Vector analysis, calculate path

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SUMMARY

The discussion focuses on calculating line integrals over three unit circles C1, C2, and C3 centered at (0,0), (-2,0), and (-1,0) respectively, within a conservative vector field H. It is established that the integrals over C1 and C2 equal zero due to the conservative nature of H, allowing for the "repair" of the hole in the middle. The integral over C3 is expressed as a combination of four integrals, I1, I2, I3, and I4, with the conclusion that I_c can be evaluated by considering the contributions from small circles S0 and S2, leading to the final expression I_c = I_3 + I_4 - I_3' - I_4'.

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  • Understanding of conservative vector fields
  • Knowledge of line integrals in vector calculus
  • Familiarity with unit circles and their properties
  • Ability to evaluate integrals along closed paths
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Hannibal123
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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)
 

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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.
 
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