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

[joemama69]

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

 

[gabbagabbahey]

a..

grad f = F,

I found f = x³ + y² - 2sin(xy)

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

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...

 

[dx]

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 \int_a^b \nabla f \cdot dr = f(b) - f(a).

EDIT: Oops, gabba beat me to it.

 

[joemama69]

b is an open curve


c is a closed curve

 

[joemama69]

so c... should be -8 - 8 = -16 what about b

 

[dx]

The curve in c is not a closed curve. It's a curve joining (-2,0) and (2,0).

 

[joemama69]

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