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Stokes' Theorem Problem watch

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    Considering a semicircle-faced cylinder with radius a and height h, and the vector field V=(x+y+z)i+zj+yzk. Perform the closed loop integral around the cross-section at z=0 in the anticlockwise direction, and verify using the surface integral.

    I keep getting two different answers for each part.

    For the closed loop integral:

    The semicircle edge:

    Convert to plane polar form:
    x=rcosθ, y=rsinθ, r=a
    dx=-asinθdθ, dy=acosθdθ

    dr=adθ(-cosθi+sinθj)
    V.dr=a^2dθ(-sinθ)(cosθ+sinθ)

    Limits: 0<θ<π
    -a^2∫(sinθ)(cosθ+sinθ)dθ=-(πa^2)/2

    The flat edge:

    dr=dxi
    V.dr=(x+y+z)dx=xdx

    Limits: -a<x<a

    As the integrand is an odd function, this yields zero.

    Thus, the closed loop integral is -(πa^2)/2.


    For the surface integral:

    The semicircle edge:

    curl(V)=(z-1)i+j-k

    Surface can be parameterised by x and z

    r=xi+sqrt(a^2-x^2)j+zk
    δr/δx=i-(x/sqrt(a^2-x^2))jδr/δz=k

    dS=[-(x/sqrt(a^2-x^2))i-j]dxdz

    curl(V).dS=[1+x(1-z)/sqrt(a^2-x^2)]dxdz

    Limits: -a<x<a, 0<z<h

    ∫∫[1+x(1-z)/sqrt(a^2-x^2)]dxdz=2ah


    The flat edge:

    y=0

    r=xi+zk
    δr/δx=iδr/δz=k

    dS=-dxdzj

    curl(V).dS=-dxdx

    Limits: -a<x<a, 0<z<h

    -∫∫dxdz=-2ah

    The surface integrals of the two edges add to give zero.


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    (Original post by Nuclear Ghost)
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    Eurgh I seem to have made a sign error somewhere, so I've got two answers - differing by a factor of -1.
    That said, you're missing one of the surfaces - the half-disk at z=h.

    Edit: Noticed my mistake, you're good apart from the additional surface.
    Also it seems more natural to parameterise the cylinder with cylindrical polars, but I guess that's a matter of taste.
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    (Original post by joostan)
    Eurgh I seem to have made a sign error somewhere, so I've got two answers - differing by a factor of -1.
    That said, you're missing one of the surfaces - the half-disk at z=h.

    Edit: Noticed my mistake, you're good apart from the additional surface.
    Also it seems more natural to parameterise the cylinder with cylindrical polars, but I guess that's a matter of taste.
    I did use cylindrical coordinates, but it did get pretty messy. Either a careless mistake somewhere or just the nature of this question. Anyways, thanx.
 
 
 
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