Line Integral Help: Evaluating F ds on a Curve in 1st Quadrant

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SUMMARY

The discussion focuses on evaluating the line integral of the vector field F = (6x²y², 4x³y + 5y⁴) along the boundary curve of the first quadrant under the curve y = 1 - x², traversed in a clockwise direction. Participants confirm that by applying Green's Theorem, the line integral evaluates to zero, as the vector field is conservative. This conclusion is reached because the partial derivatives satisfy the condition dF1/dy = dF2/dx, indicating that the vector field is the gradient of a potential function.

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  • Understanding of line integrals and vector fields
  • Familiarity with Green's Theorem
  • Knowledge of conservative vector fields
  • Basic calculus concepts, including partial derivatives
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Kuma
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Homework Statement



Trying to evaluate the following line integral:

integral F ds where F = (6(x^2)(y^2), 4(x^3)(y) + 5y^4)
and the path is the boundary curve of the first quadrant below y = 1-x^2 in a clockwise direction.

Homework Equations





The Attempt at a Solution



So since the curve is piecewise smooth closed simple and closed curve I can use greens theorem. Simply put I get an answer as 0 since dF1/dy = dF2/dx. Is that right?
 
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Kuma said:

Homework Statement



Trying to evaluate the following line integral:

integral F ds where F = (6(x^2)(y^2), 4(x^3)(y) + 5y^4)
and the path is the boundary curve of the first quadrant below y = 1-x^2 in a clockwise direction.

Homework Equations





The Attempt at a Solution



So since the curve is piecewise smooth closed simple and closed curve I can use greens theorem. Simply put I get an answer as 0 since dF1/dy = dF2/dx. Is that right?

I would agree with that.
 
Your vector field is conservative. So the line integral would be zero over any closed curve (even non-simple closed curve!). A vector field is conservative if it is the gradient of a potential function.
 

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