Surface integral over graph of a function












3












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Q: Evaluate the surface integral $iint_S mathbf{F}cdot dmathbf{S}$, where $mathbf{F}(x,y,z)= 3x^2mathbf{i}-2xymathbf{j}+8mathbf{k}$, and $S$ is the graph of the function $z=f(x,y)=2x-y$ for $0leq xleq2$ and $0leq y leq2$.



Attempt: This question is from a previous years Vector Calculus exam. When attempting to solve, I had thought to parameterize the surface, and then use $int_a^b mathbf{F}(mathbf{c}(t))cdot mathbf{c}'(t),dt$, $a leq t leq b$. In the solutions, they have $$iint_S mathbf{F}cdot dmathbf{S} = int_0^2int_0^2 (-6x^2-2xy+8),dxdy= -8$$



I'm mostly looking for an explanation as to how they arrived at the expression inside the integral ($-6x^2-2xy+8$), any help would be greatly appreciated.



Thanks.






calculus multivariable-calculus







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    3












    $begingroup$


    Q: Evaluate the surface integral $iint_S mathbf{F}cdot dmathbf{S}$, where $mathbf{F}(x,y,z)= 3x^2mathbf{i}-2xymathbf{j}+8mathbf{k}$, and $S$ is the graph of the function $z=f(x,y)=2x-y$ for $0leq xleq2$ and $0leq y leq2$.



    Attempt: This question is from a previous years Vector Calculus exam. When attempting to solve, I had thought to parameterize the surface, and then use $int_a^b mathbf{F}(mathbf{c}(t))cdot mathbf{c}'(t),dt$, $a leq t leq b$. In the solutions, they have $$iint_S mathbf{F}cdot dmathbf{S} = int_0^2int_0^2 (-6x^2-2xy+8),dxdy= -8$$



    I'm mostly looking for an explanation as to how they arrived at the expression inside the integral ($-6x^2-2xy+8$), any help would be greatly appreciated.



    Thanks.






    calculus multivariable-calculus







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      Q: Evaluate the surface integral $iint_S mathbf{F}cdot dmathbf{S}$, where $mathbf{F}(x,y,z)= 3x^2mathbf{i}-2xymathbf{j}+8mathbf{k}$, and $S$ is the graph of the function $z=f(x,y)=2x-y$ for $0leq xleq2$ and $0leq y leq2$.



      Attempt: This question is from a previous years Vector Calculus exam. When attempting to solve, I had thought to parameterize the surface, and then use $int_a^b mathbf{F}(mathbf{c}(t))cdot mathbf{c}'(t),dt$, $a leq t leq b$. In the solutions, they have $$iint_S mathbf{F}cdot dmathbf{S} = int_0^2int_0^2 (-6x^2-2xy+8),dxdy= -8$$



      I'm mostly looking for an explanation as to how they arrived at the expression inside the integral ($-6x^2-2xy+8$), any help would be greatly appreciated.



      Thanks.






      calculus multivariable-calculus







      share|cite|improve this question











      $endgroup$




      Q: Evaluate the surface integral $iint_S mathbf{F}cdot dmathbf{S}$, where $mathbf{F}(x,y,z)= 3x^2mathbf{i}-2xymathbf{j}+8mathbf{k}$, and $S$ is the graph of the function $z=f(x,y)=2x-y$ for $0leq xleq2$ and $0leq y leq2$.



      Attempt: This question is from a previous years Vector Calculus exam. When attempting to solve, I had thought to parameterize the surface, and then use $int_a^b mathbf{F}(mathbf{c}(t))cdot mathbf{c}'(t),dt$, $a leq t leq b$. In the solutions, they have $$iint_S mathbf{F}cdot dmathbf{S} = int_0^2int_0^2 (-6x^2-2xy+8),dxdy= -8$$



      I'm mostly looking for an explanation as to how they arrived at the expression inside the integral ($-6x^2-2xy+8$), any help would be greatly appreciated.



      Thanks.






      calculus multivariable-calculus




      calculus multivariable-calculus






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      share|cite|improve this question













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      edited Dec 26 '18 at 13:06









      mechanodroid

      27.6k62447




      27.6k62447










      asked Jun 13 '13 at 7:19









      user68093user68093

      603




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          1 Answer
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          $begingroup$

          The surface integral is better written as



          $$iint_S dS , hat{mathbf{n}} cdot mathbf{F}$$



          where $hat{mathbf{n}}$ is the unit surface normal. For surfaces of the form $z=f(x,y)$, the unit normal is



          $$hat{mathbf{n}} = frac{displaystyleleft(-frac{partial f}{partial x},-frac{partial f}{partial y},1right)}{displaystylesqrt{left(frac{partial f}{partial x}right)^2+left(frac{partial f}{partial y}right)^2+1}}$$



          Note that the normalizing factor in the denominator is exactly the stretching factor relating the surface area element to the differential $dx,dy$:



          $$dS = sqrt{left(frac{partial f}{partial x}right)^2+left(frac{partial f}{partial y}right)^2+1} , dx, dy$$



