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Green's theorem find circulation of vector field

Applying Green's TheoremCan Green's theorem be used in a plane other than the xy-plane?Circulation using Green's TheoremDemonstrating Green's theoremDoes Green's Theorem hold for polar coordinates?Green's Theorem with respect to a given polar region.Evaluating Line Integral with Green's TheoremDivergence theorem does not hold for this vector field?Applying Green's theorem on scalar fieldsUsing Green's Theorem to find the flux

1

$begingroup$

$4.$ [$10$ Marks] Find the circulation of the vector field $$vec F(x,y,z) = langle x^2018 -233x +ycos x, 5x +sin x +e^2018y -233 rangle$$ along the circle traced by $vec r(t) = langle 3costheta +1, 3sintheta -1 rangle$ from $theta=0$ to $theta=2pi$.

Applying Green's theorem we get
$$int^2pi_0int_0^?5rdrdtheta$$

$x^2+y^2=r^2$
$$(3costheta+1)^2+(3sintheta+1)^2 =r^2$$
But I end up with $$11 +6costheta-6sintheta = r^2$$

Can't solve for $r$. Am I not seeing something?

integration multivariable-calculus polar-coordinates

share|cite|improve this question

edited Mar 18 at 1:16

Rócherz

3,0013821

asked Mar 18 at 0:13

MasterYoshiMasterYoshi

807

$endgroup$

add a comment | 

1

$begingroup$

$4.$ [$10$ Marks] Find the circulation of the vector field $$vec F(x,y,z) = langle x^2018 -233x +ycos x, 5x +sin x +e^2018y -233 rangle$$ along the circle traced by $vec r(t) = langle 3costheta +1, 3sintheta -1 rangle$ from $theta=0$ to $theta=2pi$.

Applying Green's theorem we get
$$int^2pi_0int_0^?5rdrdtheta$$

$x^2+y^2=r^2$
$$(3costheta+1)^2+(3sintheta+1)^2 =r^2$$
But I end up with $$11 +6costheta-6sintheta = r^2$$

Can't solve for $r$. Am I not seeing something?

integration multivariable-calculus polar-coordinates

share|cite|improve this question

edited Mar 18 at 1:16

Rócherz

3,0013821

asked Mar 18 at 0:13

MasterYoshiMasterYoshi

807

$endgroup$

add a comment | 

$begingroup$

$4.$ [$10$ Marks] Find the circulation of the vector field $$vec F(x,y,z) = langle x^2018 -233x +ycos x, 5x +sin x +e^2018y -233 rangle$$ along the circle traced by $vec r(t) = langle 3costheta +1, 3sintheta -1 rangle$ from $theta=0$ to $theta=2pi$.

Applying Green's theorem we get
$$int^2pi_0int_0^?5rdrdtheta$$

$x^2+y^2=r^2$
$$(3costheta+1)^2+(3sintheta+1)^2 =r^2$$
But I end up with $$11 +6costheta-6sintheta = r^2$$

Can't solve for $r$. Am I not seeing something?

integration multivariable-calculus polar-coordinates

share|cite|improve this question

edited Mar 18 at 1:16

Rócherz

3,0013821

asked Mar 18 at 0:13

MasterYoshiMasterYoshi

807

$endgroup$

share|cite|improve this question

edited Mar 18 at 1:16

Rócherz

3,0013821

asked Mar 18 at 0:13

MasterYoshiMasterYoshi

807

share|cite|improve this question

edited Mar 18 at 1:16

Rócherz

3,0013821

asked Mar 18 at 0:13

MasterYoshiMasterYoshi

807

edited Mar 18 at 1:16

Rócherz

3,0013821

edited Mar 18 at 1:16

Rócherz

3,0013821

asked Mar 18 at 0:13

MasterYoshiMasterYoshi

807

asked Mar 18 at 0:13

MasterYoshiMasterYoshi

807

1 Answer
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active

oldest

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3

$begingroup$

The issue here is that a circunference of radius $r$ centered in $(x_0,y_0)$ has
$$r^2= (x-x_0)^2+(y-y_0)^2.$$
The parametrized curve $r(t)=(3cos(t)+1,3sin(t)-1)$ is a circunference centered in $(1,-1)$. The radius $r$ is then
$$r^2=9cos^2(t)+9sin^2(t)=9 implies r=3.$$

share|cite|improve this answer

edited Mar 18 at 1:17

Rócherz

3,0013821

answered Mar 18 at 0:54

Aquerman KuczmendaAquerman Kuczmenda

1065

$endgroup$

add a comment | 

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3

$begingroup$

The issue here is that a circunference of radius $r$ centered in $(x_0,y_0)$ has
$$r^2= (x-x_0)^2+(y-y_0)^2.$$
The parametrized curve $r(t)=(3cos(t)+1,3sin(t)-1)$ is a circunference centered in $(1,-1)$. The radius $r$ is then
$$r^2=9cos^2(t)+9sin^2(t)=9 implies r=3.$$

share|cite|improve this answer

edited Mar 18 at 1:17

Rócherz

3,0013821

answered Mar 18 at 0:54

Aquerman KuczmendaAquerman Kuczmenda

1065

$endgroup$

add a comment | 

3

$begingroup$

The issue here is that a circunference of radius $r$ centered in $(x_0,y_0)$ has
$$r^2= (x-x_0)^2+(y-y_0)^2.$$
The parametrized curve $r(t)=(3cos(t)+1,3sin(t)-1)$ is a circunference centered in $(1,-1)$. The radius $r$ is then
$$r^2=9cos^2(t)+9sin^2(t)=9 implies r=3.$$

share|cite|improve this answer

edited Mar 18 at 1:17

Rócherz

3,0013821

answered Mar 18 at 0:54

Aquerman KuczmendaAquerman Kuczmenda

1065

$endgroup$

add a comment | 

$begingroup$

The issue here is that a circunference of radius $r$ centered in $(x_0,y_0)$ has
$$r^2= (x-x_0)^2+(y-y_0)^2.$$
The parametrized curve $r(t)=(3cos(t)+1,3sin(t)-1)$ is a circunference centered in $(1,-1)$. The radius $r$ is then
$$r^2=9cos^2(t)+9sin^2(t)=9 implies r=3.$$

share|cite|improve this answer

edited Mar 18 at 1:17

Rócherz

3,0013821

answered Mar 18 at 0:54

Aquerman KuczmendaAquerman Kuczmenda

1065

$endgroup$

share|cite|improve this answer

edited Mar 18 at 1:17

Rócherz

3,0013821

answered Mar 18 at 0:54

Aquerman KuczmendaAquerman Kuczmenda

1065

edited Mar 18 at 1:17

Rócherz

3,0013821

edited Mar 18 at 1:17

Rócherz

3,0013821

answered Mar 18 at 0:54

Aquerman KuczmendaAquerman Kuczmenda

1065

answered Mar 18 at 0:54

Aquerman KuczmendaAquerman Kuczmenda

1065