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Finding the amount of work done (line integrals)

  • Thread starter Deimantas
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  • #1
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Homework Statement



Find the amount of work (ω) done by moving a point from (2;0) to (1;3) along the curve y=4-(x^2), in the effect of force F=(x-y;x).

Homework Equations





The Attempt at a Solution



ω = ∫((x-y)dx + xdy)

ω = ∫(x-4+x^2)dx + ∫√(4-y) dy

In the end, I get this: http://www.wolframalpha.com/input/?i=%28integrate+%28x-4%2Bx^2%29+from+1+to+2%29+%2B+%28integrate+sqrt%284-y%29+from+0+to+3%29
with the final answer 4,5. Is it correct?
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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No, that is not correct. In fact, your entire method is wrong. This is a one dimensional problem (a curve) and you are doing a two dimensional integral. In general, you need to write the integral in terms of a single parameter. Here, since we have y equals a function of x, we can use x itself as that parameter.

On the curve [itex]y= 4- x^2[/itex], [itex]dy= -2x dx[/itex]. So (x- y) dx= (x- 4+ x^2) dx and [itex]xdy= -2x^2 dx[/itex] That is,
[tex]\int_{(2, 0)}^{(1, 3)} (x- y)dx+ xdy= \int_{x=2}^1 (x^2+ x- 4- 2x^2)dx= \int_{x=2}^1 (-x^3+ x- 4)dx[/tex]
 
  • #3
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I thought I must have chosen a wrong method. Thank you very much for giving such a thorough answer.
 
  • #4
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No, that is not correct. In fact, your entire method is wrong. This is a one dimensional problem (a curve) and you are doing a two dimensional integral. In general, you need to write the integral in terms of a single parameter. Here, since we have y equals a function of x, we can use x itself as that parameter.

On the curve [itex]y= 4- x^2[/itex], [itex]dy= -2x dx[/itex]. So (x- y) dx= (x- 4+ x^2) dx and [itex]xdy= -2x^2 dx[/itex] That is,
[tex]\int_{(2, 0)}^{(1, 3)} (x- y)dx+ xdy= \int_{x=2}^1 (x^2+ x- 4- 2x^2)dx= \int_{x=2}^1 (-x^3+ x- 4)dx[/tex]
just wanted to ask

i'm hoping to become a regular around here and I was hoping that you could tell me how you set out your equations in those graphics?

thanks a lot
 

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