Surface integral over a cone above the xy plane












1












$begingroup$


From Schaum's vector analysis



enter image description here



My approach:



First, parametrize the equation $x^2 +y^2 = z^2 $



$ x = rho cos phi$ , $y= rho sinphi$ , $z= rho$



Then,



$vec A = 4 rho^2 cos phi hat i + rho^4 cos phi sin phi hat j + 3 rho hat k$



$vec n = nabla S = 2x hat i + 2y hat j - 2z hat k$



Then the unit normal $ hat n = frac {1}{sqrt 2} ( cos phi hat i + sin phi hat j - hat k)$



$ vec A . hat n = frac {1}{sqrt 2} ( 4 rho^2 cos^2 phi + rho^4 cos phi sin^2 phi - 3 rho)$



Surface area projection: $dS = frac {dxdy}{| hat n . hat k|} = sqrt 2 dxdy = sqrt 2 rho d rho d phi$



$ iint_S vec A . hat n dS = iint_0^4 (4 rho^3 cos^2 phi + rho^5 cos phi sin^2 phi - 3 rho^2) d phi = int_0^{2 pi} [ rho^4 cos^2 phi + frac { rho^6}{6} cos phi sin^2 phi - rho^3 ] d phi$



then,



$ iint_S vec A . hat n dS = rho^4 [ int_0^{2 pi} cos^2 phi d phi + frac { rho^6}{6} int_0^{2 pi} cos phi sin^2 phi d phi - rho^3 int_0^{2 pi} d phi$



= $ frac { rho^4}{2} [ phi + frac {sin 2 phi}{2} ] - rho^3 [phi]$ , $ phi in [0, 2pi]$



Finally,



$ iint_S vec A . hat n dS = 2 pi [ frac {256}{2} - 64 ] = 128 pi $



However the answer is $320 pi$ , why? Where did I go wrong? Thanks.



enter image description here






vector-analysis surface-integrals







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    From Schaum's vector analysis



    enter image description here



    My approach:



    First, parametrize the equation $x^2 +y^2 = z^2 $



    $ x = rho cos phi$ , $y= rho sinphi$ , $z= rho$



    Then,



    $vec A = 4 rho^2 cos phi hat i + rho^4 cos phi sin phi hat j + 3 rho hat k$



    $vec n = nabla S = 2x hat i + 2y hat j - 2z hat k$



    Then the unit normal $ hat n = frac {1}{sqrt 2} ( cos phi hat i + sin phi hat j - hat k)$



    $ vec A . hat n = frac {1}{sqrt 2} ( 4 rho^2 cos^2 phi + rho^4 cos phi sin^2 phi - 3 rho)$



    Surface area projection: $dS = frac {dxdy}{| hat n . hat k|} = sqrt 2 dxdy = sqrt 2 rho d rho d phi$



    $ iint_S vec A . hat n dS = iint_0^4 (4 rho^3 cos^2 phi + rho^5 cos phi sin^2 phi - 3 rho^2) d phi = int_0^{2 pi} [ rho^4 cos^2 phi + frac { rho^6}{6} cos phi sin^2 phi - rho^3 ] d phi$



    then,



    $ iint_S vec A . hat n dS = rho^4 [ int_0^{2 pi} cos^2 phi d phi + frac { rho^6}{6} int_0^{2 pi} cos phi sin^2 phi d phi - rho^3 int_0^{2 pi} d phi$



    = $ frac { rho^4}{2} [ phi + frac {sin 2 phi}{2} ] - rho^3 [phi]$ , $ phi in [0, 2pi]$



    Finally,



    $ iint_S vec A . hat n dS = 2 pi [ frac {256}{2} - 64 ] = 128 pi $



    However the answer is $320 pi$ , why? Where did I go wrong? Thanks.



    enter image description here






    vector-analysis surface-integrals







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      From Schaum's vector analysis



      enter image description here



      My approach:



      First, parametrize the equation $x^2 +y^2 = z^2 $



      $ x = rho cos phi$ , $y= rho sinphi$ , $z= rho$



      Then,



      $vec A = 4 rho^2 cos phi hat i + rho^4 cos phi sin phi hat j + 3 rho hat k$



      $vec n = nabla S = 2x hat i + 2y hat j - 2z hat k$



      Then the unit normal $ hat n = frac {1}{sqrt 2} ( cos phi hat i + sin phi hat j - hat k)$



