Counter-Clockwise Circulation and Parameterization: Reviewing Past Question and Answer

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In this question, I was trying to find the counter-clockwise circulation for the field $mathbfF$ and curve $C$ in this image:

As you can see, Sorfosh kindly took the time to answer my question, but I am confused by his answer:

1. The solution he got for the circulation was $oint_partial D (P, dx+Q, dy) =-4pi$. However, according to my textbook, a counter-clockwise circulation should be a positive number; negative numbers indicate clockwise circulation. Since we're trying to find the counter-clockwise circulation, shouldn't we be getting a positive value?




2. His parameterization is $x=sqrt2cos(t)$ and $y=sin(t)$. But if we let $t = 0$, then we get $x = sqrt2$ and $y = 0$, which is not the curve in the image? Should we not be parameterizing the curve $C$ in such a way as to get $x = 2$ and $y = 0$ when $t = 0$?

Thank you all for taking the time to help.

vector-analysis parametrization greens-function

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asked Jul 23 at 14:44

Wyuw

1127

• 
  
  
  

  
  

  
  
  
  
  
  

  
  So $P(x,y)=x+3y$ and $Q(x,y) = 2x-y$? Don't see in either answer where $P$ and $Q$ are defined, so it's a bit hard to follow.
  – John
  Jul 23 at 14:55
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  In the figure the tags on the $x$-axis should be $sqrt2$ instead of $2$.
  – Christian Blatter
  Jul 23 at 15:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @John Yes, it should be that $P(x,y)=x+3y$ and $Q(x,y) = 2x-y$, as per the field $mathbfF$.
  – Wyuw
  Jul 23 at 15:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @ChristianBlatter Hi Christian. Are you saying that the textbook image is incorrect? How do you know it should be $sqrt2$ instead of 2? And what about the circulation?
  – Wyuw
  Jul 23 at 15:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The picture is of the ellipse $x^2 + 4y^2 = 4.$ For the ciruclation around the curve $x^2 + 2y^2 = 2$ in the counter clockwise direction, I get $-sqrt 2 pi$
  – Doug M
  Jul 23 at 15:36
  
  

  
  
  
  

 | 
show 1 more comment

up vote
3
down vote

favorite

In this question, I was trying to find the counter-clockwise circulation for the field $mathbfF$ and curve $C$ in this image:

As you can see, Sorfosh kindly took the time to answer my question, but I am confused by his answer:

1. The solution he got for the circulation was $oint_partial D (P, dx+Q, dy) =-4pi$. However, according to my textbook, a counter-clockwise circulation should be a positive number; negative numbers indicate clockwise circulation. Since we're trying to find the counter-clockwise circulation, shouldn't we be getting a positive value?




2. His parameterization is $x=sqrt2cos(t)$ and $y=sin(t)$. But if we let $t = 0$, then we get $x = sqrt2$ and $y = 0$, which is not the curve in the image? Should we not be parameterizing the curve $C$ in such a way as to get $x = 2$ and $y = 0$ when $t = 0$?

Thank you all for taking the time to help.

vector-analysis parametrization greens-function

share|cite|improve this question

asked Jul 23 at 14:44

Wyuw

1127

• 
  
  
  

  
  

  
  
  
  
  
  

  
  So $P(x,y)=x+3y$ and $Q(x,y) = 2x-y$? Don't see in either answer where $P$ and $Q$ are defined, so it's a bit hard to follow.
  – John
  Jul 23 at 14:55
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  In the figure the tags on the $x$-axis should be $sqrt2$ instead of $2$.
  – Christian Blatter
  Jul 23 at 15:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @John Yes, it should be that $P(x,y)=x+3y$ and $Q(x,y) = 2x-y$, as per the field $mathbfF$.
  – Wyuw
  Jul 23 at 15:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @ChristianBlatter Hi Christian. Are you saying that the textbook image is incorrect? How do you know it should be $sqrt2$ instead of 2? And what about the circulation?
  – Wyuw
  Jul 23 at 15:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The picture is of the ellipse $x^2 + 4y^2 = 4.$ For the ciruclation around the curve $x^2 + 2y^2 = 2$ in the counter clockwise direction, I get $-sqrt 2 pi$
  – Doug M
  Jul 23 at 15:36
  
  

  
  
  
  

 | 
show 1 more comment

up vote
3
down vote

favorite

up vote
3
down vote

favorite

In this question, I was trying to find the counter-clockwise circulation for the field $mathbfF$ and curve $C$ in this image:

As you can see, Sorfosh kindly took the time to answer my question, but I am confused by his answer:

1. The solution he got for the circulation was $oint_partial D (P, dx+Q, dy) =-4pi$. However, according to my textbook, a counter-clockwise circulation should be a positive number; negative numbers indicate clockwise circulation. Since we're trying to find the counter-clockwise circulation, shouldn't we be getting a positive value?




