Vector analysis, calculate path

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 2K views
Hannibal123
Messages
20
Reaction score
0

Homework Statement


I have these vectorfields defined on the specified set. All of them are conservative on their set.
upload_2016-1-17_12-14-15.png


Also i have three unit circles C1, C2 and C3 centered respectively on (0,0) , (-2,0) and (-1,0)

I need to find the line integrals over all of them on H
upload_2016-1-17_12-17-23.png

(beregn = calculate, og = and. (I'm Danish))

2. The attempt at a solution
Since H is conservative i think both integrals in (b) equals zero because you can just "repair" the hole in the middle. Not sure about (c)
 

Attachments

  • upload_2016-1-17_12-13-10.png
    upload_2016-1-17_12-13-10.png
    9 KB · Views: 483
Hannibal123 said:

Homework Statement


I have these vectorfields defined on the specified set. All of them are conservative on their set.
View attachment 94407

Also i have three unit circles C1, C2 and C3 centered respectively on (0,0) , (-2,0) and (-1,0)

I need to find the line integrals over all of them on H
View attachment 94408
(beregn = calculate, og = and. (I'm Danish))

2. The attempt at a solution
Since H is conservative i think both integrals in (b) equals zero because you can just "repair" the hole in the middle. Not sure about (c)

You can draw small circles S2 of small radius r around (-2,0) and S0 of small radius r around (0,0). Suppose S2 intersects C3 at A and B (where A is nearly (-2,r) and B is nearly (-2,-r)),and S0 intersects C3 at C and D (where C is nearly (0,-r) and D is nearly (0,r)). Then your integral in (c) equals
[tex]I_c = I_1 + I_2 + I_3 + I_4,[/tex]
where
[tex]\begin{array}{l}I_1 = \int_D^A \vec{H} \cdot d\vec{r} \\<br /> I_2 = \int_B^C \vec{H} \cdot d\vec{r} \\<br /> I_3 = \int_C^D \vec{H} \cdot d\vec{r} \; \text{along C3}\\<br /> I_4 = \int_A^B \vec{H} \cdot d\vec{r} \; \text{along C3}<br /> \end{array}[/tex]
Also, let ##I_3'## be the integral from C to D along the small circle S0 (clockwise, inside the big circle C3) and let ##I_4'## be the integral from A to B along the small circle S2 (clockwise, inside C3). We have ##I_1 + I_2 + I_3' + I_4' = 0## because there are no singularities inside the corresponding closed path. Thus, the answer ##I_c = I_3+I_4-I_3' - I_4'##. We can try to evaluate ##I_3 - I_3'## directly, since inside the small circle S0 the direct path (along C3) from C to D is almost a straight line from (0,-r) to (0,r), so ##I_3 - I_3'## is an integral along a closed semicircle, where the singularity in H at (0,0) does not appear because x = 0 along the vertical line through (0,0); that is, the term ##y dx/(x^2+y^2) = 0## because dx = 0 on the vertical line, and ##x dy/(x^2+y^2) = 0## because ##x=0## on the vertical line.
 
Last edited: