Flux through the surface of an ellipsoidCalculate the flux through a closed surfaceFlux Integral in spherical coordinatesFlux through a SurfaceEvaluating surface integral (1) directly and (2) by applying Divergence Theorem give different resolutsCompute the flux of the vector field $vecF$ through the surface SThe flux of the vector field $u = x hatimath + y hatjmath + z hatk$ through the surface of the ellipsoidFlux through surface (correct or not?)Calculation of flux through sphere when the vector field is not defined at the origincalculate flux through surfaceFlux of a hemisphere

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Flux through the surface of an ellipsoid


Calculate the flux through a closed surfaceFlux Integral in spherical coordinatesFlux through a SurfaceEvaluating surface integral (1) directly and (2) by applying Divergence Theorem give different resolutsCompute the flux of the vector field $vecF$ through the surface SThe flux of the vector field $u = x hatimath + y hatjmath + z hatk$ through the surface of the ellipsoidFlux through surface (correct or not?)Calculation of flux through sphere when the vector field is not defined at the origincalculate flux through surfaceFlux of a hemisphere













1












$begingroup$


I was asked to calculate the flux of the field $$mathbf A = (1/R^2)hat r$$
where $R$ is the radius, through the surface of the ellipsoid $$left(fracx^2a^2right) + left(fracy^2b^2right) + z^2 = 1$$



Without doing the formal calculations (I have never encountered a problem like this), I neverless suspect that it's equivalent to the flux of the same field through the surface of a sphere whose radius is the mean radius of the ellipsoid (i.e. $4 pi$)...am I wrong and/or how to solve?










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$endgroup$
















    1












    $begingroup$


    I was asked to calculate the flux of the field $$mathbf A = (1/R^2)hat r$$
    where $R$ is the radius, through the surface of the ellipsoid $$left(fracx^2a^2right) + left(fracy^2b^2right) + z^2 = 1$$



    Without doing the formal calculations (I have never encountered a problem like this), I neverless suspect that it's equivalent to the flux of the same field through the surface of a sphere whose radius is the mean radius of the ellipsoid (i.e. $4 pi$)...am I wrong and/or how to solve?










    share|cite|improve this question









    $endgroup$














      1












      1








      1


      1



      $begingroup$


      I was asked to calculate the flux of the field $$mathbf A = (1/R^2)hat r$$
      where $R$ is the radius, through the surface of the ellipsoid $$left(fracx^2a^2right) + left(fracy^2b^2right) + z^2 = 1$$



      Without doing the formal calculations (I have never encountered a problem like this), I neverless suspect that it's equivalent to the flux of the same field through the surface of a sphere whose radius is the mean radius of the ellipsoid (i.e. $4 pi$)...am I wrong and/or how to solve?










      share|cite|improve this question









      $endgroup$




      I was asked to calculate the flux of the field $$mathbf A = (1/R^2)hat r$$
      where $R$ is the radius, through the surface of the ellipsoid $$left(fracx^2a^2right) + left(fracy^2b^2right) + z^2 = 1$$



      Without doing the formal calculations (I have never encountered a problem like this), I neverless suspect that it's equivalent to the flux of the same field through the surface of a sphere whose radius is the mean radius of the ellipsoid (i.e. $4 pi$)...am I wrong and/or how to solve?







      integration multivariable-calculus vector-analysis surface-integrals






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      asked Mar 13 at 10:43









      Lo ScrondoLo Scrondo

      19410




      19410




















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          $begingroup$

          The field you are asked about is the prominent gravitational / Coulomb field which main characteristics is the fact that its divergence is $0$ everywhere except for the origin point $(0,0,0)$:
          $$
          nablacdotmathbf A=4pidelta(mathbf r),
          $$

          which is easy to check.



          Therefore by divergence theorem the flux over the surface is $4pi$ since the origin point is inside the surface.






          share|cite|improve this answer











          $endgroup$












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            $begingroup$

            The field you are asked about is the prominent gravitational / Coulomb field which main characteristics is the fact that its divergence is $0$ everywhere except for the origin point $(0,0,0)$:
            $$
            nablacdotmathbf A=4pidelta(mathbf r),
            $$

            which is easy to check.



            Therefore by divergence theorem the flux over the surface is $4pi$ since the origin point is inside the surface.






            share|cite|improve this answer











            $endgroup$

















              1












              $begingroup$

              The field you are asked about is the prominent gravitational / Coulomb field which main characteristics is the fact that its divergence is $0$ everywhere except for the origin point $(0,0,0)$:
              $$
              nablacdotmathbf A=4pidelta(mathbf r),
              $$

              which is easy to check.



              Therefore by divergence theorem the flux over the surface is $4pi$ since the origin point is inside the surface.






              share|cite|improve this answer











              $endgroup$















                1












                1








                1





                $begingroup$

                The field you are asked about is the prominent gravitational / Coulomb field which main characteristics is the fact that its divergence is $0$ everywhere except for the origin point $(0,0,0)$:
                $$
                nablacdotmathbf A=4pidelta(mathbf r),
                $$

                which is easy to check.



                Therefore by divergence theorem the flux over the surface is $4pi$ since the origin point is inside the surface.






                share|cite|improve this answer











                $endgroup$



                The field you are asked about is the prominent gravitational / Coulomb field which main characteristics is the fact that its divergence is $0$ everywhere except for the origin point $(0,0,0)$:
                $$
                nablacdotmathbf A=4pidelta(mathbf r),
                $$

                which is easy to check.



                Therefore by divergence theorem the flux over the surface is $4pi$ since the origin point is inside the surface.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Mar 13 at 14:54

























                answered Mar 13 at 14:49









                useruser

                5,45411030




                5,45411030



























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