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Integrating complex equations

  1. Aug 14, 2012 #1
    I have a maths B assignment where I need to find the volume of a tunnel. We have been given two equations to describe the area of this tunnel. The teacher says that we haven't learnt how to integrate at this level, and he has also said that the chain rule and product rule do not work for integration.



    The two equations are
    y=3+2√(1-(x^2/9)
    y=3-4√((x-6)^2/9 -1)



    How do you integrate these two functions? Thanks
     
  2. jcsd
  3. Aug 14, 2012 #2

    Simon Bridge

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    What are x and y in relation to the tunnel?

    Or do you just want [itex]\int y.dx[/itex] ?

    [edit]
    Having plotted the functions it looks like they are the boundary of the cross-sectional area of the tunnel.
    The tunnel extends in the z direction.

    If the tunnel is L long, the V=LA: A=the area of the cross-section.

    if the first equation is y1 and the second is y2 then you just need to look at how the between each and the x axis relates to the entire cross-section.
    I'd start by plotting them. (which is, in fact, what I did :) )
     
    Last edited: Aug 14, 2012
  4. Aug 14, 2012 #3
    yes you're right, this is the cross section of a 1.25km tunnel.
    the x values for the first equation are -3 and 3
    the x values for the second equation are 3 and 0
    I have a diagram provided. The first equation covers the top half, the second equation joins onto the first equation. This will just need to be multiplied by 2 to get the area of the bottom half.
     
  5. Aug 14, 2012 #4

    Simon Bridge

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    OK- cool: you have understood the geometry.
    Can you express the area as a sum of definite integrals of y1 and y2?

    To help you write it here, use LaTeX, the code: A = \int_{a}^{b} y_{n}(x)dx will come out as: [tex]A = \int_{a}^{b} y_{n}(x)dx[/tex]... you just have to fill in the right values, and you'll have more than one integral. Hit the "quote" button at the bottom of this post to see what I did to do that.

    I cannot over-stress how useful this is.

    After that ... you need to learn how to handle the functions-within-a-function in an integral.
    You're teacher is not quite correct - there are equivalents to the chain rule and so on for integration though the derivative chain rule does not work.
     
  6. Aug 14, 2012 #5

    Simon Bridge

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    So - to continue:

    You will end up having to do integrals like:

    [tex]A=\int_a^b \sqrt{1-(kx)^2}dx[/tex]... you can probably integrate [itex]\sqrt{x}[/itex] right?

    You need to make a substitution, in this case we exploit trigonometry like this:

    let [itex]kx=\sin(\theta)[/itex]
    then the stuff inside the square root becomes [itex]1-\sin^2(\theta)=\cos^2(\theta)[/itex]
    sub that in and the square root vanishes.

    But now we have to integrate a function of theta (which is, in turn, a function of x) with respect to x
    ... inconvenient. So we need to sub for dx as well.

    We can do this by realizing that [itex]\frac{dx}{d\theta}[/itex] is another way to write a derivative as a ratio of dx and dθ. In this case:[tex]\frac{dx}{d\theta}=\frac{d}{d\theta}\frac{1}{k} \sin(\theta)=\frac{1}{k}\cos(\theta) \Rightarrow dx=\frac{1}{k}\cos(\theta)d\theta[/tex]

    Now you can do the indefinite integral and convert back before applying the limits of you can convert the limits directly since [itex]\theta = \sin^{-1}(kx)[/itex] so the entire integral becomes:[tex]A=\frac{1}{k}\int_{ \sin^{-1}(ka)}^{ \sin^{-1}(kb)}\cos^2(\theta)d\theta[/tex]... presumably you can integrate a power of a trig function?

    But you see why your teacher thought it was an advanced method?
    It's called "integration by substitution" and you can google for that or "chain rule for integration".
     
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