2. His parameterization is $x=sqrt2cos(t)$ and $y=sin(t)$. But if we let $t = 0$, then we get $x = sqrt2$ and $y = 0$, which is not the curve in the image? Should we not be parameterizing the curve $C$ in such a way as to get $x = 2$ and $y = 0$ when $t = 0$?

Thank you all for taking the time to help.

vector-analysis parametrization greens-function

share|cite|improve this question

asked Jul 23 at 14:44

Wyuw

1127

In this question, I was trying to find the counter-clockwise circulation for the field $mathbfF$ and curve $C$ in this image:

As you can see, Sorfosh kindly took the time to answer my question, but I am confused by his answer:

1. The solution he got for the circulation was $oint_partial D (P, dx+Q, dy) =-4pi$. However, according to my textbook, a counter-clockwise circulation should be a positive number; negative numbers indicate clockwise circulation. Since we're trying to find the counter-clockwise circulation, shouldn't we be getting a positive value?




2. His parameterization is $x=sqrt2cos(t)$ and $y=sin(t)$. But if we let $t = 0$, then we get $x = sqrt2$ and $y = 0$, which is not the curve in the image? Should we not be parameterizing the curve $C$ in such a way as to get $x = 2$ and $y = 0$ when $t = 0$?

Thank you all for taking the time to help.

vector-analysis parametrization greens-function

share|cite|improve this question

asked Jul 23 at 14:44

Wyuw

1127

share|cite|improve this question

share|cite|improve this question

asked Jul 23 at 14:44

Wyuw

1127

asked Jul 23 at 14:44

Wyuw

1127

asked Jul 23 at 14:44

Wyuw

1127

1127

• 
  
  
  

  
  

  
  
  
  
  
  

  
  So $P(x,y)=x+3y$ and $Q(x,y) = 2x-y$? Don't see in either answer where $P$ and $Q$ are defined, so it's a bit hard to follow.
  – John
  Jul 23 at 14:55
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  In the figure the tags on the $x$-axis should be $sqrt2$ instead of $2$.
  – Christian Blatter
  Jul 23 at 15:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @John Yes, it should be that $P(x,y)=x+3y$ and $Q(x,y) = 2x-y$, as per the field $mathbfF$.
  – Wyuw
  Jul 23 at 15:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @ChristianBlatter Hi Christian. Are you saying that the textbook image is incorrect? How do you know it should be $sqrt2$ instead of 2? And what about the circulation?
  – Wyuw
  Jul 23 at 15:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The picture is of the ellipse $x^2 + 4y^2 = 4.$ For the ciruclation around the curve $x^2 + 2y^2 = 2$ in the counter clockwise direction, I get $-sqrt 2 pi$
  – Doug M
  Jul 23 at 15:36
  
  

  
  
  
  

 | 
show 1 more comment

• 
  
  
  

  
  

  
  
  
  
  
  

  
  So $P(x,y)=x+3y$ and $Q(x,y) = 2x-y$? Don't see in either answer where $P$ and $Q$ are defined, so it's a bit hard to follow.
  – John
  Jul 23 at 14:55
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  In the figure the tags on the $x$-axis should be $sqrt2$ instead of $2$.
  – Christian Blatter
  Jul 23 at 15:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @John Yes, it should be that $P(x,y)=x+3y$ and $Q(x,y) = 2x-y$, as per the field $mathbfF$.
  – Wyuw
  Jul 23 at 15:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @ChristianBlatter Hi Christian. Are you saying that the textbook image is incorrect? How do you know it should be $sqrt2$ instead of 2? And what about the circulation?
  – Wyuw
  Jul 23 at 15:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The picture is of the ellipse $x^2 + 4y^2 = 4.$ For the ciruclation around the curve $x^2 + 2y^2 = 2$ in the counter clockwise direction, I get $-sqrt 2 pi$
  – Doug M
  Jul 23 at 15:36
  
  

  
  
  
  

So $P(x,y)=x+3y$ and $Q(x,y) = 2x-y$? Don't see in either answer where $P$ and $Q$ are defined, so it's a bit hard to follow.
– John
Jul 23 at 14:55

So $P(x,y)=x+3y$ and $Q(x,y) = 2x-y$? Don't see in either answer where $P$ and $Q$ are defined, so it's a bit hard to follow.
– John
Jul 23 at 14:55

