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Help evaluating a line integral!

  1. Apr 9, 2010 #1
    So, my multivariable class has just started line integrals, and I could use a little help with them. The problem I'm currently working on says:

    Evaluate the line integral, where C is the given curve:

    [tex]\int\limits_C \! xy \,ds[/tex]
    [tex]C: x=t^2, y=2t, 0 \le t \le 1[/tex]

    I realize that, by eliminating the parameter (note really necessary, but just for the sake of understanding), it is the curve [tex]x=(1/4)y^2, 0 \le y \le 2[/tex].

    I've managed to get this down to 4 times the integral, from 0 to 1, of (t^3) sqrt(t^2 + 1), but I have no idea where to go from here. Is there something I've majorly screwed up on?
  2. jcsd
  3. Apr 9, 2010 #2
    Try substitution with [itex]u=t^2+1[/itex]. Consider

    [tex]t^3 \sqrt{t^2+1}\ dt = t^2 (t^2+1)^{1/2} \cdot t\ dt[/tex]
  4. Apr 9, 2010 #3
    There must be something I'm blanking on from my days in calculus one. I feel like the deeper I go into calculus, the more prone I am to making beginner's errors.

    Why can I substitute with u = (t^2) +1? If I separate (t^3) into t(t^2), what does the (t^2) term become in terms of u?
  5. Apr 9, 2010 #4
    =) I think you might be overthinking the problem. If [itex]u=t^2+1[/itex], can you use this equation to find [itex]t^2[/itex] in terms of u?
  6. Apr 9, 2010 #5
    ... Wow. That was very obvious, haha. I'm so used to setting u as a convenient expression in terms of t that I didn't even consider subtracting 1.

    Is this new integral in terms of u correct?

    [tex]\frac{1}{2}\int_1^2 \! (u-1)\sqrt{u} \,du[/tex]

    In which case I would just multiply the function of interest through and antidifferentiate?
  7. Apr 9, 2010 #6
    Looks good to me!
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