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Math Help - Stoke's Theorem

  1. #1
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    Stoke's Theorem

    F = ( y - z + 2)i +(yz +4)j -xzk, where S is the surface of the cube x = 0, y = 0, z = 0, x = 2, y = 2, z = 2, above the xy plane

    So i have computed \nabla x F = -yi + (-1+z)j -k now i dont know what to do next
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  2. #2
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    If you insist on using Stoke's theorem here, you need to calculate the curve integral of the six squares that make up the sides of the cube. It would be a lot easier to use the divergence theorem here since dV is a cube.
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  3. #3
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    Stokes thm is

    \iint_{S} \nabla \times F \cdot {\bf n} dS = \int_C {\bf F} \cdot d{\bf r}

    Are you trying to find the LHS?
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  4. #4
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    Quote Originally Posted by Danny View Post
    Stokes thm is

    \iint_{S} \nabla \times F \cdot {\bf n} dS = \int_C {\bf F} \cdot d{\bf r}

    Are you trying to find the LHS?
    Yes
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  5. #5
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    Well, you want to compute \iint_s \nabla \times F \cdot {\bf n}dS for 5 sides of the cube (the bottom is open). Things will cancel though

    S_1: \;\;x = 2  \;\;{\bf n} = <1,0,0> \;\;\;\;\;\nabla  \times F \cdot {\bf n} = -y

    S_2: \;\;x = 0  \;\;{\bf n} = <-1,0,0> \;\;\nabla \times F \cdot {\bf n} = y

    S_3: \;\;y = 2  \;\;{\bf n} = <0,1,0> \;\;\;\;\;\nabla \times F \cdot {\bf n} = z-1

    S_4: \;\;y = 9  \;\;{\bf n} = <0,-1,0> \;\;\nabla \times F \cdot {\bf n} = -(z-1)

    S_5: \;\;z = 2  \;\;{\bf n} = <0,0,1> \;\;\;\;\;\nabla \times F \cdot {\bf n} = -1

    Now the surface integrals

    S_1:  -\int_0^2\int_0^2 y dy dz

    S_2:  \int_0^2\int_0^2 y dy dz

    S_3:  \int_0^2\int_0^2 (z-1) dx dz

    S_4:  - \int_0^2\int_0^2 (z-1) dx dz

    S_5:  \int_0^2\int_0^2 -1 dx dy
    Now the first 4 add together give zero (you don't even have to integrate) so your final answer is

    \iint_s \nabla \times F \cdot {\bf n} = -4
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