $endgroup$
            – Ng Chung Tak
            Jan 19 '17 at 11:09












          • $begingroup$
            Where are the mistakes? You just put the value of $h$ and simplified the expressions.
            $endgroup$
            – Ahmed Bilâl
            Jan 19 '17 at 11:14










          • $begingroup$
            You say $dfrac{2r}{3}$
            $endgroup$
            – Ng Chung Tak
            Jan 19 '17 at 11:17










          • $begingroup$
            Ok. I said the centroid of a semicircle of radius 2r/3 is the same point that is the centroid of a semidisk of radius r when their centers coincide. Put another way, if they have same radius, disk's centroid should be at a distance above center 2/3 times of the circle's. My english suffers a lot so sorry for that.
            $endgroup$
            – Ahmed Bilâl
            Jan 19 '17 at 11:34


















          • $begingroup$
            Yeah but how to get those centroids mathematically without using calculus or hanging them or something?
            $endgroup$
            – Ahmed Bilâl
            Jan 19 '17 at 5:58










          • $begingroup$
            I didn't claim that the centroids popped out from stone but just point out your mistakes. The key point is how to apply the Pappus' theorem. Lazy guys love wikipedia so that no needs to do everything from scratch.
            $endgroup$
            – Ng Chung Tak
            Jan 19 '17 at 11:09












          • $begingroup$
            Where are the mistakes? You just put the value of $h$ and simplified the expressions.
            $endgroup$
            – Ahmed Bilâl
            Jan 19 '17 at 11:14










          • $begingroup$
            You say $dfrac{2r}{3}$
            $endgroup$
            – Ng Chung Tak
            Jan 19 '17 at 11:17










          • $begingroup$
            Ok. I said the centroid of a semicircle of radius 2r/3 is the same point that is the centroid of a semidisk of radius r when their centers coincide. Put another way, if they have same radius, disk's centroid should be at a distance above center 2/3 times of the circle's. My english suffers a lot so sorry for that.
            $endgroup$
            – Ahmed Bilâl
            Jan 19 '17 at 11:34
















          $begingroup$
          Yeah but how to get those centroids mathematically without using calculus or hanging them or something?
          $endgroup$
          – Ahmed Bilâl
          Jan 19 '17 at 5:58




          $begingroup$
          Yeah but how to get those centroids mathematically without using calculus or hanging them or something?
          $endgroup$
          – Ahmed Bilâl
          Jan 19 '17 at 5:58












          $begingroup$
          I didn't claim that the centroids popped out from stone but just point out your mistakes. The key point is how to apply the Pappus' theorem. Lazy guys love wikipedia so that no needs to do everything from scratch.
          $endgroup$
          – Ng Chung Tak
          Jan 19 '17 at 11:09






          $begingroup$
          I didn't claim that the centroids popped out from stone but just point out your mistakes. The key point is how to apply the Pappus' theorem. Lazy guys love wikipedia so that no needs to do everything from scratch.
          $endgroup$
          – Ng Chung Tak
          Jan 19 '17 at 11:09














          $begingroup$
          Where are the mistakes? You just put the value of $h$ and simplified the expressions.
          $endgroup$
          – Ahmed Bilâl
          Jan 19 '17 at 11:14




          $begingroup$
          Where are the mistakes? You just put the value of $h$ and simplified the expressions.
          $endgroup$
          – Ahmed Bilâl
          Jan 19 '17 at 11:14












          $begingroup$
          You say $dfrac{2r}{3}$
          $endgroup$
          – Ng Chung Tak
          Jan 19 '17 at 11:17




          $begingroup$
          You say $dfrac{2r}{3}$
          $endgroup$
          – Ng Chung Tak
          Jan 19 '17 at 11:17












          $begingroup$
          Ok. I said the centroid of a semicircle of radius 2r/3 is the same point that is the centroid of a semidisk of radius r when their centers coincide. Put another way, if they have same radius, disk's centroid should be at a distance above center 2/3 times of the circle's. My english suffers a lot so sorry for that.
          $endgroup$
          – Ahmed Bilâl
          Jan 19 '17 at 11:34




          $begingroup$
          Ok. I said the centroid of a semicircle of radius 2r/3 is the same point that is the centroid of a semidisk of radius r when their centers coincide. Put another way, if they have same radius, disk's centroid should be at a distance above center 2/3 times of the circle's. My english suffers a lot so sorry for that.
          $endgroup$
          – Ahmed Bilâl
          Jan 19 '17 at 11:34


















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