      $ vec A . hat n = frac {1}{sqrt 2} ( 4 rho^2 cos^2 phi + rho^4 cos phi sin^2 phi - 3 rho)$



      Surface area projection: $dS = frac {dxdy}{| hat n . hat k|} = sqrt 2 dxdy = sqrt 2 rho d rho d phi$



      $ iint_S vec A . hat n dS = iint_0^4 (4 rho^3 cos^2 phi + rho^5 cos phi sin^2 phi - 3 rho^2) d phi = int_0^{2 pi} [ rho^4 cos^2 phi + frac { rho^6}{6} cos phi sin^2 phi - rho^3 ] d phi$



      then,



      $ iint_S vec A . hat n dS = rho^4 [ int_0^{2 pi} cos^2 phi d phi + frac { rho^6}{6} int_0^{2 pi} cos phi sin^2 phi d phi - rho^3 int_0^{2 pi} d phi$



      = $ frac { rho^4}{2} [ phi + frac {sin 2 phi}{2} ] - rho^3 [phi]$ , $ phi in [0, 2pi]$



      Finally,



      $ iint_S vec A . hat n dS = 2 pi [ frac {256}{2} - 64 ] = 128 pi $



      However the answer is $320 pi$ , why? Where did I go wrong? Thanks.



      enter image description here






      vector-analysis surface-integrals







      share|cite|improve this question









      $endgroup$




      From Schaum's vector analysis



      enter image description here



      My approach:



      First, parametrize the equation $x^2 +y^2 = z^2 $



      $ x = rho cos phi$ , $y= rho sinphi$ , $z= rho$



      Then,



      $vec A = 4 rho^2 cos phi hat i + rho^4 cos phi sin phi hat j + 3 rho hat k$



      $vec n = nabla S = 2x hat i + 2y hat j - 2z hat k$



      Then the unit normal $ hat n = frac {1}{sqrt 2} ( cos phi hat i + sin phi hat j - hat k)$



      $ vec A . hat n = frac {1}{sqrt 2} ( 4 rho^2 cos^2 phi + rho^4 cos phi sin^2 phi - 3 rho)$



      Surface area projection: $dS = frac {dxdy}{| hat n . hat k|} = sqrt 2 dxdy = sqrt 2 rho d rho d phi$



      $ iint_S vec A . hat n dS = iint_0^4 (4 rho^3 cos^2 phi + rho^5 cos phi sin^2 phi - 3 rho^2) d phi = int_0^{2 pi} [ rho^4 cos^2 phi + frac { rho^6}{6} cos phi sin^2 phi - rho^3 ] d phi$



      then,



      $ iint_S vec A . hat n dS = rho^4 [ int_0^{2 pi} cos^2 phi d phi + frac { rho^6}{6} int_0^{2 pi} cos phi sin^2 phi d phi - rho^3 int_0^{2 pi} d phi$



      = $ frac { rho^4}{2} [ phi + frac {sin 2 phi}{2} ] - rho^3 [phi]$ , $ phi in [0, 2pi]$



      Finally,



      $ iint_S vec A . hat n dS = 2 pi [ frac {256}{2} - 64 ] = 128 pi $



      However the answer is $320 pi$ , why? Where did I go wrong? Thanks.



      enter image description here






      vector-analysis surface-integrals




      vector-analysis surface-integrals






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      asked Dec 27 '18 at 15:53









      khaled014zkhaled014z

      1749




      1749






















          2 Answers
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          1












          $begingroup$

          The problem says "the entire surface". You need to calculate the surface integral over the plane $z = 4$, $rho leq 4$ and $0leq phi leq 2pi$. It is just a matter of repeating the same you did already with



          $$
          hat{n} = hat{z}
          $$



          So that



          $$
          int {bf A}cdot {rm d}^2{bf S} = int 12r{rm d}phi {rm d}rho = 192pi
          $$



          So the total integral is



          $$
          192pi + 128pi = 320pi
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you! Can you please check out math.stackexchange.com/questions/3053525/… please?
            $endgroup$
            – khaled014z
            Dec 27 '18 at 16:27



















          0












          $begingroup$

          As you probably know, you can use the divergence theorem.



