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Using Green's Theorem to Calculate the Counter-Clockwise Circulation for the Field $mathbfF$ and Curve $C$.
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I have this problem
Use Green’s Theorem to find the counter-clockwise circulation for the field $mathbfF$ and curve $C$.
with this image
Green's Theorem says that the counter-clockwise circulation is $oint_C mathbfF cdot mathbfT ds = oint_C M dx + N dy$. I will use the latter formula.
We can see from the vector field $mathbfF$ that $M = x + 3y$ and $N = 2x - y$.
The parameterization I used is $x = cos(t)$, $y = sin(t)$ $forall 0 le t le 2pi$. Therefore, the parameterized curve is $r(t) = 2cos(t) mathbfi + sin(t) mathbfj$.
Taking the derivatives, we get $dx = -sin(t) dt$ and $dy = cos(t) dt$.
So we get
$int_C M dx + N dy = int^2pi_0 (cos(t) + 3(sin(t))(-sin(t)) + (2cos(t) - sin(t))(cos(t)) dt = -pi$ by my calculations. I confirmed my calculations by using this calculator.
But I'm unsure if this is correct. Can someone please check my work?
vector-analysis line-integrals greens-theorem
asked Jul 22 at 20:48
Wyuw
1127
add a comment |Â
up vote
2
down vote
favorite
I have this problem
Use Green’s Theorem to find the counter-clockwise circulation for the field $mathbfF$ and curve $C$.
with this image
Green's Theorem says that the counter-clockwise circulation is $oint_C mathbfF cdot mathbfT ds = oint_C M dx + N dy$. I will use the latter formula.
We can see from the vector field $mathbfF$ that $M = x + 3y$ and $N = 2x - y$.
The parameterization I used is $x = cos(t)$, $y = sin(t)$ $forall 0 le t le 2pi$. Therefore, the parameterized curve is $r(t) = 2cos(t) mathbfi + sin(t) mathbfj$.
Taking the derivatives, we get $dx = -sin(t) dt$ and $dy = cos(t) dt$.
So we get
$int_C M dx + N dy = int^2pi_0 (cos(t) + 3(sin(t))(-sin(t)) + (2cos(t) - sin(t))(cos(t)) dt = -pi$ by my calculations. I confirmed my calculations by using this calculator.
But I'm unsure if this is correct. Can someone please check my work?
vector-analysis line-integrals greens-theorem
asked Jul 22 at 20:48
Wyuw
1127
Green's theorem converts the line integral to a double integral. You just calculated the line integral by parametrization, which is a valid approach as well, but not what the question asks. However, you parametrized the elipse wrong.
– Sorfosh
Jul 22 at 22:23
@Sorfosh Ok thanks. What is the correct parameterization?
– Wyuw
Jul 22 at 22:29
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I have this problem
Use Green’s Theorem to find the counter-clockwise circulation for the field $mathbfF$ and curve $C$.
with this image
Green's Theorem says that the counter-clockwise circulation is $oint_C mathbfF cdot mathbfT ds = oint_C M dx + N dy$. I will use the latter formula.
We can see from the vector field $mathbfF$ that $M = x + 3y$ and $N = 2x - y$.
The parameterization I used is $x = cos(t)$, $y = sin(t)$ $forall 0 le t le 2pi$. Therefore, the parameterized curve is $r(t) = 2cos(t) mathbfi + sin(t) mathbfj$.
Taking the derivatives, we get $dx = -sin(t) dt$ and $dy = cos(t) dt$.
So we get
$int_C M dx + N dy = int^2pi_0 (cos(t) + 3(sin(t))(-sin(t)) + (2cos(t) - sin(t))(cos(t)) dt = -pi$ by my calculations. I confirmed my calculations by using this calculator.
But I'm unsure if this is correct. Can someone please check my work?
vector-analysis line-integrals greens-theorem
asked Jul 22 at 20:48
Wyuw
1127
I have this problem
Use Green’s Theorem to find the counter-clockwise circulation for the field $mathbfF$ and curve $C$.
with this image
Green's Theorem says that the counter-clockwise circulation is $oint_C mathbfF cdot mathbfT ds = oint_C M dx + N dy$. I will use the latter formula.
We can see from the vector field $mathbfF$ that $M = x + 3y$ and $N = 2x - y$.