          Now, $partial f/partial x = 2$ and $partial f/partial y = -1$, so that



          $$iint_S dS , hat{mathbf{n}} cdot mathbf{F} = int_0^2 dx , int_0^2 dy , [-(2) cdot (3 x^2) - (-1) cdot (-2 x y) + (1)cdot (8)] $$



          which is the result you wanted to see.






          share|cite|improve this answer









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            1 Answer
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            active

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            $begingroup$

            The surface integral is better written as



            $$iint_S dS , hat{mathbf{n}} cdot mathbf{F}$$



            where $hat{mathbf{n}}$ is the unit surface normal. For surfaces of the form $z=f(x,y)$, the unit normal is



            $$hat{mathbf{n}} = frac{displaystyleleft(-frac{partial f}{partial x},-frac{partial f}{partial y},1right)}{displaystylesqrt{left(frac{partial f}{partial x}right)^2+left(frac{partial f}{partial y}right)^2+1}}$$



            Note that the normalizing factor in the denominator is exactly the stretching factor relating the surface area element to the differential $dx,dy$:



            $$dS = sqrt{left(frac{partial f}{partial x}right)^2+left(frac{partial f}{partial y}right)^2+1} , dx, dy$$



            Now, $partial f/partial x = 2$ and $partial f/partial y = -1$, so that



            $$iint_S dS , hat{mathbf{n}} cdot mathbf{F} = int_0^2 dx , int_0^2 dy , [-(2) cdot (3 x^2) - (-1) cdot (-2 x y) + (1)cdot (8)] $$



            which is the result you wanted to see.






            share|cite|improve this answer









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              3












              $begingroup$

              The surface integral is better written as



              $$iint_S dS , hat{mathbf{n}} cdot mathbf{F}$$



              where $hat{mathbf{n}}$ is the unit surface normal. For surfaces of the form $z=f(x,y)$, the unit normal is



              $$hat{mathbf{n}} = frac{displaystyleleft(-frac{partial f}{partial x},-frac{partial f}{partial y},1right)}{displaystylesqrt{left(frac{partial f}{partial x}right)^2+left(frac{partial f}{partial y}right)^2+1}}$$



              Note that the normalizing factor in the denominator is exactly the stretching factor relating the surface area element to the differential $dx,dy$:



              $$dS = sqrt{left(frac{partial f}{partial x}right)^2+left(frac{partial f}{partial y}right)^2+1} , dx, dy$$



              Now, $partial f/partial x = 2$ and $partial f/partial y = -1$, so that



              $$iint_S dS , hat{mathbf{n}} cdot mathbf{F} = int_0^2 dx , int_0^2 dy , [-(2) cdot (3 x^2) - (-1) cdot (-2 x y) + (1)cdot (8)] $$



              which is the result you wanted to see.






              share|cite|improve this answer









              $endgroup$
















                3












                3








                3





                $begingroup$

                The surface integral is better written as



                $$iint_S dS , hat{mathbf{n}} cdot mathbf{F}$$



                where $hat{mathbf{n}}$ is the unit surface normal. For surfaces of the form $z=f(x,y)$, the unit normal is



                $$hat{mathbf{n}} = frac{displaystyleleft(-frac{partial f}{partial x},-frac{partial f}{partial y},1right)}{displaystylesqrt{left(frac{partial f}{partial x}right)^2+left(frac{partial f}{partial y}right)^2+1}}$$



                Note that the normalizing factor in the denominator is exactly the stretching factor relating the surface area element to the differential $dx,dy$:



                $$dS = sqrt{left(frac{partial f}{partial x}right)^2+left(frac{partial f}{partial y}right)^2+1} , dx, dy$$



                Now, $partial f/partial x = 2$ and $partial f/partial y = -1$, so that



                $$iint_S dS , hat{mathbf{n}} cdot mathbf{F} = int_0^2 dx , int_0^2 dy , [-(2) cdot (3 x^2) - (-1) cdot (-2 x y) + (1)cdot (8)] $$



                which is the result you wanted to see.






                share|cite|improve this answer









                $endgroup$



                The surface integral is better written as



                $$iint_S dS , hat{mathbf{n}} cdot mathbf{F}$$



                where $hat{mathbf{n}}$ is the unit surface normal. For surfaces of the form $z=f(x,y)$, the unit normal is



                $$hat{mathbf{n}} = frac{displaystyleleft(-frac{partial f}{partial x},-frac{partial f}{partial y},1right)}{displaystylesqrt{left(frac{partial f}{partial x}right)^2+left(frac{partial f}{partial y}right)^2+1}}$$



                Note that the normalizing factor in the denominator is exactly the stretching factor relating the surface area element to the differential $dx,dy$:



                $$dS = sqrt{left(frac{partial f}{partial x}right)^2+left(frac{partial f}{partial y}right)^2+1} , dx, dy$$



                Now, $partial f/partial x = 2$ and $partial f/partial y = -1$, so that



                $$iint_S dS , hat{mathbf{n}} cdot mathbf{F} = int_0^2 dx , int_0^2 dy , [-(2) cdot (3 x^2) - (-1) cdot (-2 x y) + (1)cdot (8)] $$



                which is the result you wanted to see.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jun 13 '13 at 8:16









                Ron GordonRon Gordon

                122k14155265




                122k14155265






























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