Calculating \int F dr with Green's Theorem

[bugatti79]

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?

 

[bugatti79]

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

 

[SammyS]

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.

 

[bugatti79]

Yes, clockwise gives the opposite sign compared to counter-clockwise.

Thanks

 

[bugatti79]

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

 

[SammyS]

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.