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Math Help - I need Integral help please

  1. #1
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    I need Integral help please

    I need to setup and evaluate the integral that gives volume of the solid formed by revolving around the region about the y-axis.


    x=-y^2+4y

    Volume =Pi r^2 So the Intergral evaluated from 4 to 1(guessing)V=Pi (-y^2+4y)^2dy after the algebra I get y^4-8y^3+16y^2 My question is do I use y^5/5-8y^4/4+16y^3/3 to evaluate from 1 to 4?
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Darkhrse99 View Post
    I need to setup and evaluate the integral that gives volume of the solid formed by revolving around the region about the y-axis.


    x=-y^2+4y

    Volume =Pi r^2 So the Intergral evaluated from 4 to 1(guessing)V=Pi (-y^2+4y)^2dy after the algebra I get y^4-8y^3+16y^2 My question is do I use y^5/5-8y^4/4+16y^3/3 to evaluate from 1 to 4?
    You're right...except for one thing.

    It should be \pi\int_{{\color{red}0}}^4\left(4y-y^2\right)^2\,dy=\dots
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  3. #3
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    Quote Originally Posted by Chris L T521 View Post
    You're right...except for one thing.

    It should be \pi\int_{{\color{red}0}}^4\left(4y-y^2\right)^2\,dy=\dots
    So I do use the equation y5/5 ... to evaluate from 1 to 4?
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Darkhrse99 View Post
    So I do use the equation y5/5 ... to evaluate from 1 to 4?
    You evaluate \tfrac{1}{5}y^5-2y^4+\tfrac{16}{3}y^3 from 0 to 4.
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  5. #5
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    I got an answer of 6272Pi/15 is this correct?
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  6. #6
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Darkhrse99 View Post
    I got an answer of 6272Pi/15 is this correct?
    No...that's quite too big...xD

    Note that \pi\int_0^4\left(4y-y^2\right)^2\,dy=\pi\left.\left[\frac{1}{5}y^5-2y^4+\frac{16}{3}y^3\right]\right|_0^4=\pi\left[\frac{1}{5}(4)^5-2(4)^4+\frac{16}{3}(4)^3\right]=\frac{512}{15}\pi
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  7. #7
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    Thanks again Chris. I had it right the first time, then I redid it and came out with that huge answer.LOL
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