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Math Help - volume of solid

  1. #1
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    volume of solid

    What i the volume of the solid
    y=3x-x^2, x=3, y=0, revolved about the y-axis
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  2. #2
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    Talking

    What have you tried? How far have you gotten on this exercise?

    Please be complete. Thank you!
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  3. #3
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    do i use the formula
    V=(2pie)interval from a to b ((f(x)-(g(x)))dx
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  4. #4
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    If thats the right formula i get the answer to be 18pie/2
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  5. #5
    Senior Member DeMath's Avatar
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    Quote Originally Posted by chevy900ss View Post
    What i the volume of the solid
    y=3x-x^2, x=3, y=0, revolved about the y-axis
    Shell method

    {V_y} = 2\pi \int\limits_a^b {xydx}  = 2\pi \int\limits_0^3 {x\left( {3x - {x^2}} \right)dx}  = 2\pi \int\limits_0^3 {\left( {3{x^2} - {x^3}}\right)dx}  =

    = \left. {2\pi \left( {{x^3} - \frac{{{x^4}}}{4}} \right)} \right|_0^3 = 2\pi \left( {27 - \frac{{81}}{4}} \right) = 2\pi  \cdot \frac{{27}}<br />
{4} = \frac{{27}}{2}\pi .
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  6. #6
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    Ok i forgot that x in the beginning equation.
    So, with the problem y=(x^3)-(5x^2)+(6x), x=2, y=0
    the answer for this one would be 24pie/5
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  7. #7
    Senior Member DeMath's Avatar
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    Quote Originally Posted by chevy900ss View Post
    Ok i forgot that x in the beginning equation.
    So, with the problem y=(x^3)-(5x^2)+(6x), x=2, y=0
    the answer for this one would be 24pie/5
    Yes!

    {V_y} = 2\pi \int\limits_0^2 {x\left( {{x^3} - 5{x^2} + 6x} \right)dx}  =  \ldots  = \frac{{24}}{5}\pi .

    See picture

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