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Calc III Problem Integrating Sq Rt's in Arc Length Formula

  1. Feb 10, 2010 #1
    Hey guys, I'm studying for a test in calc 3 tomorrow and have run into a problem. On the practice test we have a problem "Find the length of the curve: r=theta^2, 0≤theta≤pi/2"

    I know the length of a curve in polar coordinates is int(sqrt(r^2 + (dr/dtheta)^2))dtheta...but when I get to where I have to integrate sqrt(theta^4 + 4theta^2) I become very stuck. What is the procedure for handling this square root integration?

    Thanks for any help.
  2. jcsd
  3. Feb 10, 2010 #2

    so you are trying to integrate:

    [tex]\int\sqrt{t^4 + 4t^2} dt = \int t\sqrt{t^2 + 4}dt[/tex]

    Substitute t^2+4 = u

    => dt = du/2t

    t cancels out:
    [tex]\int t\sqrt{t^2 + 4}dt = \frac{1}{2} \int \sqrt{u} du = \frac{u^{\frac{3}{2}}}{3} + c[/tex]

    [tex]\Rightarrow \int\sqrt{t^4 + 4t^2} dt = \frac{(t^2+4)^{\frac{3}{2}}}{3} + c [/tex]

    is this what you are asking for?
    Last edited: Feb 10, 2010
  4. Feb 11, 2010 #3
    Wow I feel stupid. Haha thanks a bunch.
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