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# Green's Theorem Question watch

1. 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?

http://img56.exs.cx/img56/8182/question12ro.jpg
2. 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}
3. 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...)
4. (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.
5. (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]).
6. No, you're absolutely right - the x-limits are 0->1

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