          The surface integral over the cone and the plane transforms to the volume integral over the interior of the cone. We have $operatorname{div} mathbf{A} = 4z+xz^2+3$ so
          $$intlimits_{text{surface of cone}} mathbf{A}cdot dmathbf{S} = intlimits_{text{volume of cone}}operatorname{div} mathbf{A} ,dV = intlimits_{text{volume of cone}}(4z+xz^2+3) ,dV$$
          The integral of $xz^2$ vanishes because of symmetry, and the integral of $3$ is just three times the volume of the cone, or $64pi$.



          We have
          $$intlimits_{text{volume of cone}}4z,dV = 4int_{phi = 0}^{2pi}int_{z = 0}^4int_{rho = 0}^z z, rho,drho,dz,dphi = 8pi int_{z=0}^4 frac{z^3}2,dz = 4^4pi = 256pi$$



          Hence the entire integral is $320pi$.






          share|cite|improve this answer









          $endgroup$













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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            The problem says "the entire surface". You need to calculate the surface integral over the plane $z = 4$, $rho leq 4$ and $0leq phi leq 2pi$. It is just a matter of repeating the same you did already with



            $$
            hat{n} = hat{z}
            $$



            So that



            $$
            int {bf A}cdot {rm d}^2{bf S} = int 12r{rm d}phi {rm d}rho = 192pi
            $$



            So the total integral is



            $$
            192pi + 128pi = 320pi
            $$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thank you! Can you please check out math.stackexchange.com/questions/3053525/… please?
              $endgroup$
              – khaled014z
              Dec 27 '18 at 16:27
















            1












            $begingroup$

            The problem says "the entire surface". You need to calculate the surface integral over the plane $z = 4$, $rho leq 4$ and $0leq phi leq 2pi$. It is just a matter of repeating the same you did already with



            $$
            hat{n} = hat{z}
            $$



            So that



            $$
            int {bf A}cdot {rm d}^2{bf S} = int 12r{rm d}phi {rm d}rho = 192pi
            $$



            So the total integral is



            $$
            192pi + 128pi = 320pi
            $$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thank you! Can you please check out math.stackexchange.com/questions/3053525/… please?
              $endgroup$
              – khaled014z
              Dec 27 '18 at 16:27














            1












            1








            1





            $begingroup$

            The problem says "the entire surface". You need to calculate the surface integral over the plane $z = 4$, $rho leq 4$ and $0leq phi leq 2pi$. It is just a matter of repeating the same you did already with



            $$
            hat{n} = hat{z}
            $$



            So that



            $$
            int {bf A}cdot {rm d}^2{bf S} = int 12r{rm d}phi {rm d}rho = 192pi
            $$



            So the total integral is



            $$
            192pi + 128pi = 320pi
            $$






            share|cite|improve this answer









            $endgroup$



            The problem says "the entire surface". You need to calculate the surface integral over the plane $z = 4$, $rho leq 4$ and $0leq phi leq 2pi$. It is just a matter of repeating the same you did already with



            $$
            hat{n} = hat{z}
            $$



            So that



            $$
            int {bf A}cdot {rm d}^2{bf S} = int 12r{rm d}phi {rm d}rho = 192pi
            $$



            So the total integral is



            $$
            192pi + 128pi = 320pi
            $$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 27 '18 at 16:13









            caveraccaverac

            14.6k31130




            14.6k31130












            • $begingroup$
              Thank you! Can you please check out math.stackexchange.com/questions/3053525/… please?
              $endgroup$
              – khaled014z
              Dec 27 '18 at 16:27


















            • $begingroup$
              Thank you! Can you please check out math.stackexchange.com/questions/3053525/… please?
              $endgroup$
              – khaled014z
              Dec 27 '18 at 16:27
















            $begingroup$
            Thank you! Can you please check out math.stackexchange.com/questions/3053525/… please?
            $endgroup$
            – khaled014z
            Dec 27 '18 at 16:27




            $begingroup$
            Thank you! Can you please check out math.stackexchange.com/questions/3053525/… please?
            $endgroup$
            – khaled014z
            Dec 27 '18 at 16:27











            0












            $begingroup$

            As you probably know, you can use the divergence theorem.



