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Thread: integral

  1. #1
    Aug 2007

    Exclamation integral

    Hi everyone,

    Could someone please explain to me how to find the volume under the curve for y=lnx, y=1, y=2, x=o about the y-axis?

    I first solved for x. I'm not sure if I did it correctly. I got x=e to the x(y) Is that correct?

    This is the radius, so I then I need to square it and multiply it by pie. However, all my values are in x's and y's, so how could I deceide what the points are on the interval???????

    Could someone please help me with this?

    Thank you very much
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  2. #2
    MHF Contributor
    Apr 2005
    You're looking for x. Then you square it.
    So you are using disc to find the volume.
    Yes, you then square the x, and multiply it by pi. Because area of circle is pi(r^2).

    y = ln(x)

    Use e^y = e^[ln(x)], so e^y = x.

    Or, use y = Log(e)[x]
    So, x = e^y

    dV = pi(x^2)*dy

    What are the boundaries of dy?
    dy goes from y=1 to y=2, so,
    V = INT.(1-->2)[pi(x^2)]dy

    Oops, x^2 and dy don't go together.
    We must express the x^2 into its equivalent y^2, so that we can use the dy.

    x = e^y
    Square both sides,
    x^2 = (e^y)^2 = e^(2y)
    V = INT.(1-->2)[(pi)e^(2y)]dy
    V = (pi)INT.(1-->2)[e^2y]dy ----------(i)

    Uhh, dy is not the derivative of e^(2y).
    We cannot integrate yet.

    Suppose u = e^(2y)
    Then, du = [e^(2y)](2dy)

    So, (i) becomes
    V = (pi)INT(y=1 --> y=2)[(1/2)du]
    V = (pi/2)INT.(y=1 --> y=2)[du] ------(ii)

    Er, ah, the limits of du are not from y=1 to y=2. In fact, what is y doing here? It's out of place.

    What are the boundaries of du then?
    u = e^(2y)
    So when y = 1, u = e^(2*1) = e^2
    When y = 2, u = e^(2*2) = e^4
    Hence, the boundaries of du are from e^2 to e^4
    So, (ii) becomes
    V = (pi/2)INT.(e^2 --> e^4)[du]
    Now we can integrate,
    V = (pi/2)[u]|(e^2 --> e^4)
    V = (pi/2)[e^4 -e^2] sq.units --------------answer.

    In numbers, V = (pi/2)[47.2091] = 23.6045pi = 74.1559 sq.units ----answer
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