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Math Help - Another find the volume of a solid.Can I get myanswer checked please?

  1. #1
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    Another find the volume of a solid.Can I get myanswer checked please?

    Find the volume of a solid generated by revolving the region bounded by the given lines adn curves about the x-axis.


    y=6x,y=6,y=0
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  2. #2
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    I get an answer of 12pi,is this correct?



    Thanks
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  3. #3
    Junior Member slider142's Avatar
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    The question is vague: the region bounded by those lines is infinite and could be on either side of the line y = 6x.
    The answer you gave is correct for the region bounded by the lines y = 6x, y = 0, and x = 1.
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  4. #4
    Senior Member DeMath's Avatar
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    Maybe y = 6x,{\text{ }}y = 6,{\text{ }}x = 0 ??
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  5. #5
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    Quote Originally Posted by DeMath View Post
    Maybe y = 6x,{\text{ }}y = 6,{\text{ }}x = 0 ??
    It was suppose to be x=0, not y=0.

    Sorry about that.
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  6. #6
    Senior Member DeMath's Avatar
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    So, you have {y_1} = 6x,{\text{ }}{y_2} = 6,{\text{ }}x = 0,{\text{ }}{V_x} = ?

    Find the intersection point(s) of curves

    {y_1} = {y_2} \Leftrightarrow 6x = 6 \Leftrightarrow x = 1.

    Then you have

    {V_x} = \pi \int\limits_{x = a}^{x = b} {\left( {y_2^2 - y_1^2} \right)dx}  = \pi \int\limits_0^1 {\left( {36 - 36{x^2}} \right)dx}  = 36\pi \int\limits_0^1 {\left( {1 - {x^2}} \right)dx}  =

    = 36\pi \left. {\left( {x - \frac{{{x^3}}}{3}} \right)} \right|_0^1 = 36\pi \left( {1 - \frac{1}{3}} \right) = 24\pi {\text{ }}\left( {{\text{cubic units}}} \right).

    See this picture

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  7. #7
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    Thanks for the picture. It really helps understanding it better!
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