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Thread: calc 3- line integral

  1. #1
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    Exclamation calc 3- line integral

    find the mass of a wire in the shape of the helix traced by (cost, sint, t/pi)
    from pi to 3 pi if its density at each point is proportional to the distance from the point to the xy-plane

    so i have mass= the integral from pi to 3pi of
    cost + sint + t/pi dL

    but i'm not sure what dL is..i saw it in an example in the book.
    is dL the derivative of what i have above? or is it the magnitude? or do i ignore it?
    after i clarify this, i think i'll be able to hopefully solve it..
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    $\displaystyle dL$ means that the integral is with respect to arc length. Imagine if the wire had a uniform mass density - then its mass would clearly be equal to its length times the mass density.

    You probably know that the length of a parametrized curve $\displaystyle \phi$ between $\displaystyle \phi(a)$ and $\displaystyle \phi(b)$ is

    $\displaystyle L=\int_a^b|\phi'(s)|ds$

    so $\displaystyle dL = |\phi'(s)|ds$. Therefore the integral you are looking for is $\displaystyle \int_{\pi}^{3\pi}W(\phi(s))|\phi'(s)|ds$ where $\displaystyle W(\phi(s))$ is the mass density of the wire at the point $\displaystyle \phi(s)$. You are told that this is proportional to the distance from the point $\displaystyle \phi(s)$ to the $\displaystyle XY$ plane, i.e. $\displaystyle W(\phi(s))=k \times d(\phi(s), XY\mbox{ plane})$ for some constant $\displaystyle k$. Find the expression for that, substitute it into the integral and then evaluate it.
    Last edited by Bruno J.; Oct 27th 2009 at 03:42 PM.
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  3. #3
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    i'm sorry i'm sort of confused with all of your symbols. my final answer is:
    4*sqrt(pi^2 +1)
    is that right/close?
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  4. #4
    MHF Contributor Bruno J.'s Avatar
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    I didn't do the calculation! Please post your work, and I'll tell you if it's good.
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  5. #5
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    f(x)= cost+sint+t/pi
    dL= -sint+cost+1/pi then i took the magnitude of this to get
    dL= (sqrt.(pi^2+1))/(pi)

    so i took the integral from pi to 3pi of f(x)*dL
    and got

    (sint-cost+(1/2pi)*t^2)*(sqrt(pi^2 +1))/(pi))* t

    and evaluated from 3pi to pi
    and got

    a final answer of

    18pi* sqrt(pi^2 +1)

    i found a few algebraic answers and fixed my first original answer to get this
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  6. #6
    MHF Contributor Bruno J.'s Avatar
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    Well the function isn't $\displaystyle f(t)=\cos t+\sin t+\frac{t}{\pi}$, it's $\displaystyle f(t) = (\cos t, \sin t, \frac{t}{\pi})$.

    Moreover, what did you do of the fact that the mass density at the point $\displaystyle f(t)$ is proportional to the distance between $\displaystyle f(t)$ and the $\displaystyle XY$ plane? If you do not take that into account in your solution then your solution is certainly wrong.
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