The parameterization I used is $x = cos(t)$, $y = sin(t)$ $forall 0 le t le 2pi$. Therefore, the parameterized curve is $r(t) = 2cos(t) mathbfi + sin(t) mathbfj$.
Taking the derivatives, we get $dx = -sin(t) dt$ and $dy = cos(t) dt$.
So we get
$int_C M dx + N dy = int^2pi_0 (cos(t) + 3(sin(t))(-sin(t)) + (2cos(t) - sin(t))(cos(t)) dt = -pi$ by my calculations. I confirmed my calculations by using this calculator.
But I'm unsure if this is correct. Can someone please check my work?
vector-analysis line-integrals greens-theorem
asked Jul 22 at 20:48
Wyuw
1127
asked Jul 22 at 20:48
Wyuw
1127
asked Jul 22 at 20:48
Wyuw
1127
asked Jul 22 at 20:48
Wyuw
1127
1127
Green's theorem converts the line integral to a double integral. You just calculated the line integral by parametrization, which is a valid approach as well, but not what the question asks. However, you parametrized the elipse wrong.
– Sorfosh
Jul 22 at 22:23
@Sorfosh Ok thanks. What is the correct parameterization?
– Wyuw
Jul 22 at 22:29
add a comment |Â
Green's theorem converts the line integral to a double integral. You just calculated the line integral by parametrization, which is a valid approach as well, but not what the question asks. However, you parametrized the elipse wrong.
– Sorfosh
Jul 22 at 22:23
@Sorfosh Ok thanks. What is the correct parameterization?
– Wyuw
Jul 22 at 22:29
Green's theorem converts the line integral to a double integral. You just calculated the line integral by parametrization, which is a valid approach as well, but not what the question asks. However, you parametrized the elipse wrong.
– Sorfosh
Jul 22 at 22:23
Green's theorem converts the line integral to a double integral. You just calculated the line integral by parametrization, which is a valid approach as well, but not what the question asks. However, you parametrized the elipse wrong.
– Sorfosh
Jul 22 at 22:23
@Sorfosh Ok thanks. What is the correct parameterization?
– Wyuw
Jul 22 at 22:29
@Sorfosh Ok thanks. What is the correct parameterization?
– Wyuw
Jul 22 at 22:29
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
As I mentioned in my comment Green's theorem draws a connection between the double integral over the region enclosed by the curve, and the line integral of a vector field over that curve. In symbolic terms:
$$oint_partial D (P, dx+Q, dy) = iint_D dx,dy : left ( fracpartial Qpartial x - fracpartial Ppartial yright)$$
Now in our case, $$fracpartial Qpartial x=2,fracpartial Ppartial y=3$$
Hence, We get, by the formula:
$$iint_D dx,dy : left ( fracpartial Qpartial x - fracpartial Ppartial yright)=iint_D 2-3dx,dy=-iint_D 1dx,dy$$
This is simply $-1$ times the area of the enlosed region, which we know is simply $sqrt2pi$. So:
$$oint_partial D (P, dx+Q, dy) =-sqrt2pi$$
Regarding the parametrization, it should be $x=sqrt2cos(t)$ and $y=sin(t)$ and it should give you the same answer.
answered Jul 22 at 22:35
Sorfosh
910616
Thank you for clearing this up. Can you please demonstrate how you arrived at that parameterization? I'm unsure why we use $sqrt2$. Also, doesn't a negative value for circulation mean that it is circulating clockwise instead of anti-clockwise? Is that ok in this context?
– Wyuw
Jul 22 at 22:59
@Wyuw The problem says counterclockwise, and if you look at the picture, it is indeed counterclockwise. So we got the direction right. Regarding parametrization, it is something you should memorize, it is a convienient tool. Simply note that when we plug in our parametrization into the equation of the elipse, we get $x^2+2y^2=2cos^2(t)+2sin^2(t)=2(cos^2(t)+sin^2(t))=2$. So our curve lays on the elipse.
– Sorfosh
Jul 23 at 8:52
1
Your $partial Qoverpartial x$ and $partial Poverpartial y$ are wrong. They should be $3$ and $2$.
– Christian Blatter
Jul 23 at 14:59
@ChristianBlatter oh boy, my bad. Corrected, thank you!
