calculate the surface integral in the upper hemisphere












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Calculate the surface integral $f(x,y,z)=x^2+y^2+z^2$ in the upper hemisphere of the sphere $x^2+y^2+(z-1)^2=1$



I tried to compute the value of the surface integral $iint_S{F.n} dS$ with the explicit representation z=$sqrt{-x^2-y^2-2z}$ but the term $2z$ brings me conflict.



I don´t know if it should be done like this or it is convenient to change coordinates.



I would appreciate someone helping me. Since it is for a job that is delivered today and I am stuck.






calculus surface-integrals stokes-theorem







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    0














    Calculate the surface integral $f(x,y,z)=x^2+y^2+z^2$ in the upper hemisphere of the sphere $x^2+y^2+(z-1)^2=1$



    I tried to compute the value of the surface integral $iint_S{F.n} dS$ with the explicit representation z=$sqrt{-x^2-y^2-2z}$ but the term $2z$ brings me conflict.



    I don´t know if it should be done like this or it is convenient to change coordinates.



    I would appreciate someone helping me. Since it is for a job that is delivered today and I am stuck.






    calculus surface-integrals stokes-theorem







    share|cite|improve this question

























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      Calculate the surface integral $f(x,y,z)=x^2+y^2+z^2$ in the upper hemisphere of the sphere $x^2+y^2+(z-1)^2=1$



      I tried to compute the value of the surface integral $iint_S{F.n} dS$ with the explicit representation z=$sqrt{-x^2-y^2-2z}$ but the term $2z$ brings me conflict.



      I don´t know if it should be done like this or it is convenient to change coordinates.



      I would appreciate someone helping me. Since it is for a job that is delivered today and I am stuck.






      calculus surface-integrals stokes-theorem







      share|cite|improve this question













      Calculate the surface integral $f(x,y,z)=x^2+y^2+z^2$ in the upper hemisphere of the sphere $x^2+y^2+(z-1)^2=1$



      I tried to compute the value of the surface integral $iint_S{F.n} dS$ with the explicit representation z=$sqrt{-x^2-y^2-2z}$ but the term $2z$ brings me conflict.



      I don´t know if it should be done like this or it is convenient to change coordinates.



      I would appreciate someone helping me. Since it is for a job that is delivered today and I am stuck.






      calculus surface-integrals stokes-theorem




      calculus surface-integrals stokes-theorem






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      asked Nov 27 at 11:44









      F.A. Alday

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          Hoping that by "upper" hemisphere, you mean "with respect to $z$:



          Let
          $$
          Q = int_S x^2+y^2+z^2 ~ dA
          $$          , where $S$ is the upper hemisphere of the sphere $x^2+y^2+(z-1)^2=1$.



          Substituting $u = z-1; u+1 = z; du = dz$, we get
          $$
          Q = int_T x^2+y^2+(u+1)^2 ~ dA
          $$          , where $T$ is the upper hemisphere of the sphere $x^2+y^2+u^2=1$, which I'll rewrite as
          $$
          Q = int_U x^2+y^2+(z+1)^2 ~ dA,
          $$
          where $U$ is the upper hemisphere of the sphere $x^2+y^2+z^2=1$.



          Expanding the integrand, we get
          begin{align}
          Q &= int_U x^2+y^2+z^2 + 2z + 1 ~ dA \
          &= int_U x^2+y^2+z^2 ~dA+ int_U 2z~dA + int_U 1 ~ dA \
          &= int_U 1 ~dA+ int_U 2z~dA + int_U 1 ~ dA \
          &= int_U 2 ~dA+ int_U 2z~dA\
          &= 4pi+ 2int_U z~dA
          end{align}



          The latter integral is the average height of a point on the hemisphere characterized by $z ge 0$, muktiplied by the area ($2pi$) of that hemisphere.



          Because horizontal radial projection from the $z$-axis to a cylinder surrounding the sphere, i.e., the map
          $$
          (x, y, z) mapsto (frac{x}{sqrt{x^2 + y^2}}, frac{y}{sqrt{x^2 + y^2}}, z)
          $$
          is area-preserving (which requires a little proof, but was mentioned by Boy in his dissertation somewhere around 1895-1905, so it's a pretty well-established result!), this is the same as the average-height-times-area of a point on a cylinder of radius $1$ and height $1$, i.e., $frac{1}{2} cdot 2pi = pi$.



