Rigorous Geometric Proof That dA=rdrdθ?












5












$begingroup$


I've seen the same question as above for both the geometric derivation of polar coordinates and the x-y transformation of polar coordinates. The first answer states that:



"In the geometric approach, dr2=0 as it is not only small but also symmetric (see here)." And links to a wikipedia article on exterior algebra. Could someone clarify this for me? To reiterate:



Given the polar coordinates of a 'circular wedge'
Geometrically, the exact area would be
$$frac{(r+dr)^2dθ}{2}−frac{r^2dθ}{2}$$
$$=(r+frac{dr}{2})drdθ$$
$$= r dr dtheta + frac{dr^2 dtheta}{2}$$
How do we get rid of $frac{dr^2dθ}{2}$? Why are we allowed to consider it insiginficant to the point where we can ignore it as opposed to keeping the tiny values of $dr$ and $dtheta$






calculus integration polar-coordinates







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  • 1




    $begingroup$
    at small enough $Delta r$ and $Deltatheta$ (i.e. $dr,dtheta$) we can visualise the area as a rectangle of side length $dr$ and $rdtheta$. Hence the area is $rdrdtheta$
    $endgroup$
    – Henry Lee
    11 hours ago












  • $begingroup$
    You are asking for a rigorous proof, but you haven't provided a rigorous definition of differentials. What you ask for isn't truly possible until you define differentials and their properties in a rigorous manner.
    $endgroup$
    – Brevan Ellefsen
    3 hours ago
















5












$begingroup$


I've seen the same question as above for both the geometric derivation of polar coordinates and the x-y transformation of polar coordinates. The first answer states that:



"In the geometric approach, dr2=0 as it is not only small but also symmetric (see here)." And links to a wikipedia article on exterior algebra. Could someone clarify this for me? To reiterate:



Given the polar coordinates of a 'circular wedge'
Geometrically, the exact area would be
$$frac{(r+dr)^2dθ}{2}−frac{r^2dθ}{2}$$
$$=(r+frac{dr}{2})drdθ$$
$$= r dr dtheta + frac{dr^2 dtheta}{2}$$
How do we get rid of $frac{dr^2dθ}{2}$? Why are we allowed to consider it insiginficant to the point where we can ignore it as opposed to keeping the tiny values of $dr$ and $dtheta$






calculus integration polar-coordinates







share|cite|improve this question







New contributor




user2471881 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 1




    $begingroup$
    at small enough $Delta r$ and $Deltatheta$ (i.e. $dr,dtheta$) we can visualise the area as a rectangle of side length $dr$ and $rdtheta$. Hence the area is $rdrdtheta$
    $endgroup$
    – Henry Lee
    11 hours ago












  • $begingroup$
    You are asking for a rigorous proof, but you haven't provided a rigorous definition of differentials. What you ask for isn't truly possible until you define differentials and their properties in a rigorous manner.
    $endgroup$
    – Brevan Ellefsen
    3 hours ago














5












5








5


3



$begingroup$


I've seen the same question as above for both the geometric derivation of polar coordinates and the x-y transformation of polar coordinates. The first answer states that:



"In the geometric approach, dr2=0 as it is not only small but also symmetric (see here)." And links to a wikipedia article on exterior algebra. Could someone clarify this for me? To reiterate:



Given the polar coordinates of a 'circular wedge'
Geometrically, the exact area would be
$$frac{(r+dr)^2dθ}{2}−frac{r^2dθ}{2}$$
$$=(r+frac{dr}{2})drdθ$$
$$= r dr dtheta + frac{dr^2 dtheta}{2}$$
How do we get rid of $frac{dr^2dθ}{2}$? Why are we allowed to consider it insiginficant to the point where we can ignore it as opposed to keeping the tiny values of $dr$ and $dtheta$






calculus integration polar-coordinates







share|cite|improve this question







New contributor




user2471881 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




I've seen the same question as above for both the geometric derivation of polar coordinates and the x-y transformation of polar coordinates. The first answer states that:



"In the geometric approach, dr2=0 as it is not only small but also symmetric (see here)." And links to a wikipedia article on exterior algebra. Could someone clarify this for me? To reiterate:



Given the polar coordinates of a 'circular wedge'
Geometrically, the exact area would be
$$frac{(r+dr)^2dθ}{2}−frac{r^2dθ}{2}$$
$$=(r+frac{dr}{2})drdθ$$
$$= r dr dtheta + frac{dr^2 dtheta}{2}$$
How do we get rid of $frac{dr^2dθ}{2}$? Why are we allowed to consider it insiginficant to the point where we can ignore it as opposed to keeping the tiny values of $dr$ and $dtheta$






calculus integration polar-coordinates




calculus integration polar-coordinates






share|cite|improve this question







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Check out our Code of Conduct.











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user2471881 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









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asked 11 hours ago









user2471881user2471881

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  • 1




    $begingroup$
    at small enough $Delta r$ and $Deltatheta$ (i.e. $dr,dtheta$) we can visualise the area as a rectangle of side length $dr$ and $rdtheta$. Hence the area is $rdrdtheta$
    $endgroup$
    – Henry Lee
    11 hours ago












  • $begingroup$
    You are asking for a rigorous proof, but you haven't provided a rigorous definition of differentials. What you ask for isn't truly possible until you define differentials and their properties in a rigorous manner.
    $endgroup$
    – Brevan Ellefsen
    3 hours ago














  • 1




    $begingroup$
    at small enough $Delta r$ and $Deltatheta$ (i.e. $dr,dtheta$) we can visualise the area as a rectangle of side length $dr$ and $rdtheta$. Hence the area is $rdrdtheta$
    $endgroup$
    – Henry Lee
    11 hours ago












  • $begingroup$
    You are asking for a rigorous proof, but you haven't provided a rigorous definition of differentials. What you ask for isn't truly possible until you define differentials and their properties in a rigorous manner.
    $endgroup$
    – Brevan Ellefsen
    3 hours ago








1




1




$begingroup$
at small enough $Delta r$ and $Deltatheta$ (i.e. $dr,dtheta$) we can visualise the area as a rectangle of side length $dr$ and $rdtheta$. Hence the area is $rdrdtheta$
$endgroup$
– Henry Lee
11 hours ago






$begingroup$
at small enough $Delta r$ and $Deltatheta$ (i.e. $dr,dtheta$) we can visualise the area as a rectangle of side length $dr$ and $rdtheta$. Hence the area is $rdrdtheta$
$endgroup$
– Henry Lee
11 hours ago














$begingroup$
You are asking for a rigorous proof, but you haven't provided a rigorous definition of differentials. What you ask for isn't truly possible until you define differentials and their properties in a rigorous manner.
$endgroup$
– Brevan Ellefsen
3 hours ago




$begingroup$
You are asking for a rigorous proof, but you haven't provided a rigorous definition of differentials. What you ask for isn't truly possible until you define differentials and their properties in a rigorous manner.
$endgroup$
– Brevan Ellefsen
3 hours ago










3 Answers
3






active

oldest

votes


















6












$begingroup$

The meaning of $dA=r dr dtheta$ is that when you integrate over a region with respect to $dA$, you get the same result as you do by parametrizing it in polar coordinates and integrating with respect to the differential $r dr d theta$.



Thus your question (because $dtheta$ is not a problem) reduces to showing that "integrating" against a differential like "$(dr)^2$" gives zero. What this actually means is that sums of the form $S=sum_i f(r_i) (Delta r)_i^2$ should all go to zero as you refine the partition.



For bounded $f$ for example $|S| leq MLdelta$ where $M$ is the bound on $|f|$, $L$ is the total length of the interval, and $delta$ is the longest of the $(Delta r)_i$'s. This indeed goes to zero as $delta to 0$.



