Verify Gauss divergence theorem of $F=xi+yj+zk$ over the sphere $x^{2}+y^{2}+z^{2}=a^{2}$












2












$begingroup$


Verify Gauss divergence theorem of $F=xi+yj+zk$ over the sphere $x^{2}+y^{2}+z^{2}=a^{2}$
When I evaluate taking normal $N=k$, I get the answer $2pi a^{3}$.
But when I take normal to the surface by taking out gradient of $f(x,y,z)= x^{2}+y^{2}+z^{2}$ I am able to verify theorem.



But my course instructor told we can take normal $N=k$ when whenever we are able to translate system in $xy$ plane .
And if plane is given then we have to find gradient otherwise not.



Please clarify!!!






calculus multivariable-calculus vectors vector-analysis







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$endgroup$








  • 1




    $begingroup$
    $div F =3$ you should find $4pi a^3$.
    $endgroup$
    – hamam_Abdallah
    Jan 11 at 19:11






  • 1




    $begingroup$
    You can only use $vec{k}$ as the normal vector to your surface when the surface is parallel to the $xy-$plane.
    $endgroup$
    – pwerth
    Jan 11 at 19:12
















2












$begingroup$


Verify Gauss divergence theorem of $F=xi+yj+zk$ over the sphere $x^{2}+y^{2}+z^{2}=a^{2}$
When I evaluate taking normal $N=k$, I get the answer $2pi a^{3}$.
But when I take normal to the surface by taking out gradient of $f(x,y,z)= x^{2}+y^{2}+z^{2}$ I am able to verify theorem.



But my course instructor told we can take normal $N=k$ when whenever we are able to translate system in $xy$ plane .
And if plane is given then we have to find gradient otherwise not.



Please clarify!!!






calculus multivariable-calculus vectors vector-analysis







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $div F =3$ you should find $4pi a^3$.
    $endgroup$
    – hamam_Abdallah
    Jan 11 at 19:11






  • 1




    $begingroup$
    You can only use $vec{k}$ as the normal vector to your surface when the surface is parallel to the $xy-$plane.
    $endgroup$
    – pwerth
    Jan 11 at 19:12














2












2








2





$begingroup$


Verify Gauss divergence theorem of $F=xi+yj+zk$ over the sphere $x^{2}+y^{2}+z^{2}=a^{2}$
When I evaluate taking normal $N=k$, I get the answer $2pi a^{3}$.
But when I take normal to the surface by taking out gradient of $f(x,y,z)= x^{2}+y^{2}+z^{2}$ I am able to verify theorem.



But my course instructor told we can take normal $N=k$ when whenever we are able to translate system in $xy$ plane .
And if plane is given then we have to find gradient otherwise not.



Please clarify!!!






calculus multivariable-calculus vectors vector-analysis







share|cite|improve this question











$endgroup$




Verify Gauss divergence theorem of $F=xi+yj+zk$ over the sphere $x^{2}+y^{2}+z^{2}=a^{2}$
When I evaluate taking normal $N=k$, I get the answer $2pi a^{3}$.
But when I take normal to the surface by taking out gradient of $f(x,y,z)= x^{2}+y^{2}+z^{2}$ I am able to verify theorem.



But my course instructor told we can take normal $N=k$ when whenever we are able to translate system in $xy$ plane .
And if plane is given then we have to find gradient otherwise not.



Please clarify!!!






calculus multivariable-calculus vectors vector-analysis




calculus multivariable-calculus vectors vector-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 11 at 20:41









mechanodroid

28.5k62548




28.5k62548










asked Jan 11 at 19:05









sejysejy

1629




1629








  • 1




    $begingroup$
    $div F =3$ you should find $4pi a^3$.
    $endgroup$
    – hamam_Abdallah
    Jan 11 at 19:11






  • 1




    $begingroup$
    You can only use $vec{k}$ as the normal vector to your surface when the surface is parallel to the $xy-$plane.
    $endgroup$
    – pwerth
    Jan 11 at 19:12














  • 1




    $begingroup$
    $div F =3$ you should find $4pi a^3$.
    $endgroup$
    – hamam_Abdallah
    Jan 11 at 19:11