In the figure the tags on the $x$-axis should be $sqrt2$ instead of $2$.
– Christian Blatter
Jul 23 at 15:02

In the figure the tags on the $x$-axis should be $sqrt2$ instead of $2$.
– Christian Blatter
Jul 23 at 15:02

@John Yes, it should be that $P(x,y)=x+3y$ and $Q(x,y) = 2x-y$, as per the field $mathbfF$.
– Wyuw
Jul 23 at 15:02

@John Yes, it should be that $P(x,y)=x+3y$ and $Q(x,y) = 2x-y$, as per the field $mathbfF$.
– Wyuw
Jul 23 at 15:02

@ChristianBlatter Hi Christian. Are you saying that the textbook image is incorrect? How do you know it should be $sqrt2$ instead of 2? And what about the circulation?
– Wyuw
Jul 23 at 15:09

@ChristianBlatter Hi Christian. Are you saying that the textbook image is incorrect? How do you know it should be $sqrt2$ instead of 2? And what about the circulation?
– Wyuw
Jul 23 at 15:09

The picture is of the ellipse $x^2 + 4y^2 = 4.$ For the ciruclation around the curve $x^2 + 2y^2 = 2$ in the counter clockwise direction, I get $-sqrt 2 pi$
– Doug M
Jul 23 at 15:36

The picture is of the ellipse $x^2 + 4y^2 = 4.$ For the ciruclation around the curve $x^2 + 2y^2 = 2$ in the counter clockwise direction, I get $-sqrt 2 pi$
– Doug M
Jul 23 at 15:36

 | 
show 1 more comment

1 Answer
1

active

oldest

votes

up vote
1
down vote



accepted

If we attack this directly.

$x = sqrt 2 cos t\y = sin t$

$int_0^2pi (sqrt 2cos t + 3sin t, 2sqrt 2cos t -sin t)cdot(-sqrt 2 sin t, cos t) dt\
int_0^2pi -2cos tsin t - 3sqrt 2sin^2 t + 2sqrt 2cos^2 t -cos tsin t dt \ -sqrt 2pi$

or we can use Greens theorem.

$oint P dx + Q dy = iint fracpartial Qpartial x - frac partial Ppartial y dx dy\
Q = 2x-y, frac partial Qpartial x = 2\
P = x+3y, frac partial Qpartial y = 3\
iint fracpartial Qpartial x - frac partial Ppartial y dx dy = iint -1 dx dy$

Or you can use Stoke theorem (which is essentially the same as Greens)

$oint Fcdot dr = iint nabla times F dV\
nabla times F = -1\
iint -1 dA$

And the area of the ellipse is $sqrt 2 pi$

If you reverse the direction you will change the signs of the integrals. However, that doesn't mean that the that the result will always be positive if you travel in the counter-clockwise direction. Particularly, if your vector field is curling in the opposite direction.

share|cite|improve this answer

edited Jul 23 at 16:51

answered Jul 23 at 16:30

Doug M

39.1k31749

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for the help Doug!
  – Wyuw
  Jul 23 at 16:50
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

If we attack this directly.

$x = sqrt 2 cos t\y = sin t$

$int_0^2pi (sqrt 2cos t + 3sin t, 2sqrt 2cos t -sin t)cdot(-sqrt 2 sin t, cos t) dt\
int_0^2pi -2cos tsin t - 3sqrt 2sin^2 t + 2sqrt 2cos^2 t -cos tsin t dt \ -sqrt 2pi$

or we can use Greens theorem.

$oint P dx + Q dy = iint fracpartial Qpartial x - frac partial Ppartial y dx dy\
Q = 2x-y, frac partial Qpartial x = 2\
P = x+3y, frac partial Qpartial y = 3\
iint fracpartial Qpartial x - frac partial Ppartial y dx dy = iint -1 dx dy$

Or you can use Stoke theorem (which is essentially the same as Greens)

$oint Fcdot dr = iint nabla times F dV\
nabla times F = -1\
iint -1 dA$

And the area of the ellipse is $sqrt 2 pi$

If you reverse the direction you will change the signs of the integrals. However, that doesn't mean that the that the result will always be positive if you travel in the counter-clockwise direction. Particularly, if your vector field is curling in the opposite direction.

share|cite|improve this answer

edited Jul 23 at 16:51

answered Jul 23 at 16:30

Doug M

39.1k31749

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for the help Doug!
  – Wyuw
  Jul 23 at 16:50
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

If we attack this directly.