The point can be understood without inspecting all the details by just counting how many terms there are (call that $n$) and how small the individual terms are relative to that (which is on the order of $1/n^2$).






share|cite|improve this answer











$endgroup$





















    3












    $begingroup$

    $$lim_{dtheta,drto0}frac{r,dr,dtheta+frac12dr^2dtheta}{dr,dtheta}=lim_{dtheta,drto0}(r+tfrac12dr)=r.$$






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      If we take $dr$ to be very small then $dr^2$ is so much smaller than $dr$ that it can be ignored. More rigorously we have
      $$lim_{drto 0} frac{dr^2}{dr}= lim_{drto 0} dr = 0$$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        "more rigorously we have [...]": proceeds to cancel out infinitesimals
        $endgroup$
        – FreeSalad
        1 hour ago













      Your Answer





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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      6












      $begingroup$

      The meaning of $dA=r dr dtheta$ is that when you integrate over a region with respect to $dA$, you get the same result as you do by parametrizing it in polar coordinates and integrating with respect to the differential $r dr d theta$.



      Thus your question (because $dtheta$ is not a problem) reduces to showing that "integrating" against a differential like "$(dr)^2$" gives zero. What this actually means is that sums of the form $S=sum_i f(r_i) (Delta r)_i^2$ should all go to zero as you refine the partition.



      For bounded $f$ for example $|S| leq MLdelta$ where $M$ is the bound on $|f|$, $L$ is the total length of the interval, and $delta$ is the longest of the $(Delta r)_i$'s. This indeed goes to zero as $delta to 0$.



      The point can be understood without inspecting all the details by just counting how many terms there are (call that $n$) and how small the individual terms are relative to that (which is on the order of $1/n^2$).






      share|cite|improve this answer











      $endgroup$


















        6












        $begingroup$

        The meaning of $dA=r dr dtheta$ is that when you integrate over a region with respect to $dA$, you get the same result as you do by parametrizing it in polar coordinates and integrating with respect to the differential $r dr d theta$.



        Thus your question (because $dtheta$ is not a problem) reduces to showing that "integrating" against a differential like "$(dr)^2$" gives zero. What this actually means is that sums of the form $S=sum_i f(r_i) (Delta r)_i^2$ should all go to zero as you refine the partition.



        For bounded $f$ for example $|S| leq MLdelta$ where $M$ is the bound on $|f|$, $L$ is the total length of the interval, and $delta$ is the longest of the $(Delta r)_i$'s. This indeed goes to zero as $delta to 0$.



        The point can be understood without inspecting all the details by just counting how many terms there are (call that $n$) and how small the individual terms are relative to that (which is on the order of $1/n^2$).






        share|cite|improve this answer











        $endgroup$
















          6












          6








          6





          $begingroup$

          The meaning of $dA=r dr dtheta$ is that when you integrate over a region with respect to $dA$, you get the same result as you do by parametrizing it in polar coordinates and integrating with respect to the differential $r dr d theta$.



          Thus your question (because $dtheta$ is not a problem) reduces to showing that "integrating" against a differential like "$(dr)^2$" gives zero. What this actually means is that sums of the form $S=sum_i f(r_i) (Delta r)_i^2$ should all go to zero as you refine the partition.



          For bounded $f$ for example $|S| leq MLdelta$ where $M$ is the bound on $|f|$, $L$ is the total length of the interval, and $delta$ is the longest of the $(Delta r)_i$'s. This indeed goes to zero as $delta to 0$.



          The point can be understood without inspecting all the details by just counting how many terms there are (call that $n$) and how small the individual terms are relative to that (which is on the order of $1/n^2$).






          share|cite|improve this answer











          $endgroup$



          The meaning of $dA=r dr dtheta$ is that when you integrate over a region with respect to $dA$, you get the same result as you do by parametrizing it in polar coordinates and integrating with respect to the differential $r dr d theta$.