          So the overall value of your integral is $6pi$.






          share|cite|improve this answer





















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            Hoping that by "upper" hemisphere, you mean "with respect to $z$:



            Let
            $$
            Q = int_S x^2+y^2+z^2 ~ dA
            $$            , where $S$ is the upper hemisphere of the sphere $x^2+y^2+(z-1)^2=1$.



            Substituting $u = z-1; u+1 = z; du = dz$, we get
            $$
            Q = int_T x^2+y^2+(u+1)^2 ~ dA
            $$            , where $T$ is the upper hemisphere of the sphere $x^2+y^2+u^2=1$, which I'll rewrite as
            $$
            Q = int_U x^2+y^2+(z+1)^2 ~ dA,
            $$
            where $U$ is the upper hemisphere of the sphere $x^2+y^2+z^2=1$.



            Expanding the integrand, we get
            begin{align}
            Q &= int_U x^2+y^2+z^2 + 2z + 1 ~ dA \
            &= int_U x^2+y^2+z^2 ~dA+ int_U 2z~dA + int_U 1 ~ dA \
            &= int_U 1 ~dA+ int_U 2z~dA + int_U 1 ~ dA \
            &= int_U 2 ~dA+ int_U 2z~dA\
            &= 4pi+ 2int_U z~dA
            end{align}



            The latter integral is the average height of a point on the hemisphere characterized by $z ge 0$, muktiplied by the area ($2pi$) of that hemisphere.



            Because horizontal radial projection from the $z$-axis to a cylinder surrounding the sphere, i.e., the map
            $$
            (x, y, z) mapsto (frac{x}{sqrt{x^2 + y^2}}, frac{y}{sqrt{x^2 + y^2}}, z)
            $$
            is area-preserving (which requires a little proof, but was mentioned by Boy in his dissertation somewhere around 1895-1905, so it's a pretty well-established result!), this is the same as the average-height-times-area of a point on a cylinder of radius $1$ and height $1$, i.e., $frac{1}{2} cdot 2pi = pi$.



            So the overall value of your integral is $6pi$.






            share|cite|improve this answer


























              0














              Hoping that by "upper" hemisphere, you mean "with respect to $z$:



              Let
              $$
              Q = int_S x^2+y^2+z^2 ~ dA
              $$              , where $S$ is the upper hemisphere of the sphere $x^2+y^2+(z-1)^2=1$.



              Substituting $u = z-1; u+1 = z; du = dz$, we get
              $$
              Q = int_T x^2+y^2+(u+1)^2 ~ dA
              $$              , where $T$ is the upper hemisphere of the sphere $x^2+y^2+u^2=1$, which I'll rewrite as
              $$
              Q = int_U x^2+y^2+(z+1)^2 ~ dA,
              $$
              where $U$ is the upper hemisphere of the sphere $x^2+y^2+z^2=1$.



              Expanding the integrand, we get
              begin{align}
              Q &= int_U x^2+y^2+z^2 + 2z + 1 ~ dA \
              &= int_U x^2+y^2+z^2 ~dA+ int_U 2z~dA + int_U 1 ~ dA \
              &= int_U 1 ~dA+ int_U 2z~dA + int_U 1 ~ dA \
              &= int_U 2 ~dA+ int_U 2z~dA\
              &= 4pi+ 2int_U z~dA
              end{align}



              The latter integral is the average height of a point on the hemisphere characterized by $z ge 0$, muktiplied by the area ($2pi$) of that hemisphere.



              Because horizontal radial projection from the $z$-axis to a cylinder surrounding the sphere, i.e., the map
              $$
              (x, y, z) mapsto (frac{x}{sqrt{x^2 + y^2}}, frac{y}{sqrt{x^2 + y^2}}, z)
              $$
              is area-preserving (which requires a little proof, but was mentioned by Boy in his dissertation somewhere around 1895-1905, so it's a pretty well-established result!), this is the same as the average-height-times-area of a point on a cylinder of radius $1$ and height $1$, i.e., $frac{1}{2} cdot 2pi = pi$.



