Question: Calculating Work with a Line Integral

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SUMMARY

The discussion focuses on calculating the work done by the force field F(x,y) = xy i + x² j along the curve C defined by x = y² from the point (0,0) to (1,1). The user initially parametrized the curve incorrectly as x = t and y = t², leading to an incorrect integral result of 3/4 instead of the correct answer, 3/5. The correct parametrization should be y = t and x = t², which aligns with the curve definition. This adjustment allows for the accurate calculation of the line integral.

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Homework Statement


Find the work doneby the force field F on a particle that moves along the curve C.
F(x,y)=xy i + x^2j
C: x=y^2 from (0,0) to (1,1)


Homework Equations



\intF dot dr=\int^{b}_{a}F(r(t))dotr'(t)dt

The Attempt at a Solution



Okay, so I parametrized x=t and y=t^2 (giving r(t)=ti+t^2j right?) and substituted those values in for x and y in F, dotted that with 1i+2tj because I think that it is the derivative of r, if the parametric equations for r are x=t and y=t^2. I then took the integral of the dot product i just took over the interval 0 to 1. I ended up with 3/4 but the correct answer is 3/5
 
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Breedlove said:
Okay, so I parametrized x=t and y=t^2 (giving r(t)=ti+t^2j right?)
No, the curve is x=y^2 not y=x^2 so your parametric form of y should be y=\sqrt{t}. Alternatively use: y=t and x=t^2.
 

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