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Math Help - Triple Integral and where to start

  1. #16
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    The order of integration was given to be that way too. I tried splitting the denominator and using partial fractions, but I could not find a way to split it since. Since the 1 is at the end, and the beginning, the only option are (z-1)(z+1)(-z+1)(-z-1) With that, there isn't a way to get 48z. In addition, the z^2 factors must get rid of each other.
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  2. #17
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    Quote Originally Posted by markmil2002 View Post
    The order of integration was given to be that way too. I tried splitting the denominator and using partial fractions, but I could not find a way to split it since. Since the 1 is at the end, and the beginning, the only option are (z-1)(z+1)(-z+1)(-z-1) With that, there isn't a way to get 48z. In addition, the z^2 factors must get rid of each other.
    I knew I shuold have attached the following a long while ago. Then you would understand why I've asked what I've asked and said what I've said:
    Attached Thumbnails Attached Thumbnails Triple Integral and where to start-msp22529589284828493657_541.gif  
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  3. #18
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    I can't exactly read that image, well I can read it, I just don't understand it. Is that from the integration tables?
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  4. #19
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    Quote Originally Posted by markmil2002 View Post
    I can't exactly read that image, well I can read it, I just don't understand it. Is that from the integration tables?
    Click on it twice and it will be large and clear. It is obtained using the Wolfram Mathematica Online Integrator
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  5. #20
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    I looked at the larger image, It just looks like it makes the next two integrations very difficult. I also don't understand how you get that result, and I need to understand how that result was obtained.
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  6. #21
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    Quote Originally Posted by markmil2002 View Post
    I looked at the larger image, It just looks like it makes the next two integrations very difficult. I also don't understand how you get that result, and I need to understand how that result was obtained.
    My point in attaching the image was to show you that you're attempting to do something that is most likely well beyond the scope of what you know.

    I say again: Go to the person who gave you the question and ask him/her how s/he expects you to solve it.
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  7. #22
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    It can't be beyond the scope of the course. It has to be solvable. Only thing is I can't see a way to change it to cylindrical or spherical, and I don't see a way to factor the polynomial into multiples. What I was looking at was this example.
    x^5 + x^4 - x -1 = x^4(x+1)-(x+1)
    =(x^4-1)(x+1)

    I just don't understand how the book got that.
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  8. #23
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    Quote Originally Posted by markmil2002 View Post
    It can't be beyond the scope of the course. It has to be solvable. Only thing is I can't see a way to change it to cylindrical or spherical, and I don't see a way to factor the polynomial into multiples. What I was looking at was this example.
    x^5 + x^4 - x -1 = x^4(x+1)-(x+1)
    =(x^4-1)(x+1)

    I just don't understand how the book got that.
    How does this factorisation relate to your posted question? The final line is got by taking out the common factor of (x+1).

    Re: Your original question. Is there a typo in the book? The cubic has no simple factors. For the final time: Go and ask whoever gave you the question how s/he expects you to solve it.
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  9. #24
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    Did you figure it out?

    I have that same problem on my "homework" and another person from my class tried to help, but neither of us could figure it out. Any luck yet?
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  10. #25
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    Tried some new things

    I tried some new things. First I tried changing the order of integration. No luck there, because two of the limits contain variables. This makes dzdydz, the only order of integration.

    Looks at the denominator, I turned it into

    -z(z^2 - 48) + 1 which then turns into
    -z(z+7)(z-7) + (1-z) This is still equal, and almost able to use partial fractions, I just don't see a way to get rid of the 1-z and move it as a factor into the others. I still think that partial fractions must be the way, but there is something I'm not seeing.
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  11. #26
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    factoring

    When i factored it out I got 1-z(z+7)(z-7) .. somehow you have an extra z that i didn't get. this is making me mad.
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  12. #27
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    I plugged values in yours while comparing it to the third degree polynomial and the equality doesn't exist. I also looked around for a quadratic formula similar to the one used for second degree equations and there isn't anything we haven't learned in school yet. They're all very complex formulas. I am downloading mathmatica to see if it does me any good. I tried another program and it couldn't even solve it.
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