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Evaluating line integral.

  1. May 3, 2009 #1
    Let C be the line segment from (0,0,0) to the point (1,3,-2). evualuate the line integral of f along C if f(x,y) = (x + y^2 -2z)


    so far i was able to write the parametric form x = t y=3t and z=-2t
    the square root of all the derivatives is root(14).
    so i get root(14) times integral of (t + t^2 -2t)dt.

    my answer is (11/3) times root(14)
    but the answer key says (11/2) times root(14)

    my problem is that i dont know how to set the bounds of the integral.. i just integrate from 0 to 1, but really im just guessing. how do i go about this?


    3. The attempt at a solution
     
  2. jcsd
  3. May 3, 2009 #2

    diazona

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    Hint: since you're integrating over t, what are the values of t at the starting and ending points of the line segment?
     
  4. May 3, 2009 #3
    starts at 0, ends at 3?
     
  5. May 3, 2009 #4

    diazona

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    How did you come up with that? (The "how" is always important!)
     
  6. May 3, 2009 #5
    sigh, to be honest.. somehow ive made it this far in calculus without ever knowing how to set up bounds for definate integrals whenever they are not given. i can solve them when the bounds are given, but when their not. im screwed. anyways, i dont expect anyone to straight up tell me what the bounds are. just a little desparate because i really am that stupid when it comes to this.
     
  7. May 3, 2009 #6
    the reason why i thought it was from 0 to 1 to begin with is because i did this... -t+t^2.
    convert to t(t-1), t = 0, t=1
     
  8. May 3, 2009 #7

    diazona

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    Sure, that's what everyone says, until they figure it out...

    Generally, the procedure is that you first identify the starting and ending points of the line integral (which could be points in space or points in time or whatever), then use whatever equations you have to figure out the values of your integration variable at those two points. It's not terribly complicated. In your case, you can definitely identify the starting and ending points of the integral (because they're given to you), and I know you have the equations to figure out what the value of t is at those points. They're in your first post.

    Well, that's not actually correct. You're trying to get the integration limits of t from the function f, but the integration limits have nothing to do with the function f. They only have to do with the starting and ending points, (0, 0, 0) and (1, 3, -2), and with the conversion between (x, y, z) and t.
     
  9. May 3, 2009 #8
    so do i set x=y=z to get t=3t=-2t?
     
  10. May 3, 2009 #9

    diazona

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    Not really... there's no reason to do that.

    How about starting with just one of the three coordinates. You know that at the start of the integral, x = 0. Does that tell you anything about t at the start of the integral?

    And you know that at the end of the integral, x = 1. Does that tell you something about t at the end of the integral?
     
  11. May 3, 2009 #10
    well, it just tells me that at the start... t=0 since x=t.. and at the end. if x=1, then t=1 as well. which means the bounds are 0 to 1, which is how i got my answer (11/3) times root(14), but the key says (11/2) times root(14) instead..maybe the answer key is wrong?
     
  12. May 3, 2009 #11

    diazona

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    Go back and check your math. You just explained the right way to figure out the bounds of the integral (yay!), they are indeed 0 to 1... now, I just did the problem using those bounds, and I get the same answer as the answer key.

    Perhaps you can describe in more detail exactly what steps you took to get to the answer 11/3*sqrt(14)?
     
  13. May 3, 2009 #12
    ok. i re did my math and now im getting an even worse answer.
    so integrating the function (t + t^2 -2t) i get (t^2/2) + (t^3/3) - (t^2). and plugging in from 0 to 1 i get....1/2 + 1/3 -1. common denominator is 6, so 3/6 + 2/6 - 6/6 and i get -1/6, so now my answer is root(14) times -1/6
     
  14. May 3, 2009 #13

    diazona

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    OK, well, where exactly did you get (t + t^2 - 2t) from? (hint hint ;-)
     
  15. May 3, 2009 #14
    i got it from the function x + y^2 -2z. just substituted the variables with t like my book says to do.
     
  16. May 3, 2009 #15
    OH! ok i get it!!!! ok by the way, so everytime i look for the bounds, do i do the same thing as i did with this problem? just find t where x begins, and where x ends? can i do it with y or z?
     
  17. May 3, 2009 #16

    diazona

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    Yep - you could have done the same thing with y or z instead of x. In fact, you might want to try them all and make sure you get the same value of t no matter which one you pick - it's a good way to check that you figured out the right parametric form of the curve.
     
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