2D Line Integral of a Polar Function

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We have two polar functions:



$f_1(theta) = 3+2 sin(theta)$



$f_2(theta) = 2$



and



$m(theta) = (f_1(theta)+f_2(theta))/2$



enter image description here



It is straightforward to calculate the area between the two curves as



$int_-pi/6^7pi/6 (f_1(theta)-f_2(theta)), dtheta approx 23.72657.$



I was interested, however, how this might be calculated by integrating along $m(theta)$ instead. I recall doing this in 3D in vector calculus where we'd have a curve in the $(x,y)$ plane and calculate the surface area above it. Does a 2D version of the line integral exist -- I had trouble finding anything using `2d line integral' to search.




integration polar-coordinates line-integrals




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    up vote
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    down vote

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    We have two polar functions:



    $f_1(theta) = 3+2 sin(theta)$



    $f_2(theta) = 2$



    and



    $m(theta) = (f_1(theta)+f_2(theta))/2$



    enter image description here



    It is straightforward to calculate the area between the two curves as



    $int_-pi/6^7pi/6 (f_1(theta)-f_2(theta)), dtheta approx 23.72657.$



    I was interested, however, how this might be calculated by integrating along $m(theta)$ instead. I recall doing this in 3D in vector calculus where we'd have a curve in the $(x,y)$ plane and calculate the surface area above it. Does a 2D version of the line integral exist -- I had trouble finding anything using `2d line integral' to search.




    integration polar-coordinates line-integrals




    share|cite|improve this question























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      We have two polar functions:



      $f_1(theta) = 3+2 sin(theta)$



      $f_2(theta) = 2$



      and



      $m(theta) = (f_1(theta)+f_2(theta))/2$



      enter image description here



      It is straightforward to calculate the area between the two curves as



      $int_-pi/6^7pi/6 (f_1(theta)-f_2(theta)), dtheta approx 23.72657.$



      I was interested, however, how this might be calculated by integrating along $m(theta)$ instead. I recall doing this in 3D in vector calculus where we'd have a curve in the $(x,y)$ plane and calculate the surface area above it. Does a 2D version of the line integral exist -- I had trouble finding anything using `2d line integral' to search.




      integration polar-coordinates line-integrals




      share|cite|improve this question













      We have two polar functions:



      $f_1(theta) = 3+2 sin(theta)$



      $f_2(theta) = 2$



      and



      $m(theta) = (f_1(theta)+f_2(theta))/2$



      enter image description here



      It is straightforward to calculate the area between the two curves as



      $int_-pi/6^7pi/6 (f_1(theta)-f_2(theta)), dtheta approx 23.72657.$



      I was interested, however, how this might be calculated by integrating along $m(theta)$ instead. I recall doing this in 3D in vector calculus where we'd have a curve in the $(x,y)$ plane and calculate the surface area above it. Does a 2D version of the line integral exist -- I had trouble finding anything using `2d line integral' to search.





      integration polar-coordinates line-integrals





      share|cite|improve this question












      share|cite|improve this question




      share|cite|improve this question








      edited Jul 19 at 15:09









      Adrian Keister

      3,61721533




      3,61721533









      asked Jul 19 at 15:00









      user1329307

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