Let F be the vector ﬁeld F = −yi + xj / x² + y² on R³ minus the z axis. Let C₁ be the circle in the xy-plane of radius 3 and centered at the origin, and let C₂ be the circle in the xy plane of radius 3 and centered at the point (5,0).(a) Calculate ∮c₁ F • dr by direct computation.(b) Calculate curl F.(c) If possible, use Stokes’ Theorem to calculate ∮c₁ F • dr. If it cannot be used, explain why not.(d) If possible, use Stokes’ Theorem to calculate ∮c₂ F • dr. If it cannot be used, explain why not.(e) Is F a gradient ﬁeld?. Created by Z. science-mathematics-en - mathematics-en

(a) Letting x = 3 cos t, y = 3 sin t:∫(t = 0 to 2π) (<-3 sin t, 3 cos t>/3) · <-3 sin t, cos t> dt= ∫(t = 0 to 2π) 3 dt= 6π.----------------(b) curl F = <0, 0, (∂/∂x) x/(x² + y²) - (∂/∂y) -y/(x² + y²)>...............= <0, 0, (-x² + y²)/(x² + y²) + (x² - y²)/(x² + y²)²>...............= <0, 0, 0>.-----------------(c) Since F is discontinuous at (0, 0), which is inside C, Stokes' Theorem does not apply.-----------------(d) This time, F is continuous inside and on C; so Stokes' Theorem applies.∮c₂ F • dr= ∫∫ curl F • dS= 0, by part b.----------------(e) F is a gradient field, being the gradient of arctan(y/x) [check its gradient].----------------I hope this helps!

Let F be the vector ﬁeld F = −yi + xj / x² + y² on R³....?

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May 2021 2 150 Report

Let F be the vector ﬁeld F = −yi + xj / x² + y² on R³....?

Let F be the vector ﬁeld F = −yi + xj / x² + y² on R³ minus the z axis. Let C₁ be the circle in the xy-plane of radius 3 and centered at the origin, and let C₂ be the circle in the xy plane of radius 3 and centered at the point (5,0).

(a) Calculate ∮c₁ F • dr by direct computation.

(b) Calculate curl F.

(d) If possible, use Stokes’ Theorem to calculate ∮c₂ F • dr. If it cannot be used, explain why not.

(e) Is F a gradient ﬁeld?

Science & Mathematics / Mathematics

Verified answer

(a) Letting x = 3 cos t, y = 3 sin t:

∫(t = 0 to 2π) (<-3 sin t, 3 cos t>/3) · <-3 sin t, cos t> dt

= ∫(t = 0 to 2π) 3 dt

= 6π.

----------------

(b) curl F = <0, 0, (∂/∂x) x/(x² + y²) - (∂/∂y) -y/(x² + y²)>

...............= <0, 0, (-x² + y²)/(x² + y²) + (x² - y²)/(x² + y²)²>

...............= <0, 0, 0>.

-----------------

(d) This time, F is continuous inside and on C; so Stokes' Theorem applies.

∮c₂ F • dr

= ∫∫ curl F • dS

= 0, by part b.

----------------

(e) F is a gradient field, being the gradient of arctan(y/x) [check its gradient].

----------------

I hope this helps!

Anonymous

V= <(-ok*y)/(x^2+y^2) , (ok*x)/(x^2+y^2)> (in vector notation) area C: discover del dot V, that's the sum of the partial derivatives of the guy vector aspects with comprehend to the respective coordinate. div(V)=d/dx(-ok*y)/(x^2+y^2) + d/dy(ok*x)/(x^2+y^2) =(2kxy)/(x^2+y^2)^2 + (-2kxy)/(x^2+y^2)^2 =0 area D: curl(V)=<dVj/dx - dVi/dy> that's precisely what we got here upon partly C, so all of us comprehend it is 0.