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Math Help - Intergration problem im stuck on

  1. #1
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    Intergration problem im stuck on

    I stuck in a engineering question at some intergration any help in how to do it rather than just the answer would be very helpfull as i have the asnwer but have no idea where it comes from.


    [IMG]file:///C:/Users/Sami/AppData/Local/Temp/moz-screenshot.jpg[/IMG]



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  2. #2
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    Just put z=1-\frac{y^2}{b^2}.
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  3. #3
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    sorry to appear stupid but not quite sure what that means? use substitution?
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  4. #4
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    Quote Originally Posted by samipoyo View Post
    sorry to appear stupid but not quite sure what that means? use substitution?
    Yes. Use the substitution that has been suggested.
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  5. #5
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    Same intergration, still struggling

    I have recently posted a intergration problem and with a substitutional hint i have retried and refailed, please can someone give me a few steps to help me as i simply dont want to use the answer as i will have achieved nothing
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  6. #6
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    The answer you give is not correct for the integral you give! Was the integral supposed to be \int_0^b y\sqrt{1- y^2/b^2} dy?

    If so then the substitution you were given before was probably u= 1- y^2/b^2. Show exactly how you tried that substitution.
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  7. #7
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    sorry about starting a new post,

    I sub in and end up with another intergration i cant do

    any help would be great
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  8. #8
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    Quote Originally Posted by samipoyo View Post
    sorry to appear stupid but not quite sure what that means? use substitution?
    Have you been taught the substitution method?

    z = 1 - \frac{y^2}{b} \Rightarrow \frac{dz}{dy} = - \frac{2y}{b} \Rightarrow dy = -\frac{b}{2y} \, dz.

    The integral becomes -\frac{b}{2} \int_{z = 1}^{z = 1 - b} z \, dz .

    It would be a big help if you showed your working.
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  9. #9
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    yes but havent dont any real intergration in a couple years now, will try again now thanks
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  10. #10
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    Quote Originally Posted by mr fantastic View Post
    Have you been taught the substitution method?

    z = 1 - \frac{y^2}{b} \Rightarrow \frac{dz}{dy} = - \frac{2y}{b} \Rightarrow dy = -\frac{b}{2y} \, dz.

    The integral becomes -\frac{b}{2} \int_{z = 1}^{z = 1 - b} z \, dz .

    It would be a big help if you showed your working.
    From HallsofIvy's observation, the integral probbaly should become -\frac{b}{2} \int_{z = 1}^{z = 1 - b} z^{1/2} \, dz .

    It would help if you confirmed what the actual integral is ....
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