Cardioid and Integration












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$begingroup$


How to find the integral of the function f(x,y) = y over the region D which is inside the cardioid r = 2 + 2 $cos theta$ and outside the circle r=2?



I am unable to set the limits of the integrals. Please explain. The answer in my textbook for this comes out to be 22/3.



Please show me the answer with steps along with proper integral limits set and reason for choosing those limits?






multivariable-calculus







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  • $begingroup$
    $theta$ must lie between -pi/2 to pi/2 which I can write by putting the outer limit to vary from 0 to pi/2 and then multiplying the area by 2 to cover the whole area. In my textbook it's being taken from 0 to pi/2 and I am confused for why the area is not multiplied by 2 as the cardioid is above the below the x axis. Please suggest the solution
    $endgroup$
    – Shashank Dwivedi
    Jan 30 at 21:59
















0












$begingroup$


How to find the integral of the function f(x,y) = y over the region D which is inside the cardioid r = 2 + 2 $cos theta$ and outside the circle r=2?



I am unable to set the limits of the integrals. Please explain. The answer in my textbook for this comes out to be 22/3.



Please show me the answer with steps along with proper integral limits set and reason for choosing those limits?






multivariable-calculus







share|cite|improve this question









$endgroup$












  • $begingroup$
    $theta$ must lie between -pi/2 to pi/2 which I can write by putting the outer limit to vary from 0 to pi/2 and then multiplying the area by 2 to cover the whole area. In my textbook it's being taken from 0 to pi/2 and I am confused for why the area is not multiplied by 2 as the cardioid is above the below the x axis. Please suggest the solution
    $endgroup$
    – Shashank Dwivedi
    Jan 30 at 21:59














0












0








0





$begingroup$


How to find the integral of the function f(x,y) = y over the region D which is inside the cardioid r = 2 + 2 $cos theta$ and outside the circle r=2?



I am unable to set the limits of the integrals. Please explain. The answer in my textbook for this comes out to be 22/3.



Please show me the answer with steps along with proper integral limits set and reason for choosing those limits?






multivariable-calculus







share|cite|improve this question









$endgroup$




How to find the integral of the function f(x,y) = y over the region D which is inside the cardioid r = 2 + 2 $cos theta$ and outside the circle r=2?



I am unable to set the limits of the integrals. Please explain. The answer in my textbook for this comes out to be 22/3.



Please show me the answer with steps along with proper integral limits set and reason for choosing those limits?






multivariable-calculus




multivariable-calculus






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 30 at 21:50









Shashank DwivediShashank Dwivedi

735




735












  • $begingroup$
    $theta$ must lie between -pi/2 to pi/2 which I can write by putting the outer limit to vary from 0 to pi/2 and then multiplying the area by 2 to cover the whole area. In my textbook it's being taken from 0 to pi/2 and I am confused for why the area is not multiplied by 2 as the cardioid is above the below the x axis. Please suggest the solution
    $endgroup$
    – Shashank Dwivedi
    Jan 30 at 21:59


















  • $begingroup$
    $theta$ must lie between -pi/2 to pi/2 which I can write by putting the outer limit to vary from 0 to pi/2 and then multiplying the area by 2 to cover the whole area. In my textbook it's being taken from 0 to pi/2 and I am confused for why the area is not multiplied by 2 as the cardioid is above the below the x axis. Please suggest the solution
    $endgroup$
    – Shashank Dwivedi
    Jan 30 at 21:59
















$begingroup$
$theta$ must lie between -pi/2 to pi/2 which I can write by putting the outer limit to vary from 0 to pi/2 and then multiplying the area by 2 to cover the whole area. In my textbook it's being taken from 0 to pi/2 and I am confused for why the area is not multiplied by 2 as the cardioid is above the below the x axis. Please suggest the solution
$endgroup$
– Shashank Dwivedi
Jan 30 at 21:59




$begingroup$
$theta$ must lie between -pi/2 to pi/2 which I can write by putting the outer limit to vary from 0 to pi/2 and then multiplying the area by 2 to cover the whole area. In my textbook it's being taken from 0 to pi/2 and I am confused for why the area is not multiplied by 2 as the cardioid is above the below the x axis. Please suggest the solution
$endgroup$
– Shashank Dwivedi
Jan 30 at 21:59










1 Answer
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$begingroup$

Your region is symmetric around the x-axis. The integral of $f(x,y) = y$ will be $0$ over any such interval.



Anyway,



convert to polar



$x = rcos theta\
y = rsintheta$



limits for theta.



$2 + 2costheta = 2\
theta = 0\
pm frac {pi}{2}$



$int_{-frac{pi}{2}}^{frac{pi}{2}}int_2^{2+2costheta} r^2sintheta dr dtheta\
int_{-frac{pi}{2}}^{frac{pi}{2}} frac 13 ((2+2costheta)^3 - 8)sintheta dr dtheta\
int_{-frac{pi}{2}}^{frac{pi}{2}}8costhetasintheta + 8cos^2thetasintheta + frac {8}{3}cos^3sintheta dtheta$



If we added the constraint that we are also limited to the upper half-plane, $frac {22}{3}$ is plausible.



$int_{0}^{frac{pi}{2}}8costhetasintheta + 8cos^2thetasintheta + frac {8}{3}cos^3sintheta dtheta = frac {8}{2} + frac{8}{3} + frac {8}{12} = frac {22}3$






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    $begingroup$

    Your region is symmetric around the x-axis. The integral of $f(x,y) = y$ will be $0$ over any such interval.



    Anyway,



    convert to polar



    $x = rcos theta\
    y = rsintheta$



    limits for theta.



    $2 + 2costheta = 2\
    theta = 0\
    pm frac {pi}{2}$



    $int_{-frac{pi}{2}}^{frac{pi}{2}}int_2^{2+2costheta} r^2sintheta dr dtheta\
    int_{-frac{pi}{2}}^{frac{pi}{2}} frac 13 ((2+2costheta)^3 - 8)sintheta dr dtheta\
    int_{-frac{pi}{2}}^{frac{pi}{2}}8costhetasintheta + 8cos^2thetasintheta + frac {8}{3}cos^3sintheta dtheta$



    If we added the constraint that we are also limited to the upper half-plane, $frac {22}{3}$ is plausible.



    $int_{0}^{frac{pi}{2}}8costhetasintheta + 8cos^2thetasintheta + frac {8}{3}cos^3sintheta dtheta = frac {8}{2} + frac{8}{3} + frac {8}{12} = frac {22}3$






    share|cite|improve this answer









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      0












      $begingroup$

      Your region is symmetric around the x-axis. The integral of $f(x,y) = y$ will be $0$ over any such interval.



      Anyway,



      convert to polar



      $x = rcos theta\
      y = rsintheta$



      limits for theta.



      $2 + 2costheta = 2\
      theta = 0\
      pm frac {pi}{2}$



      $int_{-frac{pi}{2}}^{frac{pi}{2}}int_2^{2+2costheta} r^2sintheta dr dtheta\
      int_{-frac{pi}{2}}^{frac{pi}{2}} frac 13 ((2+2costheta)^3 - 8)sintheta dr dtheta\
      int_{-frac{pi}{2}}^{frac{pi}{2}}8costhetasintheta + 8cos^2thetasintheta + frac {8}{3}cos^3sintheta dtheta$



      If we added the constraint that we are also limited to the upper half-plane, $frac {22}{3}$ is plausible.



      $int_{0}^{frac{pi}{2}}8costhetasintheta + 8cos^2thetasintheta + frac {8}{3}cos^3sintheta dtheta = frac {8}{2} + frac{8}{3} + frac {8}{12} = frac {22}3$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Your region is symmetric around the x-axis. The integral of $f(x,y) = y$ will be $0$ over any such interval.



        Anyway,



        convert to polar



        $x = rcos theta\
        y = rsintheta$



        limits for theta.



        $2 + 2costheta = 2\
        theta = 0\
        pm frac {pi}{2}$



        $int_{-frac{pi}{2}}^{frac{pi}{2}}int_2^{2+2costheta} r^2sintheta dr dtheta\
        int_{-frac{pi}{2}}^{frac{pi}{2}} frac 13 ((2+2costheta)^3 - 8)sintheta dr dtheta\
        int_{-frac{pi}{2}}^{frac{pi}{2}}8costhetasintheta + 8cos^2thetasintheta + frac {8}{3}cos^3sintheta dtheta$



        If we added the constraint that we are also limited to the upper half-plane, $frac {22}{3}$ is plausible.



        $int_{0}^{frac{pi}{2}}8costhetasintheta + 8cos^2thetasintheta + frac {8}{3}cos^3sintheta dtheta = frac {8}{2} + frac{8}{3} + frac {8}{12} = frac {22}3$






        share|cite|improve this answer









        $endgroup$



        Your region is symmetric around the x-axis. The integral of $f(x,y) = y$ will be $0$ over any such interval.



        Anyway,



        convert to polar



        $x = rcos theta\
        y = rsintheta$



        limits for theta.



        $2 + 2costheta = 2\
        theta = 0\
        pm frac {pi}{2}$



        $int_{-frac{pi}{2}}^{frac{pi}{2}}int_2^{2+2costheta} r^2sintheta dr dtheta\
        int_{-frac{pi}{2}}^{frac{pi}{2}} frac 13 ((2+2costheta)^3 - 8)sintheta dr dtheta\
        int_{-frac{pi}{2}}^{frac{pi}{2}}8costhetasintheta + 8cos^2thetasintheta + frac {8}{3}cos^3sintheta dtheta$



        If we added the constraint that we are also limited to the upper half-plane, $frac {22}{3}$ is plausible.



        $int_{0}^{frac{pi}{2}}8costhetasintheta + 8cos^2thetasintheta + frac {8}{3}cos^3sintheta dtheta = frac {8}{2} + frac{8}{3} + frac {8}{12} = frac {22}3$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 30 at 22:14









        Doug MDoug M

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