1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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

    User Avatar
    Science Advisor

    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

    User Avatar
    Science Advisor

    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]
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook




Loading...