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Line integrals

  1. May 1, 2009 #1
    1. The problem statement, all variables and given/known data

    say I have [tex] \vec{F} = ye^x\vec{i} + e^x\vec{j} [/tex] and C the line segment from the point (1,4,-2) to the point (6,7,-2).


    I need to find:

    [tex] \int\limits_C \vec{F} d\vec{r} [/tex]

    2. Relevant equations



    3. The attempt at a solution

    First I parametrize C as:

    [tex] \vec{r} = 6 + 5t \vec{i} + 7+3t \vec{j} -2 \vec{k} [/tex]

    Then I set up the integral as:

    [tex] \int_{-1}^{0} (7+3t)e^{6+5t} \vec{i} + e^{6+5t} \vec{j}) \cdot (5\vec{i} + 3\vec{j}) dt [/tex]

    is this correct so far?
     
  2. jcsd
  3. May 1, 2009 #2

    HallsofIvy

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    I assume you mean [itex]\vec{r}= (6+ 5t)\vec{i}+ (7+ 3t)\vec{j}- 2\vec{k}[/itex].
    When t= 0 this is [itex]\vec{r}= 6\vec{i}+ 7\vec{j}- 2\vec{k}[/itex] which is one of the given points. When t= -1 this is [itex]\vec{r}= \vec{i}+ 4\vec{j}- 2\vec{k}[/itex] which is the other point. Yes, this is a vector equation for the line- although the choice of t= -1 for one of the is peculiar!

    Yes, that is correct. You will probably wnat to "separate" [itex]e^{6+ 5t}[/itex] as [itex]e^6 e^{5t}[/itex].
     
  4. May 1, 2009 #3
    and so after doing all the computation I got:

    [tex] \frac{12(y^2-x^2)}{(x^2+y^2)^2} [/tex] being evaluated from -1 to 0. Is this correct? Reason I ask is because the integral was kind of complex
     
  5. May 1, 2009 #4

    HallsofIvy

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    Why did you change from t to x and y?
     
  6. May 1, 2009 #5
    oops.. I guess you're right
     
  7. May 2, 2009 #6
    I got [tex] \frac{9}{5} - 4e [/tex]
     
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