  • 1




    $begingroup$
    You can only use $vec{k}$ as the normal vector to your surface when the surface is parallel to the $xy-$plane.
    $endgroup$
    – pwerth
    Jan 11 at 19:12








1




1




$begingroup$
$div F =3$ you should find $4pi a^3$.
$endgroup$
– hamam_Abdallah
Jan 11 at 19:11




$begingroup$
$div F =3$ you should find $4pi a^3$.
$endgroup$
– hamam_Abdallah
Jan 11 at 19:11




1




1




$begingroup$
You can only use $vec{k}$ as the normal vector to your surface when the surface is parallel to the $xy-$plane.
$endgroup$
– pwerth
Jan 11 at 19:12




$begingroup$
You can only use $vec{k}$ as the normal vector to your surface when the surface is parallel to the $xy-$plane.
$endgroup$
– pwerth
Jan 11 at 19:12










2 Answers
2






active

oldest

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1












$begingroup$

Let $S_a$ be your sphere, and $B_a$ the enclosed ball. For each point ${bf r}in S_a$ the outwards unit normal ${bf n}$ is given by ${bf n}={{bf r}over a}$. Furthermore ${bf F}({bf r})={bf r}$. Since ${bf r}cdot{bf r}=a^2$ on $S_a$ it follows that
$$int_{S_a}{bf F}cdot{bf n}>{rm d}omega=int_{S_a}{bf r}cdot{{bf r}over a}>{rm d}omega=aint_{S_a}{rm d}omega=4pi a^3 .$$
On the other hand, ${rm div}({bf F})equiv3$, and therefore
$$int_{B_a}{rm div}({bf F})>{rm dvol}=3,{rm vol}(B_a)=4pi a^3 .$$






share|cite|improve this answer









$endgroup$





















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    $begingroup$

    Parameterize your sphere using spherical coordinates as $psi : [0,pi] times [0,2pi] to mathbb{R}^3$ given by
    $$psi(theta, phi) = a(sinthetacosphi, sinthetasinphi, costheta)$$
    Then your normal is given by
    $$mathbf{n}(theta, phi) = frac{partialpsi}{partialtheta} times frac{partialpsi}{partialphi} = a^2(sin^2thetacosphi, sin^2thetasinphi, sinthetacostheta) = (asintheta),psi(theta,phi)$$



    so $$intlimits_{S(0,a)} mathbf{F}cdot dmathbf{A} = intlimits_{[0,pi] times [0,2pi]} mathbf{F}(psi(theta, phi))cdot ,mathbf{n}(theta, phi),dtheta,dphi $$
    Now notice that $mathbf{F}$ is actually the identity function so the integral equals
    $$intlimits_{[0,pi] times [0,2pi]} (acostheta)psi(theta, phi)cdot ,psi(theta, phi),dtheta,dphi = int_{theta=0}^pi int_{phi=0}^{2pi} a^3sintheta,dtheta,dphi$$
    which is $4pi a^3$.



    On the other hand, $operatorname{div} mathbf{F} = operatorname{Tr} nablamathbf{F} = operatorname{Tr} mathbf{F} = 3$ so
    $$int_{B(0,a)} operatorname{div} mathbf{F} ,dV = 3 operatorname{vol}big(B(0,a)big) = 4pi a^3$$






    share|cite|improve this answer









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      2 Answers
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      2 Answers
      2






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      active

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      active

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      1












      $begingroup$

      Let $S_a$ be your sphere, and $B_a$ the enclosed ball. For each point ${bf r}in S_a$ the outwards unit normal ${bf n}$ is given by ${bf n}={{bf r}over a}$. Furthermore ${bf F}({bf r})={bf r}$. Since ${bf r}cdot{bf r}=a^2$ on $S_a$ it follows that
      $$int_{S_a}{bf F}cdot{bf n}>{rm d}omega=int_{S_a}{bf r}cdot{{bf r}over a}>{rm d}omega=aint_{S_a}{rm d}omega=4pi a^3 .$$
      On the other hand, ${rm div}({bf F})equiv3$, and therefore
      $$int_{B_a}{rm div}({bf F})>{rm dvol}=3,{rm vol}(B_a)=4pi a^3 .$$






