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    (Original post by apo1324)
    Hello there,

    Could someone please check if my answer is correct for the following question:

    The region bounded by the curve y = sqrtx + 4/sqrtx, the x-axis and the lines x = 1, x = 4 is rotated through four right-angles about the x-axis. Find, correct to one decimal place, the volume of the solid formed.

    I got an answer of 29.7 (1dp)?? Thank you very much. PS If it is wrong I will post my workings.
    Sorry, you're wrong.

    If that's \sqrt{x}+ \dfrac{4}{\sqrt{x}} then Wolfram says it's 168.6, and I'm inclined to believe it, as I get the same value.
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    When I did it, I got:

    y^2 = x + 16/x??

    integrating:- x^2 / 2 + 16ln(x)?? I've probably gone wrong bec I'm crap at integrating. Any help?
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    (Original post by apo1324)
    When I did it, I got:

    y^2 = x + 16/x??

    integrating:- x^2 / 2 + 16ln(x)?? I've probably gone wrong bec I'm crap at integrating. Any help?
    y^2 \not= x + \frac{16}{x} . use binomial expansion if you must, but (a+b)^2 is not a^2 + b^2
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    Cant be arsed to use latex but here we go...

    Pi multiplied by the intergral of (rootx + 4/rootx)^2 = what you want.

    Multiply out the bracket and you get x + 16/x + 8

    Intergrating gives 0.5x^2 + 16lnx + 8x

    Applying the limits gives (62.18 - 8.5)pi

    =168.64
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    (Original post by ghostwalker)
    Sorry, you're wrong.

    If that's \sqrt{x}+ \dfrac{4}{\sqrt{x}} then Wolfram says it's 168.6, and I'm inclined to believe it, as I get the same value.
    this is correct I also got 168.6 (1dp)
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    (Original post by Ollie901)
    Cant be arsed to use latex but here we go...

    Pi multiplied by the intergral of (rootx + 4/rootx)^2 = what you want.

    Multiply out the bracket and you get x + 16/x + 8

    Intergrating gives 0.5x^2 + 16lnx + 8x

    Applying the limits gives (62.18 - 8.5)pi

    =168.64
    The bold part... would you not get 'x + 16x^(3/2)' ???
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    (Original post by Mobs25)
    The bold part... would you not get 'x + 16x^(3/2)' ???
    No you wouldn't. Look again.
 
 
 
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