Green's Theorem with Singularities

[Umar]

So here's the question:

You are given that F is a conservative vector field, except for singularities
at the points (0,1), (2,0), (3,0), and (0,4). You are given the following information
about line integrals around the following closed paths:

1) Around the curve C1 given by x^2 + y^2 = 2,

$\int$ F $\cdot$ dr = 2

2) Around the curve C2 given by x^2 + y^2 = 5,

$\int$ F $\cdot$ dr = 5

3) Around the curve C3 given by x^2 + y^2 = 10,

$\int$ F $\cdot$ dr = 10

What is the $\int$ F $\cdot$ dr equal to for C4 where C4 is given by (x-2.5)^2 + y^2 = 4, the circle of radius 2 centered at (2.5,0)?

(a) 15
(b) 5
(c) 10
(d) 8
(e) 17

The answer is apparently 8. I've tried drawing out the curves and for C4, I notice it bounds the singularities (2,0) and (3,0). But since we are not given the vector field, I assume I need to use the results listed above to somehow compute the line integral for C4, though I'm not sure how exactly to go about doing this.

Any help is greatly appreciated!

 

[I like Serena]

So here's the question:

You are given that F is a conservative vector field, except for singularities
at the points (0,1), (2,0), (3,0), and (0,4). You are given the following information
about line integrals around the following closed paths:

1) Around the curve C1 given by x^2 + y^2 = 2,

$\int$ F $\cdot$ dr = 2

2) Around the curve C2 given by x^2 + y^2 = 5,

$\int$ F $\cdot$ dr = 5

3) Around the curve C3 given by x^2 + y^2 = 10,

$\int$ F $\cdot$ dr = 10

What is the $\int$ F $\cdot$ dr equal to for C4 where C4 is given by (x-2.5)^2 + y^2 = 4, the circle of radius 2 centered at (2.5,0)?

(a) 15
(b) 5
(c) 10
(d) 8
(e) 17

The answer is apparently 8. I've tried drawing out the curves and for C4, I notice it bounds the singularities (2,0) and (3,0). But since we are not given the vector field, I assume I need to use the results listed above to somehow compute the line integral for C4, though I'm not sure how exactly to go about doing this.

Any help is greatly appreciated!

Hi Umar,

This is not so much about Green's Theorem, but more about the Residue theorem.
To summarize, the line integral along a closed path is zero unless a it loops around 1 or more poles.
And if poles are involved, then we can sum the contribution of each pole separately.
That is, the contribution of a pole, when looping around it once in a counter clock wise fashion is fixed.

From the given information, we can deduce that:

\(\oint_{C_1} F\cdot dr = \oint_{\text{loop around (0,1)}} F\cdot dr = 2 \\

\oint_{C_2} F\cdot dr = \oint_{\text{loop around (0,1) and (2,0)}} F\cdot dr = \oint_{\text{loop around (0,1)}} F\cdot dr + \oint_{\text{loop around (2,0)}} F\cdot dr= 5 \\

\oint_{C_3} F\cdot dr = \oint_{\text{loop around (0,1), (2,0), and (3,0)}} F\cdot dr = \oint_{\text{loop around (0,1)}} F\cdot dr + \oint_{\text{loop around (2,0)}} F\cdot dr + \oint_{\text{loop around (3,0)}} F\cdot dr= 10
\)

and we're looking for:

\( \oint_{C_4} F\cdot dr = \oint_{\text{loop around (2,0) and (3,0)}} F\cdot dr \)

Can you find it?

 

[Umar]

Hi Umar,

This is not so much about Green's Theorem, but more about the Residue theorem.
To summarize, the line integral along a closed path is zero unless a it loops around 1 or more poles.
And if poles are involved, then we can sum the contribution of each pole separately.
That is, the contribution of a pole, when looping around it once in a counter clock wise fashion is fixed.

From the given information, we can deduce that:

\(\oint_{C_1} F\cdot dr = \oint_{\text{loop around (0,1)}} F\cdot dr = 2 \\

\oint_{C_2} F\cdot dr = \oint_{\text{loop around (0,1) and (2,0)}} F\cdot dr = \oint_{\text{loop around (0,1)}} F\cdot dr + \oint_{\text{loop around (2,0)}} F\cdot dr= 5 \\

\oint_{C_3} F\cdot dr = \oint_{\text{loop around (0,1), (2,0), and (3,0)}} F\cdot dr = \oint_{\text{loop around (0,1)}} F\cdot dr + \oint_{\text{loop around (2,0)}} F\cdot dr + \oint_{\text{loop around (3,0)}} F\cdot dr= 10
\)

and we're looking for:

\( \oint_{C_4} F\cdot dr = \oint_{\text{loop around (2,0), and (3,0)}} F\cdot dr \)

Can you find it?

Yup, I see it! Thank you for explaining it so easily (Yes). On a unrelated side note, these poles you mentioned are the singularities in the question, yes? So if we had a conservative vector field, and we wanted to compute the area using Green's Theorem over a region with a singularity, as in the question above, would we be able to draw a circle around the point and compute the line integral that way to get around the problem? I'm asking this because in my textbook, there was an example with a rectangle, that had a singularity at the point (0,0). After some derivation, it was proved that the line integral around the rectangle was equal to a line integral of a circle that was entirely within the rectangle, but surrounded the origin, and the result was computed that way. Not sure if that made sense but hopefully you can provide some insight!

 

[I like Serena]

Yup, I see it! Thank you for explaining it so easily (Yes). On a unrelated side note, these poles you mentioned are the singularities in the question, yes? So if we had a conservative vector field, and we wanted to compute the area using Green's Theorem over a region with a singularity, as in the question above, would we be able to draw a circle around the point and compute the line integral that way to get around the problem? I'm asking this because in my textbook, there was an example with a rectangle, that had a singularity at the point (0,0). After some derivation, it was proved that the line integral around the rectangle was equal to a line integral of a circle that was entirely within the rectangle, but surrounded the origin, and the result was computed that way. Not sure if that made sense but hopefully you can provide some insight!

Yes, those poles are the same as the singularities in the question.

Not sure what you mean by 'compute the area'.
Either way, the line integral along a rectangle would indeed be the same as any line integral along a path in the same direction.
So if we can easily calculate it along a circle that's the way to go!