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Let $S_a$ be your sphere, and $B_a$ the enclosed ball. For each point ${bf r}in S_a$ the outwards unit normal ${bf n}$ is given by ${bf n}={{bf r}over a}$. Furthermore ${bf F}({bf r})={bf r}$. Since ${bf r}cdot{bf r}=a^2$ on $S_a$ it follows that
        $$int_{S_a}{bf F}cdot{bf n}>{rm d}omega=int_{S_a}{bf r}cdot{{bf r}over a}>{rm d}omega=aint_{S_a}{rm d}omega=4pi a^3 .$$
        On the other hand, ${rm div}({bf F})equiv3$, and therefore
        $$int_{B_a}{rm div}({bf F})>{rm dvol}=3,{rm vol}(B_a)=4pi a^3 .$$






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Let $S_a$ be your sphere, and $B_a$ the enclosed ball. For each point ${bf r}in S_a$ the outwards unit normal ${bf n}$ is given by ${bf n}={{bf r}over a}$. Furthermore ${bf F}({bf r})={bf r}$. Since ${bf r}cdot{bf r}=a^2$ on $S_a$ it follows that
          $$int_{S_a}{bf F}cdot{bf n}>{rm d}omega=int_{S_a}{bf r}cdot{{bf r}over a}>{rm d}omega=aint_{S_a}{rm d}omega=4pi a^3 .$$
          On the other hand, ${rm div}({bf F})equiv3$, and therefore
          $$int_{B_a}{rm div}({bf F})>{rm dvol}=3,{rm vol}(B_a)=4pi a^3 .$$






          share|cite|improve this answer









          $endgroup$



          Let $S_a$ be your sphere, and $B_a$ the enclosed ball. For each point ${bf r}in S_a$ the outwards unit normal ${bf n}$ is given by ${bf n}={{bf r}over a}$. Furthermore ${bf F}({bf r})={bf r}$. Since ${bf r}cdot{bf r}=a^2$ on $S_a$ it follows that
          $$int_{S_a}{bf F}cdot{bf n}>{rm d}omega=int_{S_a}{bf r}cdot{{bf r}over a}>{rm d}omega=aint_{S_a}{rm d}omega=4pi a^3 .$$
          On the other hand, ${rm div}({bf F})equiv3$, and therefore
          $$int_{B_a}{rm div}({bf F})>{rm dvol}=3,{rm vol}(B_a)=4pi a^3 .$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 11 at 20:08









          Christian BlatterChristian Blatter

          175k8115327




          175k8115327























              1












              $begingroup$

              Parameterize your sphere using spherical coordinates as $psi : [0,pi] times [0,2pi] to mathbb{R}^3$ given by
              $$psi(theta, phi) = a(sinthetacosphi, sinthetasinphi, costheta)$$
              Then your normal is given by
              $$mathbf{n}(theta, phi) = frac{partialpsi}{partialtheta} times frac{partialpsi}{partialphi} = a^2(sin^2thetacosphi, sin^2thetasinphi, sinthetacostheta) = (asintheta),psi(theta,phi)$$



              so $$intlimits_{S(0,a)} mathbf{F}cdot dmathbf{A} = intlimits_{[0,pi] times [0,2pi]} mathbf{F}(psi(theta, phi))cdot ,mathbf{n}(theta, phi),dtheta,dphi $$
              Now notice that $mathbf{F}$ is actually the identity function so the integral equals
              $$intlimits_{[0,pi] times [0,2pi]} (acostheta)psi(theta, phi)cdot ,psi(theta, phi),dtheta,dphi = int_{theta=0}^pi int_{phi=0}^{2pi} a^3sintheta,dtheta,dphi$$
              which is $4pi a^3$.



              On the other hand, $operatorname{div} mathbf{F} = operatorname{Tr} nablamathbf{F} = operatorname{Tr} mathbf{F} = 3$ so
              $$int_{B(0,a)} operatorname{div} mathbf{F} ,dV = 3 operatorname{vol}big(B(0,a)big) = 4pi a^3$$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Parameterize your sphere using spherical coordinates as $psi : [0,pi] times [0,2pi] to mathbb{R}^3$ given by
                $$psi(theta, phi) = a(sinthetacosphi, sinthetasinphi, costheta)$$
                Then your normal is given by
                $$mathbf{n}(theta, phi) = frac{partialpsi}{partialtheta} times frac{partialpsi}{partialphi} = a^2(sin^2thetacosphi, sin^2thetasinphi, sinthetacostheta) = (asintheta),psi(theta,phi)$$



