area of part of Archimedes's spiral












6















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


















  • 2




    Can you explain why you wrote the area as $inttheta dtheta$? Are you familiar with double integrals?
    – Shubham Johri
    18 hours ago








  • 2




    Isn't the area element $r dr dtheta$ in polar?
    – coffeemath
    18 hours ago












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
















6















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


















  • 2




    Can you explain why you wrote the area as $inttheta dtheta$? Are you familiar with double integrals?
    – Shubham Johri
    18 hours ago








  • 2




    Isn't the area element $r dr dtheta$ in polar?
    – coffeemath
    18 hours ago












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














6












6








6


1






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














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 18 hours ago









jjhhjjhh

2,09611121




2,09611121








  • 2




    Can you explain why you wrote the area as $inttheta dtheta$? Are you familiar with double integrals?
    – Shubham Johri
    18 hours ago








  • 2




    Isn't the area element $r dr dtheta$ in polar?
    – coffeemath
    18 hours ago












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














  • 2




    Can you explain why you wrote the area as $inttheta dtheta$? Are you familiar with double integrals?
    – Shubham Johri
    18 hours ago








  • 2




    Isn't the area element $r dr dtheta$ in polar?
    – coffeemath
    18 hours ago












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








2




2




Can you explain why you wrote the area as $inttheta dtheta$? Are you familiar with double integrals?
– Shubham Johri
18 hours ago






Can you explain why you wrote the area as $inttheta dtheta$? Are you familiar with double integrals?
– Shubham Johri
18 hours ago






2




2




Isn't the area element $r dr dtheta$ in polar?
– coffeemath
18 hours ago






Isn't the area element $r dr dtheta$ in polar?
– coffeemath
18 hours ago














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




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










2 Answers
2






active

oldest

votes


















7














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





























    5














    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























    • An extra $pi$ there
      – Shubham Johri
      18 hours ago






    • 1




      Thanks, I fixed it, now it is correct.
      – Mohammad Riazi-Kermani
      18 hours ago











    Your Answer





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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    7














    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


























      7














      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
























        7












        7








        7






        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












        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 18 hours ago









        Shubham JohriShubham Johri

        4,651717




        4,651717























            5














            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























            • An extra $pi$ there
              – Shubham Johri
              18 hours ago






            • 1




              Thanks, I fixed it, now it is correct.
              – Mohammad Riazi-Kermani
              18 hours ago
















            5














            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























            • An extra $pi$ there
              – Shubham Johri
              18 hours ago






            • 1




              Thanks, I fixed it, now it is correct.
              – Mohammad Riazi-Kermani
              18 hours ago














            5












            5








            5






            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














            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 18 hours ago

























            answered 18 hours ago









            Mohammad Riazi-KermaniMohammad Riazi-Kermani

            41.4k42061




            41.4k42061












            • An extra $pi$ there
              – Shubham Johri
              18 hours ago






            • 1




              Thanks, I fixed it, now it is correct.
              – Mohammad Riazi-Kermani
              18 hours ago


















            • An extra $pi$ there
              – Shubham Johri
              18 hours ago






            • 1




              Thanks, I fixed it, now it is correct.
              – Mohammad Riazi-Kermani
              18 hours ago
















            An extra $pi$ there
            – Shubham Johri
            18 hours ago




            An extra $pi$ there
            – Shubham Johri
            18 hours ago




            1




            1




            Thanks, I fixed it, now it is correct.
            – Mohammad Riazi-Kermani
            18 hours ago




            Thanks, I fixed it, now it is correct.
            – Mohammad Riazi-Kermani
            18 hours ago


















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