– Sorfosh
Jul 23 at 15:11
1
@Wyuw the drawing of the ellipse $x^2+2y^2=2$ is incorrect. I used the formula rather than the picture. Regarding positive negative, it is not the case that the line integral will always be positive with positive orientation. What textbook are you using? This is plainly false.
– Sorfosh
Jul 23 at 16:16
 |Â
show 2 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
As I mentioned in my comment Green's theorem draws a connection between the double integral over the region enclosed by the curve, and the line integral of a vector field over that curve. In symbolic terms:
$$oint_partial D (P, dx+Q, dy) = iint_D dx,dy : left ( fracpartial Qpartial x - fracpartial Ppartial yright)$$
Now in our case, $$fracpartial Qpartial x=2,fracpartial Ppartial y=3$$
Hence, We get, by the formula:
$$iint_D dx,dy : left ( fracpartial Qpartial x - fracpartial Ppartial yright)=iint_D 2-3dx,dy=-iint_D 1dx,dy$$
This is simply $-1$ times the area of the enlosed region, which we know is simply $sqrt2pi$. So:
$$oint_partial D (P, dx+Q, dy) =-sqrt2pi$$
Regarding the parametrization, it should be $x=sqrt2cos(t)$ and $y=sin(t)$ and it should give you the same answer.
answered Jul 22 at 22:35
Sorfosh
910616
Thank you for clearing this up. Can you please demonstrate how you arrived at that parameterization? I'm unsure why we use $sqrt2$. Also, doesn't a negative value for circulation mean that it is circulating clockwise instead of anti-clockwise? Is that ok in this context?
– Wyuw
Jul 22 at 22:59
@Wyuw The problem says counterclockwise, and if you look at the picture, it is indeed counterclockwise. So we got the direction right. Regarding parametrization, it is something you should memorize, it is a convienient tool. Simply note that when we plug in our parametrization into the equation of the elipse, we get $x^2+2y^2=2cos^2(t)+2sin^2(t)=2(cos^2(t)+sin^2(t))=2$. So our curve lays on the elipse.
– Sorfosh
Jul 23 at 8:52
1
Your $partial Qoverpartial x$ and $partial Poverpartial y$ are wrong. They should be $3$ and $2$.
– Christian Blatter
Jul 23 at 14:59
@ChristianBlatter oh boy, my bad. Corrected, thank you!
– Sorfosh
Jul 23 at 15:11
1
@Wyuw the drawing of the ellipse $x^2+2y^2=2$ is incorrect. I used the formula rather than the picture. Regarding positive negative, it is not the case that the line integral will always be positive with positive orientation. What textbook are you using? This is plainly false.
– Sorfosh
Jul 23 at 16:16
 |Â
show 2 more comments
up vote
2
down vote
accepted
As I mentioned in my comment Green's theorem draws a connection between the double integral over the region enclosed by the curve, and the line integral of a vector field over that curve. In symbolic terms:
$$oint_partial D (P, dx+Q, dy) = iint_D dx,dy : left ( fracpartial Qpartial x - fracpartial Ppartial yright)$$
Now in our case, $$fracpartial Qpartial x=2,fracpartial Ppartial y=3$$
Hence, We get, by the formula:
$$iint_D dx,dy : left ( fracpartial Qpartial x - fracpartial Ppartial yright)=iint_D 2-3dx,dy=-iint_D 1dx,dy$$
This is simply $-1$ times the area of the enlosed region, which we know is simply $sqrt2pi$. So:
$$oint_partial D (P, dx+Q, dy) =-sqrt2pi$$
Regarding the parametrization, it should be $x=sqrt2cos(t)$ and $y=sin(t)$ and it should give you the same answer.
answered Jul 22 at 22:35
Sorfosh
910616
Thank you for clearing this up. Can you please demonstrate how you arrived at that parameterization? I'm unsure why we use $sqrt2$. Also, doesn't a negative value for circulation mean that it is circulating clockwise instead of anti-clockwise? Is that ok in this context?
– Wyuw
Jul 22 at 22:59
@Wyuw The problem says counterclockwise, and if you look at the picture, it is indeed counterclockwise. So we got the direction right. Regarding parametrization, it is something you should memorize, it is a convienient tool. Simply note that when we plug in our parametrization into the equation of the elipse, we get $x^2+2y^2=2cos^2(t)+2sin^2(t)=2(cos^2(t)+sin^2(t))=2$. So our curve lays on the elipse.
