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    The region bounded by the y-axis, the lines y=1, y=3 and the curve with equation y=ln(x^2) is rotated completely about the y axis.

    Find the volume so generated.

    Use the trapezium rule with 5 strips of equal width to find an estimate of the area of the region as given.


    I have got 5.68, the answer is 5.70. I am dubious over my method.

    Could someone kindly show the correct method with fulls steps. Thanks
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    I also get 5.68.
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    How would you integrate (ln x)^2? My mind has gone blank ...
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    (Original post by mik1w)
    How would you integrate (ln x)^2? My mind has gone blank ...
    You don't. You need to write it in the form x=f(y) as the area is bounded by the curve and the y axis.
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    So what IS the volume generated? Do you have the answers? I get e³-e. But it's probably wrong as I am not known for my great integration skills!
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    (Original post by fabz)
    So what IS the volume generated? Do you have the answers? I get e³-e. But it's probably wrong as I am not known for my great integration skills!
    Almost
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    How could I forget good old pi??? Oh well at least I got the rest right!! That's always nice! Thanks btw!
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    (Original post by mik1w)
    How would you integrate (ln x)^2? My mind has gone blank ...
    Int. (lnx)^2 dx. = Int. (lnx)(lnx) dx.
    Let u = lnx -> u' = 1/x
    Let v' = lnx -> v = xlnx - x

    -> Int. (lnx)^2 dx = lnx(xlnx - x) - Int. (xlnx - x)/x dx.
    = x(lnx)^2 - xlnx - Int. lnx - 1 dx.
    = x(lnx)^2 - xlnx - Int. lnx dx + Int. 1 dx
    = x(lnx)^2 - xlnx + x - xlnx + x + k
    = x(lnx)^2 - 2xlnx + 2x + k
    = x(lnx)^2 + 2x(1 - lnx) + k
 
 
 
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