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Math Help - NEED HELP! Very Lost!!!!! Vector Calc

  1. #1
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    NEED HELP! Very Lost!!!!! Vector Calc

    I posted some of the same questions on my other post a few days back. Mr. Fantastic helped me but i want to put them on here to start afresh. Please help me understand on how to do these problems. I have to equation and the book right here but i really dont understand it. If you could help me step by step i would really appreciate it!!!!!

    1) If F = (3z - sinx)i + (x^2 + e^y)j + (y^3 - cosz)k, use Stoke's theorem to evaluate integral with lower terminal C [ F.dr] where C is given by x=cost, y=sint, z=1; 0<= t <= 2pie

    2) Let S be the first-octant portion of the plane x + y + z =1. Verify Stokes' Theorem for the vector field F = y^2i + z^2j + x^2k

    THe part i really dont get is how to do the line integral and the conditions of each terminal. Please help
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  2. #2
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    Quote Originally Posted by davidson89 View Post
    [snip]
    1) If F = (3z - sinx)i + (x^2 + e^y)j + (y^3 - cosz)k, use Stoke's theorem to evaluate integral with lower terminal C [ F.dr] where C is given by x=cost, y=sint, z=1; 0<= t <= 2pie

    [snip]
    You should know that \int F \cdot dr = \int \left( F \cdot \frac{dr}{dt} \right) dt.

    On the curve:

    1. F = (3 - \sin ( \cos t) i + (\cos^2 t + e^{\sin t}) j + (\sin^3 t - \cos 1).

    2. \frac{dr}{dt} = (- \sin t) i + (\cos t) j + 0 k.

    Take the dot product F \cdot \frac{dr}{dt} to get a function of t, g(t) say.

    Since no surface is given in the question I don't see how Stoke's Theorem is relevant here ......

    Now evaluate \int_{0}^{2 \pi} g(t) \, dt.
    Last edited by mr fantastic; July 28th 2008 at 09:30 PM.
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  3. #3
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    Quote Originally Posted by davidson89 View Post
    [snip]
    2) Let S be the first-octant portion of the plane x + y + z =1. Verify Stokes' Theorem for the vector field F = y^2i + z^2j + x^2k

    THe part i really dont get is how to do the line integral and the conditions of each terminal. Please help
    The edges of S are:

    xy-plane (z = 0): x + y = 1.

    xz-plane (y = 0): x + z = 1.

    yz-plane (x = 0): y + z = 1.

    So the closed curve is defined by joining the line segments x + y = 1, x + z = 1 and y + z = 1. Integrate F along each of these line segments.

    Along the line segment x + y = 1:

    1. y = 1 - x and z = 0.

    2. dx + dy = 0 => dy = -dx and dz = 0.

    Therefore along the line segment x + y = 1:

    \int F \cdot dr = \int_1^0  \left( [1-x]^2 i + 0^2 j + x^2 k \right) \cdot (dx \, i - dx \, j + 0 dz \, k) =  \int_1^0 (1 - x)^2 dx = \, ......

    The integration along the other two line segments is done in a similar way.


    You have to go back and extensively revise the basics.
    Last edited by mr fantastic; July 28th 2008 at 09:31 PM.
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