                so $$intlimits_{S(0,a)} mathbf{F}cdot dmathbf{A} = intlimits_{[0,pi] times [0,2pi]} mathbf{F}(psi(theta, phi))cdot ,mathbf{n}(theta, phi),dtheta,dphi $$
                Now notice that $mathbf{F}$ is actually the identity function so the integral equals
                $$intlimits_{[0,pi] times [0,2pi]} (acostheta)psi(theta, phi)cdot ,psi(theta, phi),dtheta,dphi = int_{theta=0}^pi int_{phi=0}^{2pi} a^3sintheta,dtheta,dphi$$
                which is $4pi a^3$.



                On the other hand, $operatorname{div} mathbf{F} = operatorname{Tr} nablamathbf{F} = operatorname{Tr} mathbf{F} = 3$ so
                $$int_{B(0,a)} operatorname{div} mathbf{F} ,dV = 3 operatorname{vol}big(B(0,a)big) = 4pi a^3$$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Parameterize your sphere using spherical coordinates as $psi : [0,pi] times [0,2pi] to mathbb{R}^3$ given by
                  $$psi(theta, phi) = a(sinthetacosphi, sinthetasinphi, costheta)$$
                  Then your normal is given by
                  $$mathbf{n}(theta, phi) = frac{partialpsi}{partialtheta} times frac{partialpsi}{partialphi} = a^2(sin^2thetacosphi, sin^2thetasinphi, sinthetacostheta) = (asintheta),psi(theta,phi)$$



                  so $$intlimits_{S(0,a)} mathbf{F}cdot dmathbf{A} = intlimits_{[0,pi] times [0,2pi]} mathbf{F}(psi(theta, phi))cdot ,mathbf{n}(theta, phi),dtheta,dphi $$
                  Now notice that $mathbf{F}$ is actually the identity function so the integral equals
                  $$intlimits_{[0,pi] times [0,2pi]} (acostheta)psi(theta, phi)cdot ,psi(theta, phi),dtheta,dphi = int_{theta=0}^pi int_{phi=0}^{2pi} a^3sintheta,dtheta,dphi$$
                  which is $4pi a^3$.



                  On the other hand, $operatorname{div} mathbf{F} = operatorname{Tr} nablamathbf{F} = operatorname{Tr} mathbf{F} = 3$ so
                  $$int_{B(0,a)} operatorname{div} mathbf{F} ,dV = 3 operatorname{vol}big(B(0,a)big) = 4pi a^3$$






                  share|cite|improve this answer









                  $endgroup$



                  Parameterize your sphere using spherical coordinates as $psi : [0,pi] times [0,2pi] to mathbb{R}^3$ given by
                  $$psi(theta, phi) = a(sinthetacosphi, sinthetasinphi, costheta)$$
                  Then your normal is given by
                  $$mathbf{n}(theta, phi) = frac{partialpsi}{partialtheta} times frac{partialpsi}{partialphi} = a^2(sin^2thetacosphi, sin^2thetasinphi, sinthetacostheta) = (asintheta),psi(theta,phi)$$



                  so $$intlimits_{S(0,a)} mathbf{F}cdot dmathbf{A} = intlimits_{[0,pi] times [0,2pi]} mathbf{F}(psi(theta, phi))cdot ,mathbf{n}(theta, phi),dtheta,dphi $$
                  Now notice that $mathbf{F}$ is actually the identity function so the integral equals
                  $$intlimits_{[0,pi] times [0,2pi]} (acostheta)psi(theta, phi)cdot ,psi(theta, phi),dtheta,dphi = int_{theta=0}^pi int_{phi=0}^{2pi} a^3sintheta,dtheta,dphi$$
                  which is $4pi a^3$.



                  On the other hand, $operatorname{div} mathbf{F} = operatorname{Tr} nablamathbf{F} = operatorname{Tr} mathbf{F} = 3$ so
                  $$int_{B(0,a)} operatorname{div} mathbf{F} ,dV = 3 operatorname{vol}big(B(0,a)big) = 4pi a^3$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 11 at 20:38









                  mechanodroidmechanodroid

                  28.5k62548




                  28.5k62548






























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