          Thus your question (because $dtheta$ is not a problem) reduces to showing that "integrating" against a differential like "$(dr)^2$" gives zero. What this actually means is that sums of the form $S=sum_i f(r_i) (Delta r)_i^2$ should all go to zero as you refine the partition.



          For bounded $f$ for example $|S| leq MLdelta$ where $M$ is the bound on $|f|$, $L$ is the total length of the interval, and $delta$ is the longest of the $(Delta r)_i$'s. This indeed goes to zero as $delta to 0$.



          The point can be understood without inspecting all the details by just counting how many terms there are (call that $n$) and how small the individual terms are relative to that (which is on the order of $1/n^2$).







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 8 hours ago

























          answered 11 hours ago









          IanIan

          68.3k25388




          68.3k25388























              3












              $begingroup$

              $$lim_{dtheta,drto0}frac{r,dr,dtheta+frac12dr^2dtheta}{dr,dtheta}=lim_{dtheta,drto0}(r+tfrac12dr)=r.$$






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                $$lim_{dtheta,drto0}frac{r,dr,dtheta+frac12dr^2dtheta}{dr,dtheta}=lim_{dtheta,drto0}(r+tfrac12dr)=r.$$






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  $$lim_{dtheta,drto0}frac{r,dr,dtheta+frac12dr^2dtheta}{dr,dtheta}=lim_{dtheta,drto0}(r+tfrac12dr)=r.$$






                  share|cite|improve this answer









                  $endgroup$



                  $$lim_{dtheta,drto0}frac{r,dr,dtheta+frac12dr^2dtheta}{dr,dtheta}=lim_{dtheta,drto0}(r+tfrac12dr)=r.$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 11 hours ago









                  Yves DaoustYves Daoust

                  128k675227




                  128k675227























                      0












                      $begingroup$

                      If we take $dr$ to be very small then $dr^2$ is so much smaller than $dr$ that it can be ignored. More rigorously we have
                      $$lim_{drto 0} frac{dr^2}{dr}= lim_{drto 0} dr = 0$$






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        "more rigorously we have [...]": proceeds to cancel out infinitesimals
                        $endgroup$
                        – FreeSalad
                        1 hour ago


















                      0












                      $begingroup$

                      If we take $dr$ to be very small then $dr^2$ is so much smaller than $dr$ that it can be ignored. More rigorously we have
                      $$lim_{drto 0} frac{dr^2}{dr}= lim_{drto 0} dr = 0$$






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        "more rigorously we have [...]": proceeds to cancel out infinitesimals
                        $endgroup$
                        – FreeSalad
                        1 hour ago
















                      0












                      0








                      0





                      $begingroup$

                      If we take $dr$ to be very small then $dr^2$ is so much smaller than $dr$ that it can be ignored. More rigorously we have
                      $$lim_{drto 0} frac{dr^2}{dr}= lim_{drto 0} dr = 0$$






                      share|cite|improve this answer









                      $endgroup$



                      If we take $dr$ to be very small then $dr^2$ is so much smaller than $dr$ that it can be ignored. More rigorously we have
                      $$lim_{drto 0} frac{dr^2}{dr}= lim_{drto 0} dr = 0$$







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered 11 hours ago









                      Peter ForemanPeter Foreman

                      2,17713




                      2,17713












                      • $begingroup$
                        "more rigorously we have [...]": proceeds to cancel out infinitesimals
                        $endgroup$
                        – FreeSalad
                        1 hour ago




















                      • $begingroup$
                        "more rigorously we have [...]": proceeds to cancel out infinitesimals
                        $endgroup$
                        – FreeSalad
                        1 hour ago


















                      $begingroup$
                      "more rigorously we have [...]": proceeds to cancel out infinitesimals
                      $endgroup$
                      – FreeSalad
                      1 hour ago






                      $begingroup$
                      "more rigorously we have [...]": proceeds to cancel out infinitesimals
                      $endgroup$
                      – FreeSalad
                      1 hour ago












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