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Math Help - V of solid rev..........

  1. #1
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    V of solid rev..........

    Hi,
    I have done the question on the picture attached
    I don't believe it could be the correct answer, can someone help me please

    Thanks
    Attached Thumbnails Attached Thumbnails V of solid rev..........-solid-re.jpg  
    Last edited by wolfhound; February 26th 2010 at 09:36 AM.
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  2. #2
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    actually should my answer be 1114?
    Last edited by wolfhound; February 26th 2010 at 09:36 AM.
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  3. #3
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    Quote Originally Posted by wolfhound View Post
    Hi,
    I have done the question on the picture attached
    I don't believe it could be the correct answer, can someone help me please

    Thanks
    (x^2+4x)^2=x^4+8x^3+16x^2
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  4. #4
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    Silly me,
    Hows it looking now please?
    still wrong I think
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  5. #5
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    Quote Originally Posted by wolfhound View Post
    Can someone help me with this please.....
    That's correct to the nearest whole number.
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  6. #6
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    but is it a rule to make it into a fraction and keep the pi(sign) or is it ok to do like I did
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  7. #7
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    V = \pi \int_0^3 (x^2+4x)^2 \, dx<br />

    V = \pi \int_0^3 x^4 + 8x^3 + 16x^2 \, dx<br />

    V = \pi \left[\frac{x^5}{5} + 2x^4 + \frac{16x^3}{3}\right]_0^3<br />

    V = \pi \left[\frac{3^5}{5} + 2 \cdot 3^4 + \frac{16 \cdot 3^3}{3}\right]<br />

    V = \frac{1773\pi}{5} \approx 1114
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  8. #8
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    Quote Originally Posted by wolfhound View Post
    Is anyone there to help me with this annoying solid of revolution??????????
    PLEASE!
    Hi wolfhound.

    It always helps to understand how the situation looks on a diagram.

    The volume of revolution is the sum of the areas of a series of "wafer-thin" discs (an "infinite" sum of them, evaluated as their widths "deta-x" goes to zero).

    The radius of the discs in this case is f(x) because the curve is being rotated around the x-axis.

    the area of a disc is {\pi}r^2={\pi}[f(x)]^2

    All of these disc areas are integrated from x=0 to x=3.
    This gives the volume of revolution.

    You need to get used to how to do this notationally,
    as your written work needs to be redone.

    Disc\ area={\pi}\left(x^2+4x\right)^2={\pi}\left(x^4+8x^  3+16x^2\right)

    Therefore, the volume of revolution is

    \int_{x=0}^{x=3}{\pi}\left(x^4+8x^3+16x^2\right)dx

    ={\pi}\left[\frac{x^5}{5}+\frac{8x^4}{4}+\frac{16x^3}{3}\right] from x=0 to 3

    I've shown the disc from an angle (hence it looks like an ellipse!)
    Attached Thumbnails Attached Thumbnails V of solid rev..........-vor5.jpg  
    Last edited by Archie Meade; February 27th 2010 at 08:36 AM. Reason: typo
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  9. #9
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    Smile

    Thanks for the help, I must make more sense when I write it out..
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