area of part of Archimedes's spiral












6












$begingroup$



Find the area of region inside the "first loop" of the Archimedes spiral (that is, the spiral for $0 le theta le 2pi$) and to the left of the $y$-axis.




The area the question wants is between $theta = pi/2$ and $theta = 3pi/2$ for the graph $r=theta$. Therefore, I computed the integral $int_{pi/2}^{3pi/2} theta ,dtheta = pi^2$. I even checked it with a graphing calculator to make sure that the integral was computed correctly. However, apparently, this is not a correct answer. Could someone help me see why?






calculus integration







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$endgroup$








  • 2




    $begingroup$
    Can you explain why you wrote the area as $inttheta dtheta$? Are you familiar with double integrals?
    $endgroup$
    – Shubham Johri
    2 days ago








  • 2




    $begingroup$
    Isn't the area element $r dr dtheta$ in polar?
    $endgroup$
    – coffeemath
    2 days ago












  • $begingroup$
    The integral is $int int r dtheta dr = frac{1}{2}int theta^2 dtheta$, you forgot to integrate $r$ first from zero to $theta$.
    $endgroup$
    – WarreG
    2 days ago
















6












$begingroup$



Find the area of region inside the "first loop" of the Archimedes spiral (that is, the spiral for $0 le theta le 2pi$) and to the left of the $y$-axis.




The area the question wants is between $theta = pi/2$ and $theta = 3pi/2$ for the graph $r=theta$. Therefore, I computed the integral $int_{pi/2}^{3pi/2} theta ,dtheta = pi^2$. I even checked it with a graphing calculator to make sure that the integral was computed correctly. However, apparently, this is not a correct answer. Could someone help me see why?






calculus integration







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Can you explain why you wrote the area as $inttheta dtheta$? Are you familiar with double integrals?
    $endgroup$
    – Shubham Johri
    2 days ago








  • 2




    $begingroup$
    Isn't the area element $r dr dtheta$ in polar?
    $endgroup$
    – coffeemath
    2 days ago












  • $begingroup$
    The integral is $int int r dtheta dr = frac{1}{2}int theta^2 dtheta$, you forgot to integrate $r$ first from zero to $theta$.
    $endgroup$
    – WarreG
    2 days ago














6












6








6


1



$begingroup$



Find the area of region inside the "first loop" of the Archimedes spiral (that is, the spiral for $0 le theta le 2pi$) and to the left of the $y$-axis.




The area the question wants is between $theta = pi/2$ and $theta = 3pi/2$ for the graph $r=theta$. Therefore, I computed the integral $int_{pi/2}^{3pi/2} theta ,dtheta = pi^2$. I even checked it with a graphing calculator to make sure that the integral was computed correctly. However, apparently, this is not a correct answer. Could someone help me see why?






calculus integration







share|cite|improve this question









$endgroup$





Find the area of region inside the "first loop" of the Archimedes spiral (that is, the spiral for $0 le theta le 2pi$) and to the left of the $y$-axis.




The area the question wants is between $theta = pi/2$ and $theta = 3pi/2$ for the graph $r=theta$. Therefore, I computed the integral $int_{pi/2}^{3pi/2} theta ,dtheta = pi^2$. I even checked it with a graphing calculator to make sure that the integral was computed correctly. However, apparently, this is not a correct answer. Could someone help me see why?






calculus integration




calculus integration






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 2 days ago









jjhhjjhh

2,09611121




2,09611121








  • 2




    $begingroup$
    Can you explain why you wrote the area as $inttheta dtheta$? Are you familiar with double integrals?
    $endgroup$
    – Shubham Johri
    2 days ago








  • 2




    $begingroup$
    Isn't the area element $r dr dtheta$ in polar?
    $endgroup$
    – coffeemath
    2 days ago












  • $begingroup$
    The integral is $int int r dtheta dr = frac{1}{2}int theta^2 dtheta$, you forgot to integrate $r$ first from zero to $theta$.
    $endgroup$
    – WarreG
    2 days ago














  • 2




    $begingroup$
    Can you explain why you wrote the area as $inttheta dtheta$? Are you familiar with double integrals?
    $endgroup$
    – Shubham Johri
    2 days ago








  • 2




    $begingroup$
    Isn't the area element $r dr dtheta$ in polar?
    $endgroup$
    – coffeemath
    2 days ago












  • $begingroup$
    The integral is $int int r dtheta dr = frac{1}{2}int theta^2 dtheta$, you forgot to integrate $r$ first from zero to $theta$.
    $endgroup$
    – WarreG
    2 days ago








2




2




$begingroup$
Can you explain why you wrote the area as $inttheta dtheta$? Are you familiar with double integrals?
$endgroup$
– Shubham Johri
2 days ago






$begingroup$
Can you explain why you wrote the area as $inttheta dtheta$? Are you familiar with double integrals?
$endgroup$
– Shubham Johri
2 days ago






2




2




$begingroup$
Isn't the area element $r dr dtheta$ in polar?
$endgroup$
– coffeemath
2 days ago






$begingroup$
Isn't the area element $r dr dtheta$ in polar?
$endgroup$
– coffeemath
2 days ago














$begingroup$
The integral is $int int r dtheta dr = frac{1}{2}int theta^2 dtheta$, you forgot to integrate $r$ first from zero to $theta$.
$endgroup$
– WarreG
2 days ago




$begingroup$
The integral is $int int r dtheta dr = frac{1}{2}int theta^2 dtheta$, you forgot to integrate $r$ first from zero to $theta$.
$endgroup$
– WarreG
2 days ago










2 Answers
2






active

oldest

votes


















7












$begingroup$

Recall that the area element in polar coordinates is given by $r dr dtheta$. For any value of $theta,r$ ranges from $0totheta$. Further, $theta$ ranges from $pi/2to3pi/2$.



The answer is $$int_{pi/2}^{3pi/2}int_0^theta r dr dtheta=frac{13pi^3}{24}$$



Graph






share|cite|improve this answer









$endgroup$





















    5












    $begingroup$

    The integral for finding the area in polar coordinate is different from what you have.



    Please use the correct formula and you will get the correct answer.



    $$ A = (1/2) int r^2 dtheta $$ where in your case $ r=theta $






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      An extra $pi$ there
      $endgroup$
      – Shubham Johri
      2 days ago






    • 1




      $begingroup$
      Thanks, I fixed it, now it is correct.
      $endgroup$
      – Mohammad Riazi-Kermani
      2 days ago











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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    7












    $begingroup$

    Recall that the area element in polar coordinates is given by $r dr dtheta$. For any value of $theta,r$ ranges from $0totheta$. Further, $theta$ ranges from $pi/2to3pi/2$.



    The answer is $$int_{pi/2}^{3pi/2}int_0^theta r dr dtheta=frac{13pi^3}{24}$$



    Graph






    share|cite|improve this answer









    $endgroup$


















      7












      $begingroup$

      Recall that the area element in polar coordinates is given by $r dr dtheta$. For any value of $theta,r$ ranges from $0totheta$. Further, $theta$ ranges from $pi/2to3pi/2$.



      The answer is $$int_{pi/2}^{3pi/2}int_0^theta r dr dtheta=frac{13pi^3}{24}$$



      Graph






      share|cite|improve this answer









      $endgroup$
















        7












        7








        7





        $begingroup$

        Recall that the area element in polar coordinates is given by $r dr dtheta$. For any value of $theta,r$ ranges from $0totheta$. Further, $theta$ ranges from $pi/2to3pi/2$.



        The answer is $$int_{pi/2}^{3pi/2}int_0^theta r dr dtheta=frac{13pi^3}{24}$$



        Graph






        share|cite|improve this answer









        $endgroup$



        Recall that the area element in polar coordinates is given by $r dr dtheta$. For any value of $theta,r$ ranges from $0totheta$. Further, $theta$ ranges from $pi/2to3pi/2$.



        The answer is $$int_{pi/2}^{3pi/2}int_0^theta r dr dtheta=frac{13pi^3}{24}$$



        Graph







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 days ago









        Shubham JohriShubham Johri

        4,666717




        4,666717























            5












            $begingroup$

            The integral for finding the area in polar coordinate is different from what you have.



            Please use the correct formula and you will get the correct answer.



            $$ A = (1/2) int r^2 dtheta $$ where in your case $ r=theta $






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              An extra $pi$ there
              $endgroup$
              – Shubham Johri
              2 days ago






            • 1




              $begingroup$
              Thanks, I fixed it, now it is correct.
              $endgroup$
              – Mohammad Riazi-Kermani
              2 days ago
















            5












            $begingroup$

            The integral for finding the area in polar coordinate is different from what you have.



            Please use the correct formula and you will get the correct answer.



            $$ A = (1/2) int r^2 dtheta $$ where in your case $ r=theta $






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              An extra $pi$ there
              $endgroup$
              – Shubham Johri
              2 days ago






            • 1




              $begingroup$
              Thanks, I fixed it, now it is correct.
              $endgroup$
              – Mohammad Riazi-Kermani
              2 days ago














            5












            5








            5





            $begingroup$

            The integral for finding the area in polar coordinate is different from what you have.



            Please use the correct formula and you will get the correct answer.



            $$ A = (1/2) int r^2 dtheta $$ where in your case $ r=theta $






            share|cite|improve this answer











            $endgroup$



            The integral for finding the area in polar coordinate is different from what you have.



            Please use the correct formula and you will get the correct answer.



            $$ A = (1/2) int r^2 dtheta $$ where in your case $ r=theta $







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 2 days ago

























            answered 2 days ago









            Mohammad Riazi-KermaniMohammad Riazi-Kermani

            41.5k42061




            41.5k42061












            • $begingroup$
              An extra $pi$ there
              $endgroup$
              – Shubham Johri
              2 days ago






            • 1




              $begingroup$
              Thanks, I fixed it, now it is correct.
              $endgroup$
              – Mohammad Riazi-Kermani
              2 days ago


















            • $begingroup$
              An extra $pi$ there
              $endgroup$
              – Shubham Johri
              2 days ago






            • 1




              $begingroup$
              Thanks, I fixed it, now it is correct.
              $endgroup$
              – Mohammad Riazi-Kermani
              2 days ago
















            $begingroup$
            An extra $pi$ there
            $endgroup$
            – Shubham Johri
            2 days ago




            $begingroup$
            An extra $pi$ there
            $endgroup$
            – Shubham Johri
            2 days ago




            1




            1




            $begingroup$
            Thanks, I fixed it, now it is correct.
            $endgroup$
            – Mohammad Riazi-Kermani
            2 days ago




            $begingroup$
            Thanks, I fixed it, now it is correct.
            $endgroup$
            – Mohammad Riazi-Kermani
            2 days ago


















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