Calculating \int F dr with Green's Theorem

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Homework Help Overview

The discussion revolves around applying Green's Theorem to evaluate a line integral of a vector field defined by F(x,y) = (√x + y³)i + (x² + √y)j, over a specific path C. The path consists of an arc of y = sin(x) from (0,0) to (π,0) and a straight line from (π,0) back to (0,0).

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of Green's Theorem, particularly the sign change required when the orientation of the curve is clockwise instead of counterclockwise. There are questions about whether the integral setup is correct and how the theorem's application might differ based on the direction of integration.

Discussion Status

Some participants have confirmed the sign change associated with the clockwise orientation. There is ongoing exploration regarding the correctness of the integral setup, with at least one participant expressing confidence in the formulation presented.

Contextual Notes

Participants are considering the implications of the curve's orientation on the application of Green's Theorem and are verifying the setup of the double integral involved in the calculation.

bugatti79
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Homework Statement



Use Green's Theorem to calculate \int F dr

Homework Equations



F(x,y)= (\sqrt x +y^3) i + (x^2+ \sqrt y) j where C is the arc of y=sin x from (0,0) to ( pi,0) followed by line from (pi,o) to (0,0).



The Attempt at a Solution



We have \int f dx + g dy = \int \int_R (g_x-f_y) dA for counterclockwise rotation, but the question is given in clockwise rotation so does green's theorem become

- \int \int_R (g_x-f_y) dA...? Ie, a sign change?
 
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bugatti79 said:

Homework Statement



Use Green's Theorem to calculate \int F dr

Homework Equations



F(x,y)= (\sqrt x +y^3) i + (x^2+ \sqrt y) j where C is the arc of y=sin x from (0,0) to ( pi,0) followed by line from (pi,o) to (0,0).



The Attempt at a Solution



We have \int f dx + g dy = \int \int_R (g_x-f_y) dA for counterclockwise rotation, but the question is given in clockwise rotation so does green's theorem become

- \int \int_R (g_x-f_y) dA...? Ie, a sign change?

Any clues on this one?

Thanks
 
bugatti79 said:

Homework Statement



Use Green's Theorem to calculate \int F dr

Homework Equations



F(x,y)= (\sqrt x +y^3) i + (x^2+ \sqrt y) j where C is the arc of y=sin x from (0,0) to ( pi,0) followed by line from (pi,o) to (0,0).

The Attempt at a Solution



We have \int f dx + g dy = \int \int_R (g_x-f_y) dA for counterclockwise rotation, but the question is given in clockwise rotation so does green's theorem become

- \int \int_R (g_x-f_y) dA...? Ie, a sign change?
Yes, clockwise gives the opposite sign compared to counter-clockwise.
 
SammyS said:
Yes, clockwise gives the opposite sign compared to counter-clockwise.

Thanks
 
bugatti79 said:

Homework Statement



Use Green's Theorem to calculate \int F dr

Homework Equations



F(x,y)= (\sqrt x +y^3) i + (x^2+ \sqrt y) j where C is the arc of y=sin x from (0,0) to ( pi,0) followed by line from (pi,o) to (0,0).



The Attempt at a Solution



We have \int f dx + g dy = \int \int_R (g_x-f_y) dA for counterclockwise rotation, but the question is given in clockwise rotation so does green's theorem become

- \int \int_R (g_x-f_y) dA...? Ie, a sign change?

Can anyone confirm this integral is set up correctly?

\displaystyle \int_0^ {\pi} \int_0^ {sin x} (3y^2-2x) dy dx
 
bugatti79 said:
Can anyone confirm this integral is set up correctly?

\displaystyle \int_0^ {\pi} \int_0^ {sin x} (3y^2-2x) dy dx

That looks OK to me.
 

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