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Homework Help: Line integral of a curve

  1. Dec 27, 2017 #1
    1. The problem statement, all variables and given/known data
    upload_2017-12-28_0-5-32.png

    2. Relevant equations


    3. The attempt at a solution
    Line integral of a curve

    ## I = \int_{ }^{ } yz dx + \int_{ }^{ } zx dy + \int_{ }^{ } xy dz ## with proper limits.

    ## I = \int_{\frac { \pi }4}^{ \frac { 3 \pi} 4} abc ( \cos^2 t - \sin^2 t ) dt = -abc ##

    |I| = abc

    So, the answer is option (a).

    Is this correct?
     
  2. jcsd
  3. Dec 27, 2017 #2

    kuruman

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    Except for the negative sign the answer correct. You just can't strip the negative sign because it doesn't match one of the choices. What if ##-abc## were a sixth choice? A better way to do this would be to see if ##\vec{F}## can be derived from some scalar function U, such that ##\vec{F}=-\vec{\nabla}U## (Hint: It can). Think of ##\vec{F}## as a conservative force and ##U## as the potential from which it is derived. Then the integral depends on the end points, i.e. ##I=U(t_2)-U(t_1)##.
     
  4. Dec 27, 2017 #3
    I had the impression that value of I = |I| (which is wrong according to your above comment). So, I removed the negative sign.

    I did the calculation again and I got -abc.
    ## I = \int_{\frac { \pi }4}^{ \frac { 3 \pi} 4} abc ( \cos^2 t - \sin^2 t ) dt =##
    ## \frac { \sin{(2t)}} 2 |_ {\frac { \pi }4} ^{ \frac { 3 \pi} 4} abc = -abc ##


    I do not see what is wrong in the above calculation.

    Taking ## \vec F = - \nabla U ##

    I have to calculate ## U = - \int \vec F \cdot d \vec x ## ## = -abc = \frac { \sin{ (2t)}} 2##


    Then, I have to calculate ## \int \vec F \cdot d \vec x ## for t going from ## \frac { \pi} 4 ~ to ~ \frac { 3 \pi }4 ## .

    This gives ## U(\frac { \pi} 4) – U(\frac {3 \pi} 4) ## , which is again –abc.

    I am not getting what mistake I am doing here, but I am coming towards the same answer.
     
  5. Dec 27, 2017 #4

    Mark44

    Staff: Mentor

    This is what I get, as well. It's not unheard of for posted answers to have typos.
    The line integral calculation should be pretty straightforward. I don't see why you would need to do this.
     
  6. Dec 27, 2017 #5

    kuruman

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    I also get this answer after discovering an excess of a negative sign.
     
  7. Dec 28, 2017 #6

    Mark44

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    So I'm thinking that there's a typo in the answer.
     
  8. Dec 28, 2017 #7
    So, I = -abc
    And value of I = |I| = abc
    Is this correct?
     
  9. Dec 28, 2017 #8

    kuruman

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    I think so too. I did it two ways, using OP's method and using the scalar function but introduced an extraneous negative sign in the latter which got me going for a short while.
    The value of the integral is I = - abc as we all agree by now. Why do you feel you should take the absolute value?
     
  10. Dec 28, 2017 #9

    Mark44

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    I agree that I = -abc, but ##I \ne |I|##. You can't just arbitrarily take the absolute value.
     
  11. Dec 28, 2017 #10

    Mark44

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    I would say to force the computed answer to agree with the posted answer. This is not a good reason to do so.
     
  12. Dec 28, 2017 #11
    Actually, I am having problem with the word " value".
    I thought value of I = |I|...(1)
    But,(1) is wrong.
    value of I = the integral which I get after the calculation
    absolute value of I = |I|
     
  13. Dec 28, 2017 #12

    kuruman

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    Values can be positive or negative. Absolute values are positive only. The question does not specify "absolute".
     
  14. Dec 28, 2017 #13

    Mark44

    Staff: Mentor

    No. The value of I is whatever it is -- positive, negative, or zero. I = |I| if and only if I >= 0.
     
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