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Trignometric Substiution Problem

  1. Feb 9, 2013 #1
    1. The problem statement, all variables and given/known data

    [itex]\displaystyle \int _0 ^a 4x^2 \sqrt{a^2-x^2} dx[/itex]

    2. Relevant equations

    [itex]\cos^2 x =1 - \sin^2 x[/itex]

    3. The attempt at a solution

    Substuting

    Substituting [itex]x=a \sin(\theta)[/itex]

    [itex]dx=a\cos( \theta) d\theta[/itex]

    x=0
    asin[itex]\theta[/itex]=0
    sin[itex]\theta[/itex]=0
    [itex]\theta[/itex]=0

    [itex]\displaystyle \int_{0}^{a} 4x^2 \sqrt{a^2-x^2} dx =[/itex]

    [itex]\displaystyle \int_{0}^{\pi/2} 4(a\sin\theta)^2 \sqrt{a^2-(a\sin\theta)^2} a\cos\theta d\theta=[/itex]

    [itex]\displaystyle \int_{0}^{\pi/2} 4a^2sin^2\theta \sqrt{a^2-a^2sin^2\theta}a\cos\theta d\theta=[/itex]

    [itex]\displaystyle \int_{0}^{\pi/2} 4a^2sin^2\theta a \sqrt{1-sin^2\theta}a\cos\theta d\theta=[/itex]

    [itex]\displaystyle \int_{0}^{\pi/2} 4a^2sin^2\theta a cos\theta a\cos\theta d\theta=[/itex]

    [itex]\displaystyle 4a^4\int_{0}^{\pi/2} sin \theta cos^2 \theta d\theta=[/itex]

    U substitution

    [itex]u=cos\theta[/itex]

    [itex]-du= sin\theta d\theta[/itex]

    [itex]\displaystyle 4a^4\int u^2 du=[/itex]
    When I put in the bounds it yields from 1 to 0. What did I do wrong
     
    Last edited: Feb 9, 2013
  2. jcsd
  3. Feb 9, 2013 #2
    So the code is messed up basically

    Integral from 0 to a [4x^2 squareroot(a^2 - x^2)]dx =

    4a^4 integral from 0 to pi/2 [sin theta cos^2 theta] dtheta

    I'm stuck trying to change the bounds with a u substitution
     
    Last edited: Feb 9, 2013
  4. Feb 9, 2013 #3

    SammyS

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    I fixed your LaTeX code up to this point.

    Here is the Code for some of the lines I fixed:

    Code (Text):
    [itex]\displaystyle \int_{0}^{a} 4x^2 \sqrt{a^2-x^2} dx =[/itex]

    [itex]\displaystyle \int_{0}^{\pi/2}  4(a\sin\theta)^2 \sqrt{a^2-(a\sin\theta)^2} a\cos\theta d\theta=[/itex]
     
    The "\displaystyle" modifier allows the integral symbol to be larger than otherwise.

    I'll let you edit the Original Post .
     
    Last edited: Feb 9, 2013
  5. Feb 9, 2013 #4
    Ok, I fixed it thank you very much. I didn't know how to put a bound into the intergral code.
     
  6. Feb 9, 2013 #5

    SammyS

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    So you get

    [itex]\displaystyle -4a^4\int_{1}^{0} u^2\,du[/itex]

    Right?
     
  7. Feb 9, 2013 #6
    Yes but the how can the bound be from a higher number to a smaller one.
     
  8. Feb 9, 2013 #7

    SammyS

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    What does switching the order do to the integral?
     
  9. Feb 9, 2013 #8
    I solved it I forgot to square the sin at the end. I did't need to use a u substitution.

    Any regardless if you have that bound it yields a negative answer.

    Thanks for helping though.

    The answer is

    [itex] \pi a^4/4 [/itex]
     
  10. Feb 9, 2013 #9

    SammyS

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    Yes. It works out just fine either way.

    Switching the limits of integration switches the sign. That's good because it got rid of that leading negative sign.

    ... and you learned a bit more of LaTeX coding .
     
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