- #1
kingwinner
- 1,270
- 0
Q: Let F= (-y/(x2+y2), x/(x2+y2), z) be a vector field and let U be the interior of the torus obtained by rotating the circle (x-2)2 + z2 = 1, y=0 about the z-axis.
a) Show that curl F=0 but
∫ F . dx = 2pi where dx=(dx,dy,dz)
C
and C (contained in U) is the circle x2+y2=4, z=0. Therefore F is NOT conservative.
b) Determine the possible values of
(0,2,0)
∫ F . dx 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) + z2/2 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 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!
a) Show that curl F=0 but
∫ F . dx = 2pi where dx=(dx,dy,dz)
C
and C (contained in U) is the circle x2+y2=4, z=0. Therefore F is NOT conservative.
b) Determine the possible values of
(0,2,0)
∫ F . dx 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) + z2/2 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 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!
Last edited: