1. Not finding help here? Sign up for a free 30min 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!

Line integral, problems with substitution (probably!)

  1. Mar 26, 2009 #1
    1. The problem statement, all variables and given/known data

    I am trying to solve a line integral (bear with me, I am new to calculus!) and my basic skills of integration seem to fail me. I am sure the mistake is quite obvious, as I keep getting the wrong answer, 2, when it should be ~2.69

    2. Relevant equations
    [tex]\int_C 4x^3dS[/tex]

    C is the curve given by [tex]x=t, y=t^3-1[/tex] and [tex]0 \leq t \leq 1[/tex]


    3. The attempt at a solution
    [tex]\int_C 4x^3 dS = \int_0^1 4t^3 \sqrt{1^2+(3t^2)^2} dt = \int_0^1 4t^3 \sqrt{9t^4+1} dt[/tex]
    I attempt substitution in order to solve the integral:
    [tex]u=9t^4+1 \Rightarrow \frac{1}{36}du=t^3 dt[/tex].
    The limits are now [tex]9 \cdot 0^4=0[/tex] and [tex]9 \cdot 1^4=9[/tex], so by substitution we have:
    [tex]\frac{4}{36} \int_0^9 \sqrt{u} \, du = \frac{1}{9} \left[\frac{2}{3} u^{\frac{3}{2}}\right]_0^9 = \frac{2}{27} \left(9^{\frac{3}{2}} - 0\right)[/tex]

    Which equals.... Two! Well, there's a mistake in there somewhere, so, no it doesn't. Hopefully a non-mathematician will have mercy on me so that the punishment isn't too hard. o:)

    -- Sarah
     
    Last edited: Mar 26, 2009
  2. jcsd
  3. Mar 26, 2009 #2
    Nevermind, I found the problem. I did not correctly calculate the new limits, they should be 10 and 1 respectively. :biggrin:

    -- Sarah
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Line integral, problems with substitution (probably!)
Loading...