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Q: Let

a) Show that curl

∫

C

and C (contained in U) is the circle x

b) Determine the possible values of

(0,2,0)

∫

(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

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

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

(0,2,0)

∫

(2,0,0)

Final Answer:

(0,2,0)

∫

(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 is

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

**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 curl

**F**=**0**but∫

**F**. d**x**= 2pi where d**x**=(dx,dy,dz)C

and C (contained in U) is the circle x

^{2}+y^{2}=4, z=0. Therefore**F**is NOT conservative.b) Determine the possible values of

(0,2,0)

∫

**F**. d**x**on 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**. d**x**= 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**. d**x**= 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 is

*not 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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