$x = sqrt 2 cos t\y = sin t$

$int_0^2pi (sqrt 2cos t + 3sin t, 2sqrt 2cos t -sin t)cdot(-sqrt 2 sin t, cos t) dt\
int_0^2pi -2cos tsin t - 3sqrt 2sin^2 t + 2sqrt 2cos^2 t -cos tsin t dt \ -sqrt 2pi$

or we can use Greens theorem.

$oint P dx + Q dy = iint fracpartial Qpartial x - frac partial Ppartial y dx dy\
Q = 2x-y, frac partial Qpartial x = 2\
P = x+3y, frac partial Qpartial y = 3\
iint fracpartial Qpartial x - frac partial Ppartial y dx dy = iint -1 dx dy$

Or you can use Stoke theorem (which is essentially the same as Greens)

$oint Fcdot dr = iint nabla times F dV\
nabla times F = -1\
iint -1 dA$

And the area of the ellipse is $sqrt 2 pi$

If you reverse the direction you will change the signs of the integrals. However, that doesn't mean that the that the result will always be positive if you travel in the counter-clockwise direction. Particularly, if your vector field is curling in the opposite direction.

share|cite|improve this answer

edited Jul 23 at 16:51

answered Jul 23 at 16:30

Doug M

39.1k31749

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for the help Doug!
  – Wyuw
  Jul 23 at 16:50
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

If we attack this directly.

$x = sqrt 2 cos t\y = sin t$

$int_0^2pi (sqrt 2cos t + 3sin t, 2sqrt 2cos t -sin t)cdot(-sqrt 2 sin t, cos t) dt\
int_0^2pi -2cos tsin t - 3sqrt 2sin^2 t + 2sqrt 2cos^2 t -cos tsin t dt \ -sqrt 2pi$

or we can use Greens theorem.

$oint P dx + Q dy = iint fracpartial Qpartial x - frac partial Ppartial y dx dy\
Q = 2x-y, frac partial Qpartial x = 2\
P = x+3y, frac partial Qpartial y = 3\
iint fracpartial Qpartial x - frac partial Ppartial y dx dy = iint -1 dx dy$

Or you can use Stoke theorem (which is essentially the same as Greens)

$oint Fcdot dr = iint nabla times F dV\
nabla times F = -1\
iint -1 dA$

And the area of the ellipse is $sqrt 2 pi$

If you reverse the direction you will change the signs of the integrals. However, that doesn't mean that the that the result will always be positive if you travel in the counter-clockwise direction. Particularly, if your vector field is curling in the opposite direction.

share|cite|improve this answer

edited Jul 23 at 16:51

answered Jul 23 at 16:30

Doug M

39.1k31749

If we attack this directly.

$x = sqrt 2 cos t\y = sin t$

$int_0^2pi (sqrt 2cos t + 3sin t, 2sqrt 2cos t -sin t)cdot(-sqrt 2 sin t, cos t) dt\
int_0^2pi -2cos tsin t - 3sqrt 2sin^2 t + 2sqrt 2cos^2 t -cos tsin t dt \ -sqrt 2pi$

or we can use Greens theorem.

$oint P dx + Q dy = iint fracpartial Qpartial x - frac partial Ppartial y dx dy\
Q = 2x-y, frac partial Qpartial x = 2\
P = x+3y, frac partial Qpartial y = 3\
iint fracpartial Qpartial x - frac partial Ppartial y dx dy = iint -1 dx dy$

Or you can use Stoke theorem (which is essentially the same as Greens)

$oint Fcdot dr = iint nabla times F dV\
nabla times F = -1\
iint -1 dA$

And the area of the ellipse is $sqrt 2 pi$

If you reverse the direction you will change the signs of the integrals. However, that doesn't mean that the that the result will always be positive if you travel in the counter-clockwise direction. Particularly, if your vector field is curling in the opposite direction.

share|cite|improve this answer

edited Jul 23 at 16:51

answered Jul 23 at 16:30

Doug M

39.1k31749

share|cite|improve this answer

share|cite|improve this answer

edited Jul 23 at 16:51

edited Jul 23 at 16:51

edited Jul 23 at 16:51

answered Jul 23 at 16:30

Doug M

39.1k31749

answered Jul 23 at 16:30

Doug M

39.1k31749

answered Jul 23 at 16:30

Doug M

39.1k31749

39.1k31749

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for the help Doug!
  – Wyuw
  Jul 23 at 16:50
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for the help Doug!
  – Wyuw
  Jul 23 at 16:50
  

  
  
  
  

Thank you for the help Doug!
– Wyuw
Jul 23 at 16:50

Thank you for the help Doug!
– Wyuw
Jul 23 at 16:50

add a comment | 

 

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