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    Having done the very first part of this question and then staring at it for quite some time I have resigned myself to the fact that I can't do it. Can anyone help?

    Link to question:
    http://img56.exs.cx/img56/8182/question12ro.jpg
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    From Green's Theorem in the plane

    Int[C] F.dr =

    Int[R] curlF.dA =

    Int[R] (0,0,2xsiny+xsiny).(0,0,1)dxdy =

    Int[x:1->2] Int[y:x^2+1->2] 3xsiny dy dx =

    3 Int[x:1->2] x [-cosy] [y:x^2+1->2] =

    3 Int[x:1->2] x (cos(x^2+1)-cos2) =

    3 {-cos2 + [1/2 sin(x^2+1)] [x:1->2]} =

    3 {-cos2 + (sin5-sin1)/2}
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    Thanks

    Is there a method that doesn't involve curlF? (I haven't covered it yet, so if there's no other way I shall be having words with the person who gave me this question...)
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    (Original post by flyinghorse)
    Thanks

    Is there a method that doesn't involve curlF? (I haven't covered it yet, so if there's no other way I shall be having words with the person who gave me this question...)
    I've just used Green's Theorem in the plane - which involves curl even if you've not written it that way.

    Compare what I did with the statement of Green't Theorem that you have.

    I also didn't bother to write down the three path integrals that make up the boundary.
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    (Original post by RichE)
    From Green's Theorem in the plane

    Int[C] F.dr =

    Int[R] curlF.dA =

    Int[R] (0,0,2xsiny+xsiny).(0,0,1)dxdy =

    Int[x:1->2] Int[y:x^2+1->2] 3xsiny dy dx =

    3 Int[x:1->2] x [-cosy] [y:x^2+1->2] =

    3 Int[x:1->2] x (cos(x^2+1)-cos2) =

    3 {-cos2 + [1/2 sin(x^2+1)] [x:1->2]} =

    3 {-cos2 + (sin5-sin1)/2}

    I think a slight correction is needed to otherwise perfect solution by RichE:

    3 Int[x:0->1] x (cos(x^2+1)-cos2) =

    3 {[-(x^2/2)cos2 + 1/2 sin(x^2+1)] [x:0->1]} =

    -3 {(cos2-sin2+sin1)/2}

    maybe I am wrong about the limits of integration here([x:0->1], not [x:1->2]).
    if so, please explain why?
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    No, you're absolutely right - the x-limits are 0->1
 
 
 
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