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Math Help - Line Integral and Stokes Theorem

  1. #1
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    Line Integral and Stokes Theorem

    For the following problem, I'm suppose to Stokes Theorem and then line integral but I'm getting different results.

    The Problem:
    Let S be the triangular region with vertices A(0,0,0), B(0,1,2), C(2,2,6). Let C = \partial S oriented counterclockwise as seen from above. Let F= (y,-x,y). Find \oint_{c} F dS

    First, I find the equation of the plane and that comes out to z=x+2y.
    Then when I parametrize the surface, I get S(x,y) = (x,y,x+2y).
    Because it's counterclockwise, I need Sx X Sy = (-1,-2,1).

    The curl of F is (1,0,-2)

    By Stokes Theorem:

    \oint_{c} F dS = \int \int_{S} curl F \bullet dS

    = \int \int_{D}(1,0,-2) \bullet (-1,-2,1) dA

    = \int \int_{D} -3 dA

    = -3 Area of D
    = -3

    By Line Integral:

    Parametrization

    Vector AC: c_{1}(t) = (2t,2t,6t) for 0\leq t\leq 1
    Vector CB: c_{2}(t) = (2-2t,2-t,6-4t) for 0\leq t\leq 1
    Vector BA: c_{3}(t) = (0,1-t,2-2t) for 0\leq t\leq 1

    Keeping in mind that F= (y,-x,y):

    c_{1}:

    \int^1_{0} (2t,-2t,2t)\bullet(2,2,6)dS

    = \int^1_{0} 4t-4t+12t dS

    = 12

    c_{2}:

    \int^1_{0} (2-t,-2+2t,2-t)\bullet(-2,-1,-4)dS

    = \int^1_{0} -4+2t+2-2t-8+4t dS

    = -8

    c_{3}:

    \int^1_{0} (1-t,0,1-t)\bullet(0,-1,-2)dS

    = \int^1_{0} -2+2t dS

    = -1

    c_{1}+c_{2}+c_{3} = 12 -8-1 = 3


    With Stokes theorem, I get -3. With line integral, I get 3. I should be getting the same results though.
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  2. #2
    MHF Contributor

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    Re: Line Integral and Stokes Theorem

    Quote Originally Posted by ivinew View Post
    For the following problem, I'm suppose to Stokes Theorem and then line integral but I'm getting different results.

    The Problem:
    Let S be the triangular region with vertices A(0,0,0), B(0,1,2), C(2,2,6). Let C = \partial S oriented counterclockwise as seen from above. Let F= (y,-x,y). Find \oint_{c} F dS

    First, I find the equation of the plane and that comes out to z=x+2y.
    Then when I parametrize the surface, I get S(x,y) = (x,y,x+2y).
    Because it's counterclockwise, I need Sx X Sy = (-1,-2,1).

    The curl of F is (1,0,-2)

    By Stokes Theorem:

    \oint_{c} F dS = \int \int_{S} curl F \bullet dS

    = \int \int_{D}(1,0,-2) \bullet (-1,-2,1) dA

    = \int \int_{D} -3 dA

    = -3 Area of D
    = -3

    By Line Integral:

    Parametrization

    Vector AC: c_{1}(t) = (2t,2t,6t) for 0\leq t\leq 1
    Vector CB: c_{2}(t) = (2-2t,2-t,6-4t) for 0\leq t\leq 1
    Vector BA: c_{3}(t) = (0,1-t,2-2t) for 0\leq t\leq 1

    Keeping in mind that F= (y,-x,y):

    c_{1}:

    \int^1_{0} (2t,-2t,2t)\bullet(2,2,6)dS

    = \int^1_{0} 4t-4t+12t dS

    = 12
    No. \int_0^1 12t dt= 6t^2|_0^1= 6.

    c_{2}:

    \int^1_{0} (2-t,-2+2t,2-t)\bullet(-2,-1,-4)dS

    = \int^1_{0} -4+2t+2-2t-8+4t dS

    = -8

    c_{3}:

    \int^1_{0} (1-t,0,1-t)\bullet(0,-1,-2)dS

    = \int^1_{0} -2+2t dS

    = -1

    c_{1}+c_{2}+c_{3} = 12 -8-1 = 3


    With Stokes theorem, I get -3. With line integral, I get 3. I should be getting the same results though.
    6- 8- 1= -3, not 3. Your first line integral was wrong.
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