– Sorfosh
Jul 23 at 8:52
1
Your $partial Qoverpartial x$ and $partial Poverpartial y$ are wrong. They should be $3$ and $2$.
– Christian Blatter
Jul 23 at 14:59
@ChristianBlatter oh boy, my bad. Corrected, thank you!
– Sorfosh
Jul 23 at 15:11
1
@Wyuw the drawing of the ellipse $x^2+2y^2=2$ is incorrect. I used the formula rather than the picture. Regarding positive negative, it is not the case that the line integral will always be positive with positive orientation. What textbook are you using? This is plainly false.
– Sorfosh
Jul 23 at 16:16
 |Â
show 2 more comments
up vote
2
down vote
accepted
up vote
2
down vote
accepted
As I mentioned in my comment Green's theorem draws a connection between the double integral over the region enclosed by the curve, and the line integral of a vector field over that curve. In symbolic terms:
$$oint_partial D (P, dx+Q, dy) = iint_D dx,dy : left ( fracpartial Qpartial x - fracpartial Ppartial yright)$$
Now in our case, $$fracpartial Qpartial x=2,fracpartial Ppartial y=3$$
Hence, We get, by the formula:
$$iint_D dx,dy : left ( fracpartial Qpartial x - fracpartial Ppartial yright)=iint_D 2-3dx,dy=-iint_D 1dx,dy$$
This is simply $-1$ times the area of the enlosed region, which we know is simply $sqrt2pi$. So:
$$oint_partial D (P, dx+Q, dy) =-sqrt2pi$$
Regarding the parametrization, it should be $x=sqrt2cos(t)$ and $y=sin(t)$ and it should give you the same answer.
answered Jul 22 at 22:35
Sorfosh
910616
As I mentioned in my comment Green's theorem draws a connection between the double integral over the region enclosed by the curve, and the line integral of a vector field over that curve. In symbolic terms:
$$oint_partial D (P, dx+Q, dy) = iint_D dx,dy : left ( fracpartial Qpartial x - fracpartial Ppartial yright)$$
Now in our case, $$fracpartial Qpartial x=2,fracpartial Ppartial y=3$$
Hence, We get, by the formula:
$$iint_D dx,dy : left ( fracpartial Qpartial x - fracpartial Ppartial yright)=iint_D 2-3dx,dy=-iint_D 1dx,dy$$
This is simply $-1$ times the area of the enlosed region, which we know is simply $sqrt2pi$. So:
$$oint_partial D (P, dx+Q, dy) =-sqrt2pi$$
Regarding the parametrization, it should be $x=sqrt2cos(t)$ and $y=sin(t)$ and it should give you the same answer.
answered Jul 22 at 22:35
Sorfosh
910616
edited Jul 23 at 16:13
answered Jul 22 at 22:35
Sorfosh
910616
answered Jul 22 at 22:35
Sorfosh
910616
answered Jul 22 at 22:35
Sorfosh
910616
910616
Thank you for clearing this up. Can you please demonstrate how you arrived at that parameterization? I'm unsure why we use $sqrt2$. Also, doesn't a negative value for circulation mean that it is circulating clockwise instead of anti-clockwise? Is that ok in this context?
– Wyuw
Jul 22 at 22:59
@Wyuw The problem says counterclockwise, and if you look at the picture, it is indeed counterclockwise. So we got the direction right. Regarding parametrization, it is something you should memorize, it is a convienient tool. Simply note that when we plug in our parametrization into the equation of the elipse, we get $x^2+2y^2=2cos^2(t)+2sin^2(t)=2(cos^2(t)+sin^2(t))=2$. So our curve lays on the elipse.
– Sorfosh
Jul 23 at 8:52
1
Your $partial Qoverpartial x$ and $partial Poverpartial y$ are wrong. They should be $3$ and $2$.
– Christian Blatter
Jul 23 at 14:59
@ChristianBlatter oh boy, my bad. Corrected, thank you!
– Sorfosh
Jul 23 at 15:11
1
@Wyuw the drawing of the ellipse $x^2+2y^2=2$ is incorrect. I used the formula rather than the picture. Regarding positive negative, it is not the case that the line integral will always be positive with positive orientation. What textbook are you using? This is plainly false.
