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Homework Help: Tricky to 2 variable line/path integral

  1. Nov 9, 2012 #1
    1. The problem statement, all variables and given/known data
    Compute ∫f ds for [tex]f(x,y)=\frac {y^3}{x^7}[/tex] [tex]y=\frac {x^4}{4}[/tex] for [itex]1≤x≤2[/itex]

    2. Relevant equations

    ∫f ds= ∫f(c(t))||c'(t)||

    ||c'(t)|| is the magnitude of ∇c'(t)

    3. The attempt at a solution

    From this... I gathered the following, by saying x=t
    [tex]c(t)= <t,\frac {t^4}{4}>[/tex]
    [itex]c'(t)= <1, t^3>[/itex]

    That gave me a magnitude of [tex]\sqrt{1+t^6}[/tex]

    so now I get that ∫f ds... through plugging in the magnitude and c(t) in the integral... this

    [itex]∫\frac{t^5 \sqrt{1+t^6}}{64}[/itex] where [itex] x=t\, and\, 1≤t≤2 [/itex]

    This is the part where I'm particularly stuck... am I missing a step or did I make a math error??? How do I integrate this?
    Last edited: Nov 9, 2012
  2. jcsd
  3. Nov 9, 2012 #2


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    Let ##u = 1+t^6##.
  4. Nov 9, 2012 #3
    I am completely lost... my anti-derivative and integration is shaky at best.
    u= 1+t^6
    so do I just take the anti-derivative of t^5... and get t^6/(6*64) and then take the and then the anti-derivative of u which would [tex]\frac{2u^\frac{3}{2}}{3} [/tex] then anti-derivative of U which would t+t^7/7
  5. Nov 9, 2012 #4
    no wait, there would be a product rule... too wouldn't it.

    or am I wrong
  6. Nov 9, 2012 #5


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    Ignoring the ##\frac 1 {64}##, you have$$
    \int t^5\sqrt{1+t^6}\, dt$$If you let ##u = 1+t^6## then ##du = 6t^5dt##. That gives you$$
    \frac 1 6\int u^{\frac 1 2}\, du$$You integrate that and substitute ##u = 1+t^6## in your answer. Then put in your limits.
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