Evaluate Integral Using Stokes Theorem

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Using Stokes theorem evaluate



$$int_Gamma(z-y),dx-(x+z),dy-(x+y),dz$$



where $Gamma$ is the intersection of $x^2+y^2+z^2=4$ with the plane $z=y$ with anticlockwise orientation when looking on $z$ from the positive side



If we take $nablatimes F=(0,2,0)$ but how can we find $hatn$




calculus multivariable-calculus stokes-theorem




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  • Are you sure the curl is 0?
    – Davey
    Aug 6 at 18:06










  • @Davey correct, edited
    – newhere
    Aug 6 at 18:18














up vote
2
down vote

favorite












Using Stokes theorem evaluate



$$int_Gamma(z-y),dx-(x+z),dy-(x+y),dz$$



where $Gamma$ is the intersection of $x^2+y^2+z^2=4$ with the plane $z=y$ with anticlockwise orientation when looking on $z$ from the positive side



If we take $nablatimes F=(0,2,0)$ but how can we find $hatn$




calculus multivariable-calculus stokes-theorem




share|cite|improve this question





















  • Are you sure the curl is 0?
    – Davey
    Aug 6 at 18:06










  • @Davey correct, edited
    – newhere
    Aug 6 at 18:18












up vote
2
down vote

favorite









up vote
2
down vote

favorite











Using Stokes theorem evaluate



$$int_Gamma(z-y),dx-(x+z),dy-(x+y),dz$$



where $Gamma$ is the intersection of $x^2+y^2+z^2=4$ with the plane $z=y$ with anticlockwise orientation when looking on $z$ from the positive side



If we take $nablatimes F=(0,2,0)$ but how can we find $hatn$




calculus multivariable-calculus stokes-theorem




share|cite|improve this question













Using Stokes theorem evaluate



$$int_Gamma(z-y),dx-(x+z),dy-(x+y),dz$$



where $Gamma$ is the intersection of $x^2+y^2+z^2=4$ with the plane $z=y$ with anticlockwise orientation when looking on $z$ from the positive side



If we take $nablatimes F=(0,2,0)$ but how can we find $hatn$





calculus multivariable-calculus stokes-theorem





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Aug 6 at 18:18
























asked Aug 6 at 17:58









newhere

786310




786310











  • Are you sure the curl is 0?
    – Davey
    Aug 6 at 18:06










  • @Davey correct, edited
    – newhere
    Aug 6 at 18:18
















  • Are you sure the curl is 0?
    – Davey
    Aug 6 at 18:06










  • @Davey correct, edited
    – newhere
    Aug 6 at 18:18















Are you sure the curl is 0?
– Davey
Aug 6 at 18:06




Are you sure the curl is 0?
– Davey
Aug 6 at 18:06












@Davey correct, edited
– newhere
Aug 6 at 18:18




@Davey correct, edited
– newhere
Aug 6 at 18:18










2 Answers
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Since $Gamma$ is a circle in a plane, we can apply Stokes' Theorem to the disk $D$ that it encloses to get that $int_Gamma F cdot ds = int_D$ curl $F cdot dA$. At each point of $D$, the normal to the disk is the same as the normal to the plane. In general, a normal to a plane is a vector $v$ such that the equation of the plane can be written in the form $(x - x_0, y - y_0, z - z_0) cdot v = 0$, where $(x_0, y_0, z_o)$ is any fixed point on the plane. In this case, we can take $(x_0, y_0, z_0)$ to be the origin. So you just need to find a unit vector $v$ with this property, pointing in whichever direction agrees with the orientation of $Gamma$.






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    $vec n$ is normal to plane. So, $ vec n = <0,1,-1>/sqrt 2 \ iint Curl (vec F). vec n dS = iint <0,2,0> cdot (<0,1,-1>/sqrt 2) dS \ = iint
    sqrt 2 dS = sqrt 2 (Area space of space circle) = sqrt 2 pi(2^2) \ = 4 sqrt 2pi$






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      2 Answers
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      2 Answers
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      Since $Gamma$ is a circle in a plane, we can apply Stokes' Theorem to the disk $D$ that it encloses to get that $int_Gamma F cdot ds = int_D$ curl $F cdot dA$. At each point of $D$, the normal to the disk is the same as the normal to the plane. In general, a normal to a plane is a vector $v$ such that the equation of the plane can be written in the form $(x - x_0, y - y_0, z - z_0) cdot v = 0$, where $(x_0, y_0, z_o)$ is any fixed point on the plane. In this case, we can take $(x_0, y_0, z_0)$ to be the origin. So you just need to find a unit vector $v$ with this property, pointing in whichever direction agrees with the orientation of $Gamma$.






      share|cite|improve this answer

























        up vote
        2
        down vote



        accepted










        Since $Gamma$ is a circle in a plane, we can apply Stokes' Theorem to the disk $D$ that it encloses to get that $int_Gamma F cdot ds = int_D$ curl $F cdot dA$. At each point of $D$, the normal to the disk is the same as the normal to the plane. In general, a normal to a plane is a vector $v$ such that the equation of the plane can be written in the form $(x - x_0, y - y_0, z - z_0) cdot v = 0$, where $(x_0, y_0, z_o)$ is any fixed point on the plane. In this case, we can take $(x_0, y_0, z_0)$ to be the origin. So you just need to find a unit vector $v$ with this property, pointing in whichever direction agrees with the orientation of $Gamma$.






        share|cite|improve this answer























          up vote
          2
          down vote



          accepted







          up vote
          2
          down vote



          accepted






          Since $Gamma$ is a circle in a plane, we can apply Stokes' Theorem to the disk $D$ that it encloses to get that $int_Gamma F cdot ds = int_D$ curl $F cdot dA$. At each point of $D$, the normal to the disk is the same as the normal to the plane. In general, a normal to a plane is a vector $v$ such that the equation of the plane can be written in the form $(x - x_0, y - y_0, z - z_0) cdot v = 0$, where $(x_0, y_0, z_o)$ is any fixed point on the plane. In this case, we can take $(x_0, y_0, z_0)$ to be the origin. So you just need to find a unit vector $v$ with this property, pointing in whichever direction agrees with the orientation of $Gamma$.






          share|cite|improve this answer













          Since $Gamma$ is a circle in a plane, we can apply Stokes' Theorem to the disk $D$ that it encloses to get that $int_Gamma F cdot ds = int_D$ curl $F cdot dA$. At each point of $D$, the normal to the disk is the same as the normal to the plane. In general, a normal to a plane is a vector $v$ such that the equation of the plane can be written in the form $(x - x_0, y - y_0, z - z_0) cdot v = 0$, where $(x_0, y_0, z_o)$ is any fixed point on the plane. In this case, we can take $(x_0, y_0, z_0)$ to be the origin. So you just need to find a unit vector $v$ with this property, pointing in whichever direction agrees with the orientation of $Gamma$.







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Aug 6 at 18:36









          Davey

          43827




          43827




















              up vote
              1
              down vote













              $vec n$ is normal to plane. So, $ vec n = <0,1,-1>/sqrt 2 \ iint Curl (vec F). vec n dS = iint <0,2,0> cdot (<0,1,-1>/sqrt 2) dS \ = iint
              sqrt 2 dS = sqrt 2 (Area space of space circle) = sqrt 2 pi(2^2) \ = 4 sqrt 2pi$






              share|cite|improve this answer

























                up vote
                1
                down vote













                $vec n$ is normal to plane. So, $ vec n = <0,1,-1>/sqrt 2 \ iint Curl (vec F). vec n dS = iint <0,2,0> cdot (<0,1,-1>/sqrt 2) dS \ = iint
                sqrt 2 dS = sqrt 2 (Area space of space circle) = sqrt 2 pi(2^2) \ = 4 sqrt 2pi$






                share|cite|improve this answer























                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  $vec n$ is normal to plane. So, $ vec n = <0,1,-1>/sqrt 2 \ iint Curl (vec F). vec n dS = iint <0,2,0> cdot (<0,1,-1>/sqrt 2) dS \ = iint
                  sqrt 2 dS = sqrt 2 (Area space of space circle) = sqrt 2 pi(2^2) \ = 4 sqrt 2pi$






                  share|cite|improve this answer













                  $vec n$ is normal to plane. So, $ vec n = <0,1,-1>/sqrt 2 \ iint Curl (vec F). vec n dS = iint <0,2,0> cdot (<0,1,-1>/sqrt 2) dS \ = iint
                  sqrt 2 dS = sqrt 2 (Area space of space circle) = sqrt 2 pi(2^2) \ = 4 sqrt 2pi$







                  share|cite|improve this answer













                  share|cite|improve this answer



                  share|cite|improve this answer











                  answered Aug 6 at 19:30









                  Magneto

                  787213




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