– Sorfosh
Jul 23 at 16:16
 |Â
show 2 more comments
Thank you for clearing this up. Can you please demonstrate how you arrived at that parameterization? I'm unsure why we use $sqrt2$. Also, doesn't a negative value for circulation mean that it is circulating clockwise instead of anti-clockwise? Is that ok in this context?
– Wyuw
Jul 22 at 22:59
@Wyuw The problem says counterclockwise, and if you look at the picture, it is indeed counterclockwise. So we got the direction right. Regarding parametrization, it is something you should memorize, it is a convienient tool. Simply note that when we plug in our parametrization into the equation of the elipse, we get $x^2+2y^2=2cos^2(t)+2sin^2(t)=2(cos^2(t)+sin^2(t))=2$. So our curve lays on the elipse.
– Sorfosh
Jul 23 at 8:52
1
Your $partial Qoverpartial x$ and $partial Poverpartial y$ are wrong. They should be $3$ and $2$.
– Christian Blatter
Jul 23 at 14:59
@ChristianBlatter oh boy, my bad. Corrected, thank you!
– Sorfosh
Jul 23 at 15:11
1
@Wyuw the drawing of the ellipse $x^2+2y^2=2$ is incorrect. I used the formula rather than the picture. Regarding positive negative, it is not the case that the line integral will always be positive with positive orientation. What textbook are you using? This is plainly false.
– Sorfosh
Jul 23 at 16:16
Thank you for clearing this up. Can you please demonstrate how you arrived at that parameterization? I'm unsure why we use $sqrt2$. Also, doesn't a negative value for circulation mean that it is circulating clockwise instead of anti-clockwise? Is that ok in this context?
– Wyuw
Jul 22 at 22:59
Thank you for clearing this up. Can you please demonstrate how you arrived at that parameterization? I'm unsure why we use $sqrt2$. Also, doesn't a negative value for circulation mean that it is circulating clockwise instead of anti-clockwise? Is that ok in this context?
– Wyuw
Jul 22 at 22:59
@Wyuw The problem says counterclockwise, and if you look at the picture, it is indeed counterclockwise. So we got the direction right. Regarding parametrization, it is something you should memorize, it is a convienient tool. Simply note that when we plug in our parametrization into the equation of the elipse, we get $x^2+2y^2=2cos^2(t)+2sin^2(t)=2(cos^2(t)+sin^2(t))=2$. So our curve lays on the elipse.
– Sorfosh
Jul 23 at 8:52
@Wyuw The problem says counterclockwise, and if you look at the picture, it is indeed counterclockwise. So we got the direction right. Regarding parametrization, it is something you should memorize, it is a convienient tool. Simply note that when we plug in our parametrization into the equation of the elipse, we get $x^2+2y^2=2cos^2(t)+2sin^2(t)=2(cos^2(t)+sin^2(t))=2$. So our curve lays on the elipse.
– Sorfosh
Jul 23 at 8:52
1
1
Your $partial Qoverpartial x$ and $partial Poverpartial y$ are wrong. They should be $3$ and $2$.
– Christian Blatter
Jul 23 at 14:59
Your $partial Qoverpartial x$ and $partial Poverpartial y$ are wrong. They should be $3$ and $2$.
– Christian Blatter
Jul 23 at 14:59
@ChristianBlatter oh boy, my bad. Corrected, thank you!
– Sorfosh
Jul 23 at 15:11
@ChristianBlatter oh boy, my bad. Corrected, thank you!
– Sorfosh
Jul 23 at 15:11
1
1
@Wyuw the drawing of the ellipse $x^2+2y^2=2$ is incorrect. I used the formula rather than the picture. Regarding positive negative, it is not the case that the line integral will always be positive with positive orientation. What textbook are you using? This is plainly false.
– Sorfosh
Jul 23 at 16:16
@Wyuw the drawing of the ellipse $x^2+2y^2=2$ is incorrect. I used the formula rather than the picture. Regarding positive negative, it is not the case that the line integral will always be positive with positive orientation. What textbook are you using? This is plainly false.
– Sorfosh
Jul 23 at 16:16
 |Â
show 2 more comments
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Green's theorem converts the line integral to a double integral. You just calculated the line integral by parametrization, which is a valid approach as well, but not what the question asks. However, you parametrized the elipse wrong.
– Sorfosh
Jul 22 at 22:23
@Sorfosh Ok thanks. What is the correct parameterization?
– Wyuw
Jul 22 at 22:29