Vector analysis, calculate path

[Hannibal123]

Homework Statement

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

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

(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)

 

[Ray Vickson]

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
I_c = I_1 + I_2 + I_3 + I_4,
where
\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}
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.