              So the overall value of your integral is $6pi$.






              share|cite|improve this answer
























                0












                0








                0






                Hoping that by "upper" hemisphere, you mean "with respect to $z$:



                Let
                $$
                Q = int_S x^2+y^2+z^2 ~ dA
                $$                , where $S$ is the upper hemisphere of the sphere $x^2+y^2+(z-1)^2=1$.



                Substituting $u = z-1; u+1 = z; du = dz$, we get
                $$
                Q = int_T x^2+y^2+(u+1)^2 ~ dA
                $$                , where $T$ is the upper hemisphere of the sphere $x^2+y^2+u^2=1$, which I'll rewrite as
                $$
                Q = int_U x^2+y^2+(z+1)^2 ~ dA,
                $$
                where $U$ is the upper hemisphere of the sphere $x^2+y^2+z^2=1$.



                Expanding the integrand, we get
                begin{align}
                Q &= int_U x^2+y^2+z^2 + 2z + 1 ~ dA \
                &= int_U x^2+y^2+z^2 ~dA+ int_U 2z~dA + int_U 1 ~ dA \
                &= int_U 1 ~dA+ int_U 2z~dA + int_U 1 ~ dA \
                &= int_U 2 ~dA+ int_U 2z~dA\
                &= 4pi+ 2int_U z~dA
                end{align}



                The latter integral is the average height of a point on the hemisphere characterized by $z ge 0$, muktiplied by the area ($2pi$) of that hemisphere.



                Because horizontal radial projection from the $z$-axis to a cylinder surrounding the sphere, i.e., the map
                $$
                (x, y, z) mapsto (frac{x}{sqrt{x^2 + y^2}}, frac{y}{sqrt{x^2 + y^2}}, z)
                $$
                is area-preserving (which requires a little proof, but was mentioned by Boy in his dissertation somewhere around 1895-1905, so it's a pretty well-established result!), this is the same as the average-height-times-area of a point on a cylinder of radius $1$ and height $1$, i.e., $frac{1}{2} cdot 2pi = pi$.



                So the overall value of your integral is $6pi$.






                share|cite|improve this answer












                Hoping that by "upper" hemisphere, you mean "with respect to $z$:



                Let
                $$
                Q = int_S x^2+y^2+z^2 ~ dA
                $$                , where $S$ is the upper hemisphere of the sphere $x^2+y^2+(z-1)^2=1$.



                Substituting $u = z-1; u+1 = z; du = dz$, we get
                $$
                Q = int_T x^2+y^2+(u+1)^2 ~ dA
                $$                , where $T$ is the upper hemisphere of the sphere $x^2+y^2+u^2=1$, which I'll rewrite as
                $$
                Q = int_U x^2+y^2+(z+1)^2 ~ dA,
                $$
                where $U$ is the upper hemisphere of the sphere $x^2+y^2+z^2=1$.



                Expanding the integrand, we get
                begin{align}
                Q &= int_U x^2+y^2+z^2 + 2z + 1 ~ dA \
                &= int_U x^2+y^2+z^2 ~dA+ int_U 2z~dA + int_U 1 ~ dA \
                &= int_U 1 ~dA+ int_U 2z~dA + int_U 1 ~ dA \
                &= int_U 2 ~dA+ int_U 2z~dA\
                &= 4pi+ 2int_U z~dA
                end{align}



                The latter integral is the average height of a point on the hemisphere characterized by $z ge 0$, muktiplied by the area ($2pi$) of that hemisphere.



                Because horizontal radial projection from the $z$-axis to a cylinder surrounding the sphere, i.e., the map
                $$
                (x, y, z) mapsto (frac{x}{sqrt{x^2 + y^2}}, frac{y}{sqrt{x^2 + y^2}}, z)
                $$
                is area-preserving (which requires a little proof, but was mentioned by Boy in his dissertation somewhere around 1895-1905, so it's a pretty well-established result!), this is the same as the average-height-times-area of a point on a cylinder of radius $1$ and height $1$, i.e., $frac{1}{2} cdot 2pi = pi$.



                So the overall value of your integral is $6pi$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 27 at 13:42









                John Hughes

                62.3k24090




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