Line Integrals: Gradient Field and Calculations for -2,0 to 2,0 Points

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Homework Help Overview

The discussion revolves around a vector field F defined as F = (3x² + 2y cos(xy))i + (2y + 2x cos(xy))j. Participants are tasked with showing that F is a gradient field, calculating the line integral of F along specified paths, and determining the value of the integral for any curve connecting the points (-2,0) and (2,0).

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the conditions under which F can be considered a gradient field and question the implications of the curl of F being zero. There is an exploration of the differences between open and closed curves in relation to Stokes' theorem and the fundamental theorem of gradients. Some participants express confusion over the nature of the curves described in the problem.

Discussion Status

There is ongoing clarification regarding the nature of the curves involved in parts b and c, with some participants suggesting the use of the fundamental theorem of gradients for open curves. Multiple interpretations of the problem are being explored, particularly concerning the definitions and properties of the curves.

Contextual Notes

Participants note that the problem may involve confusion about whether the curves are open or closed, which affects the applicability of certain theorems. There is also mention of a potential sign error in the expression for f, which may impact the calculations discussed.

joemama69
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Homework Statement



F = (3x2 + 2y cos(xy))i + (2y + 2x cos(xy))j

a - show that F is a gradient field

b - calculate the integral of F dot dr where c includes the points -2,0 and 2,0

c - determine the value of the integral of F dot dr where c is any curve joining -2,0 and 2,0

Homework Equations





The Attempt at a Solution




a..

grad f = F,

I found f = x3 + y2 - 2sin(xy)


b...

curlf F = 0, therefore the integral F dot dr = curl F dot dA = 0

c...

wouldnt that be the same as the above b
 
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joemama69 said:
a..

grad f = F,

I found f = x3 + y2 - 2sin(xy)

looks like you're off by a negative sign; [itex]\frac{d}{dx}\sin(xy)=+y\cos(xy)[/itex]


b...

curlf F = 0, therefore the integral F dot dr = curl F dot dA = 0

You must have a closed path to use Stoke's theorem. Is the curve in (b) closed? It is not clear from your description of the problem...

c...

wouldnt that be the same as the above b

It sounds like you have an open curve from (-2,0) to (2,0) and so you can't use stokes theorem (an open curve does not bound a surface) try using the fundamental theorem of gradients instead...
 
Should be f = x3 + y2 + 2sin(xy) (you got the sign wrong).

For b, the integral will be zero only if C is a closed curve. Does it say that in the question?

For c, no it won't. If F = ∇f, then [itex]\int_a^b \nabla f \cdot dr = f(b) - f(a)[/itex].

EDIT: Oops, gabba beat me to it.
 
b is an open curve


c is a closed curve
 
so c... should be -8 - 8 = -16 what about b
 
The curve in c is not a closed curve. It's a curve joining (-2,0) and (2,0).
 
Ok i will explain this a little better

Part B Calculate integral F dot dr where C is the picture included

Part C Determine the value of integral F dot dr where C is anycurve joining -2,0 to 2,0. Explain Reasoning


Whats the difference in the question being asked
 

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