Q: Let(adsbygoogle = window.adsbygoogle || []).push({}); F= (-y/(x^{2}+y^{2}), x/(x^{2}+y^{2}), z) be a vector field and let U be the interior of the torus obtained by rotating the circle (x-2)^{2}+ z^{2}= 1, y=0 about the z-axis.

a) Show that curlF=0but

∫F. dx= 2pi where dx=(dx,dy,dz)

C

and C (contained in U) is the circle x^{2}+y^{2}=4, z=0. ThereforeFis NOT conservative.

b) Determine the possible values of

(0,2,0)

∫F. dxon a path in U.

(2,0,0)

===================

I am OK with part a. But part b is tricky!!

Solution to part b:

f(x,y,z)=arctan(y/x) + z^{2}/2 is a function such that grad f =F

=> f(x,y,z)=arctan(rsin[tex]\theta[/tex]/rcos[tex]\theta[/tex]) (since z=0)

=> f=[tex]\theta[/tex]

(0,2,0)

∫F. dx= f([tex]\theta[/tex]=pi/2) - f([tex]\theta[/tex]=0) = pi/2 by F.T.C. for line integrals

(2,0,0)

Final Answer:

(0,2,0)

∫F. dx= pi/2 + (2n)pi

(2,0,0)

==============

Now, I don't understand the parts in red.

Why is z=0? This is NOT true for everywhere in U

Why + (2n)pi ? What if the path isnot a circle? This is possible because the torus has some "width" and "height"

Can someone please explain the tough part (part b)? I would really appreciate it!

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# Homework Help: Conservative vector fields

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