            The surface integral over the cone and the plane transforms to the volume integral over the interior of the cone. We have $operatorname{div} mathbf{A} = 4z+xz^2+3$ so
            $$intlimits_{text{surface of cone}} mathbf{A}cdot dmathbf{S} = intlimits_{text{volume of cone}}operatorname{div} mathbf{A} ,dV = intlimits_{text{volume of cone}}(4z+xz^2+3) ,dV$$
            The integral of $xz^2$ vanishes because of symmetry, and the integral of $3$ is just three times the volume of the cone, or $64pi$.



            We have
            $$intlimits_{text{volume of cone}}4z,dV = 4int_{phi = 0}^{2pi}int_{z = 0}^4int_{rho = 0}^z z, rho,drho,dz,dphi = 8pi int_{z=0}^4 frac{z^3}2,dz = 4^4pi = 256pi$$



            Hence the entire integral is $320pi$.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              As you probably know, you can use the divergence theorem.



              The surface integral over the cone and the plane transforms to the volume integral over the interior of the cone. We have $operatorname{div} mathbf{A} = 4z+xz^2+3$ so
              $$intlimits_{text{surface of cone}} mathbf{A}cdot dmathbf{S} = intlimits_{text{volume of cone}}operatorname{div} mathbf{A} ,dV = intlimits_{text{volume of cone}}(4z+xz^2+3) ,dV$$
              The integral of $xz^2$ vanishes because of symmetry, and the integral of $3$ is just three times the volume of the cone, or $64pi$.



              We have
              $$intlimits_{text{volume of cone}}4z,dV = 4int_{phi = 0}^{2pi}int_{z = 0}^4int_{rho = 0}^z z, rho,drho,dz,dphi = 8pi int_{z=0}^4 frac{z^3}2,dz = 4^4pi = 256pi$$



              Hence the entire integral is $320pi$.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                As you probably know, you can use the divergence theorem.



                The surface integral over the cone and the plane transforms to the volume integral over the interior of the cone. We have $operatorname{div} mathbf{A} = 4z+xz^2+3$ so
                $$intlimits_{text{surface of cone}} mathbf{A}cdot dmathbf{S} = intlimits_{text{volume of cone}}operatorname{div} mathbf{A} ,dV = intlimits_{text{volume of cone}}(4z+xz^2+3) ,dV$$
                The integral of $xz^2$ vanishes because of symmetry, and the integral of $3$ is just three times the volume of the cone, or $64pi$.



                We have
                $$intlimits_{text{volume of cone}}4z,dV = 4int_{phi = 0}^{2pi}int_{z = 0}^4int_{rho = 0}^z z, rho,drho,dz,dphi = 8pi int_{z=0}^4 frac{z^3}2,dz = 4^4pi = 256pi$$



                Hence the entire integral is $320pi$.






                share|cite|improve this answer









                $endgroup$



                As you probably know, you can use the divergence theorem.



                The surface integral over the cone and the plane transforms to the volume integral over the interior of the cone. We have $operatorname{div} mathbf{A} = 4z+xz^2+3$ so
                $$intlimits_{text{surface of cone}} mathbf{A}cdot dmathbf{S} = intlimits_{text{volume of cone}}operatorname{div} mathbf{A} ,dV = intlimits_{text{volume of cone}}(4z+xz^2+3) ,dV$$
                The integral of $xz^2$ vanishes because of symmetry, and the integral of $3$ is just three times the volume of the cone, or $64pi$.



                We have
                $$intlimits_{text{volume of cone}}4z,dV = 4int_{phi = 0}^{2pi}int_{z = 0}^4int_{rho = 0}^z z, rho,drho,dz,dphi = 8pi int_{z=0}^4 frac{z^3}2,dz = 4^4pi = 256pi$$



                Hence the entire integral is $320pi$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 29 '18 at 18:34









                mechanodroidmechanodroid

                27.6k62447




                27.6k62447






























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