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Homework Help: Evaluating a line integral

  1. Apr 9, 2013 #1
    1. The problem statement, all variables and given/known data
    P is the part of the curve 9y²=4x³ between the points (1,-2/3) and (1,2/3).

    Evaluate the integral $$\int_P x ds $$

    2. Relevant equations

    3. The attempt at a solution

    $$\int_P x ds = \int_P x |r'(t)| dt $$

    My problem is that I cannot find a right parametrization r(t) for this.

    I tried y=t, then x=((3/2)t)^(2/3)

    but thent goes from -2/3 to 2/3, but that gives a complex x for t=-2/3

    Tried it with x=3t² and y=2sqrt(3)t³, I get the same problem.

    Any tips ? Thanks!
  2. jcsd
  3. Apr 9, 2013 #2


    Staff: Mentor

    How about x = t and y = ±(2/3)t3/2?
  4. Apr 9, 2013 #3
    so split the integral up in 2 parts?

    First along the path

    $$r(t)=t\hat{x} -\frac{2}{3}t^{\frac{3}{2}}\hat{y} $$ from t=1 to t=0, then

    $$r(t)=t\hat{x} + \frac{2}{3}t^{\frac{3}{2}}\hat{y} $$ from t=0 to t=1

    $$ |r'(t)| = \sqrt{1+t} $$

    So the integral becomes

    $$ \int^1_0 t \sqrt{1+t} dt + \int^0_1 t \sqrt{1+t} dt = 0$$

    Is this correct? Zero?
  5. Apr 9, 2013 #4


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    I would be inclined to use the parametrization, [itex]\displaystyle \ y=t\,,\ \text{ and }\ x=\sqrt[3]{\frac{9t^2}{4}}\ .[/itex]

    ... but I'm not sure how messy the resulting integral is.
  6. Apr 9, 2013 #5
    Could zero be the answer? It doesn't feel right,but is it possible for a line integral like that to be zero?
  7. Apr 9, 2013 #6


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    Yes, in general it is possible for ds line integrals to come out to zero. But not in this case. Your integrand is nonnegative and ##ds## is always positive. The problem is that when you use the formula ##ds =|\vec r'(t)|dt## you are assuming the curve is parameterized in the positive direction. Since that is not the case in your second integral, it needs a negative sign in front or, what is the same thing, the limits reversed.
  8. Apr 9, 2013 #7
    Ahh, thanks.

    So the integral becomes

    $$2\int^1_0 t\sqrt{1+t} dt $$

    u= 1+t, du= dt, t=u-1

    $$2\int^{u(1)}_{u(0)} (u^{\frac{3}{2}}-u^{\frac{1}{2}}) du = [\frac{4}{5}u^{\frac{5}{2}}-\frac{4}{3}u^{\frac{3}{2}}]^{{2}}_{1} = $$

    Is this alright then?
  9. Apr 9, 2013 #8


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    Yes, except the final simplified answer seems to be missing...
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